It’s a cold day in Antarctica. In fact, it’s always a cold day in Antarctica. Earth’s southernmost continent, Antarctica experiences the coldest, driest, and windiest conditions known. The coldest temperature ever recorded, over one hundred degrees below zero on the Celsius scale, was recorded by remote satellite. It is no surprise then, that no native human population can survive the harsh conditions. Only explorers and scientists brave the environment for any length of time.
Measuring and recording the characteristics of weather conditions in in Antarctica requires a use of different kinds of numbers. Calculating with them and using them to make predictions requires an understanding of relationships among numbers. In this chapter, we will review sets of numbers and properties of operations used to manipulate numbers. This understanding will serve as prerequisite knowledge throughout our study of algebra and trigonometry.
]]>It is often said that mathematics is the language of science. If this is true, then an essential part of the language of mathematics is numbers. The earliest use of numbers occurred 100 centuries ago in the Middle East to count, or enumerate items. Farmers, cattlemen, and tradesmen used tokens, stones, or markers to signify a single quantity—a sheaf of grain, a head of livestock, or a fixed length of cloth, for example. Doing so made commerce possible, leading to improved communications and the spread of civilization.
Three to four thousand years ago, Egyptians introduced fractions. They first used them to show reciprocals. Later, they used them to represent the amount when a quantity was divided into equal parts.
But what if there were no cattle to trade or an entire crop of grain was lost in a flood? How could someone indicate the existence of nothing? From earliest times, people had thought of a “base state” while counting and used various symbols to represent this null condition. However, it was not until about the fifth century A.D. in India that zero was added to the number system and used as a numeral in calculations.
Clearly, there was also a need for numbers to represent loss or debt. In India, in the seventh century A.D., negative numbers were used as solutions to mathematical equations and commercial debts. The opposites of the counting numbers expanded the number system even further.
Because of the evolution of the number system, we can now perform complex calculations using these and other categories of real numbers. In this section, we will explore sets of numbers, calculations with different kinds of numbers, and the use of numbers in expressions.
The numbers we use for counting, or enumerating items, are the natural numbers: 1, 2, 3, 4, 5, and so on. We describe them in set notation as[latex]\,\left\{1,2,3,...\right\}\,[/latex]where the ellipsis (…) indicates that the numbers continue to infinity. The natural numbers are, of course, also called the counting numbers. Any time we enumerate the members of a team, count the coins in a collection, or tally the trees in a grove, we are using the set of natural numbers. The set of whole numbers is the set of natural numbers plus zero:[latex]\,\left\{0,1,2,3,...\right\}.[/latex]
The set of integers adds the opposites of the natural numbers to the set of whole numbers:[latex]\,\left\{...,-3,-2,-1,0,1,2,3,...\right\}.[/latex]It is useful to note that the set of integers is made up of three distinct subsets: negative integers, zero, and positive integers. In this sense, the positive integers are just the natural numbers. Another way to think about it is that the natural numbers are a subset of the integers.
The set of rational numbers is written as[latex]\,\left\{\frac{m}{n}\,|m\text{ and }n\text{ are integers and }n\ne 0\right\}.\,[/latex]Notice from the definition that rational numbers are fractions (or quotients) containing integers in both the numerator and the denominator, and the denominator is never 0. We can also see that every natural number, whole number, and integer is a rational number with a denominator of 1.
Because they are fractions, any rational number can also be expressed in decimal form. Any rational number can be represented as either:
We use a line drawn over the repeating block of numbers instead of writing the group multiple times.
Write each of the following as a rational number.
Write a fraction with the integer in the numerator and 1 in the denominator.
Write each of the following as a rational number.
Write each of the following rational numbers as either a terminating or repeating decimal.
Write each fraction as a decimal by dividing the numerator by the denominator.
Write each of the following rational numbers as either a terminating or repeating decimal.
At some point in the ancient past, someone discovered that not all numbers are rational numbers. A builder, for instance, may have found that the diagonal of a square with unit sides was not 2 or even[latex]\,\frac{3}{2},[/latex]but was something else. Or a garment maker might have observed that the ratio of the circumference to the diameter of a roll of cloth was a little bit more than 3, but still not a rational number. Such numbers are said to be irrational because they cannot be written as fractions. These numbers make up the set of irrational numbers. Irrational numbers cannot be expressed as a fraction of two integers. It is impossible to describe this set of numbers by a single rule except to say that a number is irrational if it is not rational. So we write this as shown.
Determine whether each of the following numbers is rational or irrational. If it is rational, determine whether it is a terminating or repeating decimal.
So,[latex]\,\frac{33}{9}\,[/latex]is rational and a repeating decimal.
So,[latex]\,\frac{17}{34}\,[/latex]is rational and a terminating decimal.
Determine whether each of the following numbers is rational or irrational. If it is rational, determine whether it is a terminating or repeating decimal.
Given any number n, we know that n is either rational or irrational. It cannot be both. The sets of rational and irrational numbers together make up the set of real numbers. As we saw with integers, the real numbers can be divided into three subsets: negative real numbers, zero, and positive real numbers. Each subset includes fractions, decimals, and irrational numbers according to their algebraic sign (+ or –). Zero is considered neither positive nor negative.
The real numbers can be visualized on a horizontal number line with an arbitrary point chosen as 0, with negative numbers to the left of 0 and positive numbers to the right of 0. A fixed unit distance is then used to mark off each integer (or other basic value) on either side of 0. Any real number corresponds to a unique position on the number line.The converse is also true: Each location on the number line corresponds to exactly one real number. This is known as a one-to-one correspondence. We refer to this as the real number line as shown in (Figure).
Classify each number as either positive or negative and as either rational or irrational. Does the number lie to the left or the right of 0 on the number line?
Classify each number as either positive or negative and as either rational or irrational. Does the number lie to the left or the right of 0 on the number line?
Beginning with the natural numbers, we have expanded each set to form a larger set, meaning that there is a subset relationship between the sets of numbers we have encountered so far. These relationships become more obvious when seen as a diagram, such as (Figure).
When we multiply a number by itself, we square it or raise it to a power of 2. For example,[latex]\,{4}^{2}=4\cdot 4=16.\,[/latex]We can raise any number to any power. In general, the exponential notation[latex]\,{a}^{n}\,[/latex]means that the number or variable[latex]\,a\,[/latex]is used as a factor[latex]\,n\,[/latex]times.
In this notation,[latex]\,{a}^{n}\,[/latex]is read as the nth power of[latex]\,a,\,[/latex]where[latex]\,a\,[/latex]is called the base and[latex]\,n\,[/latex]is called the exponent. A term in exponential notation may be part of a mathematical expression, which is a combination of numbers and operations. For example,[latex]\,24+6\cdot \frac{2}{3}-{4}^{2}\,[/latex]is a mathematical expression.
To evaluate a mathematical expression, we perform the various operations. However, we do not perform them in any random order. We use the order of operations. This is a sequence of rules for evaluating such expressions.
Recall that in mathematics we use parentheses ( ), brackets [ ], and braces { } to group numbers and expressions so that anything appearing within the symbols is treated as a unit. Additionally, fraction bars, radicals, and absolute value bars are treated as grouping symbols. When evaluating a mathematical expression, begin by simplifying expressions within grouping symbols.
The next step is to address any exponents or radicals. Afterward, perform multiplication and division from left to right and finally addition and subtraction from left to right.
Let’s take a look at the expression provided.
There are no grouping symbols, so we move on to exponents or radicals. The number 4 is raised to a power of 2, so simplify[latex]\,{4}^{2}\,[/latex]as 16.
Next, perform multiplication or division, left to right.
Lastly, perform addition or subtraction, left to right.
Therefore,[latex]\,24+6\cdot \frac{2}{3}-{4}^{2}=12.[/latex]
For some complicated expressions, several passes through the order of operations will be needed. For instance, there may be a radical expression inside parentheses that must be simplified before the parentheses are evaluated. Following the order of operations ensures that anyone simplifying the same mathematical expression will get the same result.
Operations in mathematical expressions must be evaluated in a systematic order, which can be simplified using the acronym PEMDAS:
P(arentheses)
E(xponents) M(ultiplication) and D(ivision) A(ddition) and S(ubtraction)Given a mathematical expression, simplify it using the order of operations.
Use the order of operations to evaluate each of the following expressions.
Note that in the first step, the radical is treated as a grouping symbol, like parentheses. Also, in the third step, the fraction bar is considered a grouping symbol so the numerator is considered to be grouped.
In this example, the fraction bar separates the numerator and denominator, which we simplify separately until the last step.
Use the order of operations to evaluate each of the following expressions.
For some activities we perform, the order of certain operations does not matter, but the order of other operations does. For example, it does not make a difference if we put on the right shoe before the left or vice-versa. However, it does matter whether we put on shoes or socks first. The same thing is true for operations in mathematics.
The commutative property of addition states that numbers may be added in any order without affecting the sum.
We can better see this relationship when using real numbers.
Similarly, the commutative property of multiplication states that numbers may be multiplied in any order without affecting the product.
Again, consider an example with real numbers.
It is important to note that neither subtraction nor division is commutative. For example,[latex]\,17-5\,[/latex]is not the same as[latex]\,5-17.\,[/latex]Similarly,[latex]\,20÷5\ne 5÷20.[/latex]
The associative property of multiplication tells us that it does not matter how we group numbers when multiplying. We can move the grouping symbols to make the calculation easier, and the product remains the same.
Consider this example.
The associative property of addition tells us that numbers may be grouped differently without affecting the sum.
This property can be especially helpful when dealing with negative integers. Consider this example.
Are subtraction and division associative? Review these examples.
As we can see, neither subtraction nor division is associative.
The distributive property states that the product of a factor times a sum is the sum of the factor times each term in the sum.
This property combines both addition and multiplication (and is the only property to do so). Let us consider an example.
Note that 4 is outside the grouping symbols, so we distribute the 4 by multiplying it by 12, multiplying it by –7, and adding the products.
To be more precise when describing this property, we say that multiplication distributes over addition. The reverse is not true, as we can see in this example.
A special case of the distributive property occurs when a sum of terms is subtracted.
For example, consider the difference[latex]\,12-\left(5+3\right).\,[/latex]We can rewrite the difference of the two terms 12 and[latex]\,\left(5+3\right)\,[/latex]by turning the subtraction expression into addition of the opposite. So instead of subtracting[latex]\,\left(5+3\right),[/latex]we add the opposite.
Now, distribute[latex]\,-1\,[/latex]and simplify the result.
This seems like a lot of trouble for a simple sum, but it illustrates a powerful result that will be useful once we introduce algebraic terms. To subtract a sum of terms, change the sign of each term and add the results. With this in mind, we can rewrite the last example.
The identity property of addition states that there is a unique number, called the additive identity (0) that, when added to a number, results in the original number.
The identity property of multiplication states that there is a unique number, called the multiplicative identity (1) that, when multiplied by a number, results in the original number.
For example, we have[latex]\,\left(-6\right)+0=-6\,[/latex]and[latex]\,23\cdot 1=23.\,[/latex]There are no exceptions for these properties; they work for every real number, including 0 and 1.
The inverse property of addition states that, for every real number a, there is a unique number, called the additive inverse (or opposite), denoted−a, that, when added to the original number, results in the additive identity, 0.
For example, if[latex]\,a=-8,[/latex]the additive inverse is 8, since[latex]\,\left(-8\right)+8=0.[/latex]
The inverse property of multiplication holds for all real numbers except 0 because the reciprocal of 0 is not defined. The property states that, for every real number a, there is a unique number, called the multiplicative inverse (or reciprocal), denoted[latex]\,\frac{1}{a},[/latex]that, when multiplied by the original number, results in the multiplicative identity, 1.
For example, if[latex]\,a=-\frac{2}{3},[/latex]the reciprocal, denoted[latex]\,\frac{1}{a},[/latex]is[latex]\,-\frac{3}{2}\,[/latex] because
The following properties hold for real numbers a, b, and c.
Addition | Multiplication | |
Commutative Property | [latex]a+b=b+a[/latex] | [latex]a\cdot b=b\cdot a[/latex] |
Associative Property | [latex]a+\left(b+c\right)=\left(a+b\right)+c[/latex] | [latex]a\left(bc\right)=\left(ab\right)c[/latex] |
Distributive Property | [latex]a\cdot \left(b+c\right)=a\cdot b+a\cdot c[/latex] | |
Identity Property | There exists a unique real number called the additive identity, 0, such that, for any real number a
[latex]a+0=a[/latex] |
There exists a unique real number called the multiplicative identity, 1, such that, for any real number a
[latex]a\cdot 1=a[/latex] |
Inverse Property | Every real number a has an additive inverse, or opposite, denoted –a, such that
[latex]a+\left(-a\right)=0[/latex] |
Every nonzero real number a has a multiplicative inverse, or reciprocal, denoted[latex]\,\frac{1}{a},[/latex]such that
[latex]a\cdot \left(\frac{1}{a}\right)=1[/latex] |
Use the properties of real numbers to rewrite and simplify each expression. State which properties apply.
Use the properties of real numbers to rewrite and simplify each expression. State which properties apply.
So far, the mathematical expressions we have seen have involved real numbers only. In mathematics, we may see expressions such as[latex]\,x+5,\frac{4}{3}\pi {r}^{3},[/latex]or[latex]\,\sqrt{2{m}^{3}{n}^{2}}.\,[/latex]In the expression[latex]\,x+5,[/latex]5 is called a constant because it does not vary and x is called a variable because it does. (In naming the variable, ignore any exponents or radicals containing the variable.) An algebraic expression is a collection of constants and variables joined together by the algebraic operations of addition, subtraction, multiplication, and division.
We have already seen some real number examples of exponential notation, a shorthand method of writing products of the same factor. When variables are used, the constants and variables are treated the same way.
In each case, the exponent tells us how many factors of the base to use, whether the base consists of constants or variables.
Any variable in an algebraic expression may take on or be assigned different values. When that happens, the value of the algebraic expression changes. To evaluate an algebraic expression means to determine the value of the expression for a given value of each variable in the expression. Replace each variable in the expression with the given value, then simplify the resulting expression using the order of operations. If the algebraic expression contains more than one variable, replace each variable with its assigned value and simplify the expression as before.
List the constants and variables for each algebraic expression.
Constants | Variables | |
---|---|---|
a. x + 5 | 5 | x |
b.[latex]\,\frac{4}{3}\pi {r}^{3}[/latex] | [latex]\frac{4}{3},\pi [/latex] | [latex]r[/latex] |
c.[latex]\,\sqrt{2{m}^{3}{n}^{2}}[/latex] | 2 | [latex]m,n[/latex] |
List the constants and variables for each algebraic expression.
Constants | Variables | |
---|---|---|
a.[latex]2\pi r\left(r+h\right)[/latex] | [latex]2,\pi [/latex] | [latex]r,h[/latex] |
b. 2(L + W) | 2 | L, W |
c.[latex]\,4{y}^{3}+y[/latex] | 4 | [latex]y[/latex] |
Evaluate the expression[latex]\,2x-7\,[/latex]for each value for x.
Evaluate the expression[latex]\,11-3y\,[/latex]for each value for y.
Evaluate each expression for the given values.
Evaluate each expression for the given values.
An equation is a mathematical statement indicating that two expressions are equal. The expressions can be numerical or algebraic. The equation is not inherently true or false, but only a proposition. The values that make the equation true, the solutions, are found using the properties of real numbers and other results. For example, the equation[latex]\,2x+1=7\,[/latex]has the unique solution of 3[latex][/latex] because when we substitute 3 for[latex]\,x\,[/latex]in the equation, we obtain the true statement[latex]\,2\left(3\right)+1=7.[/latex]
A formula is an equation expressing a relationship between constant and variable quantities. Very often, the equation is a means of finding the value of one quantity (often a single variable) in terms of another or other quantities. One of the most common examples is the formula for finding the area[latex]\,A\,[/latex]of a circle in terms of the radius[latex]\,r\,[/latex]of the circle:[latex]\,A=\pi {r}^{2}.\,[/latex]For any value of[latex]\,r,[/latex]the area[latex]\,A\,[/latex]can be found by evaluating the expression[latex]\,\pi {r}^{2}.[/latex]
A right circular cylinder with radius[latex]\,r\,[/latex]and height[latex]\,h\,[/latex]has the surface area[latex]\,S\,[/latex](in square units) given by the formula[latex]\,S=2\pi r\left(r+h\right).\,[/latex]See (Figure). Find the surface area of a cylinder with radius 6 in. and height 9 in. Leave the answer in terms of[latex]\,\pi .[/latex]
Evaluate the expression[latex]\,2\pi r\left(r+h\right)\,[/latex]for[latex]\,r=6\,[/latex]and[latex]\,h=9.[/latex]
A photograph with length L and width W is placed in a matte of width 8 centimeters (cm). The area of the matte (in square centimeters, or cm^{2}) is found to be[latex]\,A=\left(L+16\right)\left(W+16\right)-L\cdot W.\,[/latex]See (Figure). Find the area of a matte for a photograph with length 32 cm and width 24 cm.
1,152 cm2
[/hidden-answer]Sometimes we can simplify an algebraic expression to make it easier to evaluate or to use in some other way. To do so, we use the properties of real numbers. We can use the same properties in formulas because they contain algebraic expressions.
Simplify each algebraic expression.
Simplify each algebraic expression.
A rectangle with length[latex]\,L\,[/latex]and width[latex]\,W\,[/latex]has a perimeter[latex]\,P\,[/latex]given by[latex]\,P=L+W+L+W.\,[/latex]Simplify this expression.
If the amount[latex]\,P\,[/latex]is deposited into an account paying simple interest[latex]\,r\,[/latex]for time[latex]\,t,[/latex]the total value of the deposit[latex]\,A\,[/latex]is given by[latex]\,A=P+Prt.\,[/latex]Simplify the expression. (This formula will be explored in more detail later in the course.)
[latex]A=P\left(1+rt\right)[/latex]
[/hidden-answer]Access these online resources for additional instruction and practice with real numbers.
Is[latex]\,\sqrt{2}\,[/latex]an example of a rational terminating, rational repeating, or irrational number? Tell why it fits that category.
irrational number. The square root of two does not terminate, and it does not repeat a pattern. It cannot be written as a quotient of two integers, so it is irrational.
[/hidden-answer]What is the order of operations? What acronym is used to describe the order of operations, and what does it stand for?
What do the Associative Properties allow us to do when following the order of operations? Explain your answer.
The Associative Properties state that the sum or product of multiple numbers can be grouped differently without affecting the result. This is because the same operation is performed (either addition or subtraction), so the terms can be re-ordered.
[/hidden-answer]For the following exercises, simplify the given expression.
[latex]10+2\,×\,\left(5-3\right)[/latex]
[latex]6÷2-\left(81÷{3}^{2}\right)[/latex]
[latex]-6[/latex]
[/hidden-answer][latex]18+{\left(6-8\right)}^{3}[/latex]
[latex]-2\,×\,{\left[16÷{\left(8-4\right)}^{2}\right]}^{2}[/latex]
[latex]-2[/latex]
[/hidden-answer][latex]4-6+2\,×\,7[/latex]
[latex]3\left(5-8\right)[/latex]
[latex]-9[/latex]
[/hidden-answer][latex]4+6-10÷2[/latex]
[latex]12÷\left(36÷9\right)+6[/latex]
9
[/hidden-answer][latex]{\left(4+5\right)}^{2}÷3[/latex]
[latex]3-12\,×\,2+19[/latex]
-2
[/hidden-answer][latex]2+8\,×\,7÷4[/latex]
[latex]5+\left(6+4\right)-11[/latex]
4
[/hidden-answer][latex]9-18÷{3}^{2}[/latex]
[latex]14\,×\,3÷7-6[/latex]
0
[/hidden-answer][latex]9-\left(3+11\right)\,×\,2[/latex]
[latex]6+2\,×\,2-1[/latex]
9
[/hidden-answer][latex]64÷\left(8+4\,×\,2\right)[/latex]
[latex]9+4\left({2}^{2}\right)[/latex]
25
[/hidden-answer][latex]{\left(12÷3\,×\,3\right)}^{2}[/latex]
[latex]25÷{5}^{2}-7[/latex]
[latex]-6[/latex]
[/hidden-answer][latex]\left(15-7\right)\,×\,\left(3-7\right)[/latex]
[latex]2\,×\,4-9\left(-1\right)[/latex]
17
[/hidden-answer][latex]{4}^{2}-25\,×\,\frac{1}{5}[/latex]
[latex]12\left(3-1\right)÷6[/latex]
4
[/hidden-answer]For the following exercises, solve for the variable.
[latex]8\left(x+3\right)=64[/latex]
[latex]4y+8=2y[/latex]
[latex]-4[/latex]
[/hidden-answer][latex]\left(11a+3\right)-18a=-4[/latex]
[latex]4z-2z\left(1+4\right)=36[/latex]
[latex]-6[/latex]
[/hidden-answer][latex]4y{\left(7-2\right)}^{2}=-200[/latex]
[latex]-{\left(2x\right)}^{2}+1=-3[/latex]
[latex]±1[/latex]
[/hidden-answer][latex]8\left(2+4\right)-15b=b[/latex]
[latex]2\left(11c-4\right)=36[/latex]
2
[/hidden-answer][latex]4\left(3-1\right)x=4[/latex]
[latex]\frac{1}{4}\left(8w-{4}^{2}\right)=0[/latex]
2
[/hidden-answer]For the following exercises, simplify the expression.
[latex]4x+x\left(13-7\right)[/latex]
[latex]2y-{\left(4\right)}^{2}y-11[/latex]
[latex]-14y-11[/latex]
[/hidden-answer][latex]\frac{a}{{2}^{3}}\left(64\right)-12a÷6[/latex]
[latex]8b-4b\left(3\right)+1[/latex]
[latex]-4b+1[/latex]
[/hidden-answer][latex]5l÷3l\,×\,\left(9-6\right)[/latex]
[latex]7z-3+z\,×\,{6}^{2}[/latex]
[latex]43z-3[/latex]
[/hidden-answer][latex]4\,×\,3+18x÷9-12[/latex]
[latex]9\left(y+8\right)-27[/latex]
[latex]9y+45[/latex]
[/hidden-answer][latex]\left(\frac{9}{6}t-4\right)2[/latex]
[latex]6+12b-3\,×\,6b[/latex]
[latex]-6b+6[/latex]
[/hidden-answer][latex]18y-2\left(1+7y\right)[/latex]
[latex]{\left(\frac{4}{9}\right)}^{2}\,×\,27x[/latex]
[latex]\frac{16x}{3}[/latex]
[/hidden-answer][latex]8\left(3-m\right)+1\left(-8\right)[/latex]
[latex]9x+4x\left(2+3\right)-4\left(2x+3x\right)[/latex]
[latex]9x[/latex]
[/hidden-answer][latex]{5}^{2}-4\left(3x\right)[/latex]
For the following exercises, consider this scenario: Fred earns $40 mowing lawns. He spends $10 on mp3s, puts half of what is left in a savings account, and gets another $5 for washing his neighbor’s car.
Write the expression that represents the number of dollars Fred keeps (and does not put in his savings account). Remember the order of operations.
[latex]\frac{1}{2}\left(40-10\right)+5[/latex]
[/hidden-answer]How much money does Fred keep?
For the following exercises, solve the given problem.
According to the U.S. Mint, the diameter of a quarter is 0.955 inches. The circumference of the quarter would be the diameter multiplied by[latex]\,\pi .\,[/latex]Is the circumference of a quarter a whole number, a rational number, or an irrational number?
irrational number
[/hidden-answer]Jessica and her roommate, Adriana, have decided to share a change jar for joint expenses. Jessica put her loose change in the jar first, and then Adriana put her change in the jar. We know that it does not matter in which order the change was added to the jar. What property of addition describes this fact?
For the following exercises, consider this scenario: There is a mound of[latex]\,g\,[/latex]pounds of gravel in a quarry. Throughout the day, 400 pounds of gravel is added to the mound. Two orders of 600 pounds are sold and the gravel is removed from the mound. At the end of the day, the mound has 1,200 pounds of gravel.
Write the equation that describes the situation.
[latex]g+400-2\left(600\right)=1200[/latex]
[/hidden-answer]Solve for g.
For the following exercise, solve the given problem.
Ramon runs the marketing department at his company. His department gets a budget every year, and every year, he must spend the entire budget without going over. If he spends less than the budget, then his department gets a smaller budget the following year. At the beginning of this year, Ramon got $2.5 million for the annual marketing budget. He must spend the budget such that[latex]\,2,500,000-x=0.\,[/latex]What property of addition tells us what the value of x must be?
inverse property of addition
[/hidden-answer]For the following exercises, use a graphing calculator to solve for x. Round the answers to the nearest hundredth.
[latex]0.5{\left(12.3\right)}^{2}-48x=\frac{3}{5}[/latex]
[latex]{\left(0.25-0.75\right)}^{2}x-7.2=9.9[/latex]
68.4
[/hidden-answer]If a whole number is not a natural number, what must the number be?
Determine whether the statement is true or false: The multiplicative inverse of a rational number is also rational.
true
[/hidden-answer]Determine whether the statement is true or false: The product of a rational and irrational number is always irrational.
Determine whether the simplified expression is rational or irrational:[latex]\,\sqrt{-18-4\left(5\right)\left(-1\right)}.[/latex]
irrational
[/hidden-answer]Determine whether the simplified expression is rational or irrational:[latex]\,\sqrt{-16+4\left(5\right)+5}.[/latex]
The division of two whole numbers will always result in what type of number?
rational
[/hidden-answer]What property of real numbers would simplify the following expression:[latex]\,4+7\left(x-1\right)?[/latex]
Mathematicians, scientists, and economists commonly encounter very large and very small numbers. But it may not be obvious how common such figures are in everyday life. For instance, a pixel is the smallest unit of light that can be perceived and recorded by a digital camera. A particular camera might record an image that is 2,048 pixels by 1,536 pixels, which is a very high resolution picture. It can also perceive a color depth (gradations in colors) of up to 48 bits per frame, and can shoot the equivalent of 24 frames per second. The maximum possible number of bits of information used to film a one-hour (3,600-second) digital film is then an extremely large number.
Using a calculator, we enter[latex]\,2,048\,×\,1,536\,×\,48\,×\,24\,×\,3,600\,[/latex]and press ENTER. The calculator displays 1.304596316E13. What does this mean? The “E13” portion of the result represents the exponent 13 of ten, so there are a maximum of approximately[latex]\,1.3\,×\,{10}^{13}\,[/latex]bits of data in that one-hour film. In this section, we review rules of exponents first and then apply them to calculations involving very large or small numbers.
Consider the product[latex]\,{x}^{3}\cdot {x}^{4}.\,[/latex]Both terms have the same base, x, but they are raised to different exponents. Expand each expression, and then rewrite the resulting expression.
The result is that[latex]\,{x}^{3}\cdot {x}^{4}={x}^{3+4}={x}^{7}.[/latex]
Notice that the exponent of the product is the sum of the exponents of the terms. In other words, when multiplying exponential expressions with the same base, we write the result with the common base and add the exponents. This is the product rule of exponents.
Now consider an example with real numbers.
We can always check that this is true by simplifying each exponential expression. We find that[latex]\,{2}^{3}\,[/latex]is 8,[latex]\,{2}^{4}\,[/latex]is 16, and[latex]\,{2}^{7}\,[/latex]is 128. The product[latex]\,8\cdot 16\,[/latex]equals 128, so the relationship is true. We can use the product rule of exponents to simplify expressions that are a product of two numbers or expressions with the same base but different exponents.
For any real number[latex]\,a\,[/latex]and natural numbers[latex]\,m\,[/latex]and[latex]\,n,[/latex]the product rule of exponents states that
Write each of the following products with a single base. Do not simplify further.
Use the product rule to simplify each expression.
At first, it may appear that we cannot simplify a product of three factors. However, using the associative property of multiplication, begin by simplifying the first two.
Notice we get the same result by adding the three exponents in one step.
Write each of the following products with a single base. Do not simplify further.
The quotient rule of exponents allows us to simplify an expression that divides two numbers with the same base but different exponents. In a similar way to the product rule, we can simplify an expression such as[latex]\,\frac{{y}^{m}}{{y}^{n}},[/latex]where[latex]\,m>n.\,[/latex]Consider the example[latex]\,\frac{{y}^{9}}{{y}^{5}}.\,[/latex]Perform the division by canceling common factors.
Notice that the exponent of the quotient is the difference between the exponents of the divisor and dividend.
In other words, when dividing exponential expressions with the same base, we write the result with the common base and subtract the exponents.
For the time being, we must be aware of the condition[latex]\,m>n.\,[/latex]Otherwise, the difference[latex]\,m-n\,[/latex]could be zero or negative. Those possibilities will be explored shortly. Also, instead of qualifying variables as nonzero each time, we will simplify matters and assume from here on that all variables represent nonzero real numbers.
For any real number[latex]\,a\,[/latex]and natural numbers[latex]\,m\,[/latex]and[latex]\,n,[/latex]such that[latex]\,m>n,[/latex]the quotient rule of exponents states that
Write each of the following products with a single base. Do not simplify further.
Use the quotient rule to simplify each expression.
Write each of the following products with a single base. Do not simplify further.
Suppose an exponential expression is raised to some power. Can we simplify the result? Yes. To do this, we use the power rule of exponents. Consider the expression[latex]\,{\left({x}^{2}\right)}^{3}.\,[/latex]The expression inside the parentheses is multiplied twice because it has an exponent of 2. Then the result is multiplied three times because the entire expression has an exponent of 3.
The exponent of the answer is the product of the exponents:[latex]\,{\left({x}^{2}\right)}^{3}={x}^{2\cdot 3}={x}^{6}.\,[/latex]In other words, when raising an exponential expression to a power, we write the result with the common base and the product of the exponents.
Be careful to distinguish between uses of the product rule and the power rule. When using the product rule, different terms with the same bases are raised to exponents. In this case, you add the exponents. When using the power rule, a term in exponential notation is raised to a power. In this case, you multiply the exponents.
For any real number[latex]\,a\,[/latex]and positive integers[latex]\,m\,[/latex]and[latex]\,n,[/latex]the power rule of exponents states that
Write each of the following products with a single base. Do not simplify further.
Use the power rule to simplify each expression.
Write each of the following products with a single base. Do not simplify further.
Return to the quotient rule. We made the condition that[latex]\,m>n\,[/latex]so that the difference[latex]\,m-n\,[/latex]would never be zero or negative. What would happen if[latex]\,m=n?[/latex]In this case, we would use the zero exponent rule of exponents to simplify the expression to 1. To see how this is done, let us begin with an example.
If we were to simplify the original expression using the quotient rule, we would have
If we equate the two answers, the result is[latex]\,{t}^{0}=1.\,[/latex]This is true for any nonzero real number, or any variable representing a real number.
The sole exception is the expression[latex]\,{0}^{0}.\,[/latex]This appears later in more advanced courses, but for now, we will consider the value to be undefined.
For any nonzero real number[latex]\,a,[/latex]the zero exponent rule of exponents states that
Simplify each expression using the zero exponent rule of exponents.
Use the zero exponent and other rules to simplify each expression.
Simplify each expression using the zero exponent rule of exponents.
Another useful result occurs if we relax the condition that[latex]\,m>n\,[/latex]in the quotient rule even further. For example, can we simplify[latex]\,\frac{{h}^{3}}{{h}^{5}}?\,[/latex]When[latex]\,m<n[/latex]—that is, where the difference[latex]\,m-n\,[/latex]is negative—we can use the negative rule of exponents to simplify the expression to its reciprocal.
Divide one exponential expression by another with a larger exponent. Use our example,[latex]\,\frac{{h}^{3}}{{h}^{5}}.[/latex]
If we were to simplify the original expression using the quotient rule, we would have
Putting the answers together, we have[latex]\,{h}^{-2}=\frac{1}{{h}^{2}}.\,[/latex]This is true for any nonzero real number, or any variable representing a nonzero real number.
A factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar—from numerator to denominator or vice versa.
We have shown that the exponential expression[latex]\,{a}^{n}\,[/latex]is defined when[latex]\,n\,[/latex]is a natural number, 0, or the negative of a natural number. That means that[latex]\,{a}^{n}\,[/latex]is defined for any integer[latex]\,n.\,[/latex]Also, the product and quotient rules and all of the rules we will look at soon hold for any integer[latex]\,n.[/latex]
For any nonzero real number[latex]\,a\,[/latex]and natural number[latex]\,n,[/latex]the negative rule of exponents states that
Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.
Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.
Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.
Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.
To simplify the power of a product of two exponential expressions, we can use the power of a product rule of exponents, which breaks up the power of a product of factors into the product of the powers of the factors. For instance, consider[latex]\,{\left(pq\right)}^{3}.\,[/latex]We begin by using the associative and commutative properties of multiplication to regroup the factors.
In other words,[latex]\,{\left(pq\right)}^{3}={p}^{3}\cdot {q}^{3}.[/latex]
For any real numbers[latex]\,a\,[/latex]and[latex]\,b\,[/latex]and any integer[latex]\,n,[/latex]the power of a product rule of exponents states that
Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.
Use the product and quotient rules and the new definitions to simplify each expression.
Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.
To simplify the power of a quotient of two expressions, we can use the power of a quotient rule, which states that the power of a quotient of factors is the quotient of the powers of the factors. For example, let’s look at the following example.
Let’s rewrite the original problem differently and look at the result.
It appears from the last two steps that we can use the power of a product rule as a power of a quotient rule.
For any real numbers[latex]\,a\,[/latex]and[latex]\,b\,[/latex]and any integer[latex]\,n,[/latex]the power of a quotient rule of exponents states that
Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.
Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.
Recall that to simplify an expression means to rewrite it by combing terms or exponents; in other words, to write the expression more simply with fewer terms. The rules for exponents may be combined to simplify expressions.
Simplify each expression and write the answer with positive exponents only.
Simplify each expression and write the answer with positive exponents only.
Recall at the beginning of the section that we found the number[latex]\,1.3\,×\,{10}^{13}\,[/latex]when describing bits of information in digital images. Other extreme numbers include the width of a human hair, which is about 0.00005 m, and the radius of an electron, which is about 0.00000000000047 m. How can we effectively work read, compare, and calculate with numbers such as these?
A shorthand method of writing very small and very large numbers is called scientific notation, in which we express numbers in terms of exponents of 10. To write a number in scientific notation, move the decimal point to the right of the first digit in the number. Write the digits as a decimal number between 1 and 10. Count the number of places n that you moved the decimal point. Multiply the decimal number by 10 raised to a power of n. If you moved the decimal left as in a very large number,[latex]\,n\,[/latex]is positive. If you moved the decimal right as in a small large number,[latex]\,n\,[/latex]is negative.
For example, consider the number 2,780,418. Move the decimal left until it is to the right of the first nonzero digit, which is 2.
We obtain 2.780418 by moving the decimal point 6 places to the left. Therefore, the exponent of 10 is 6, and it is positive because we moved the decimal point to the left. This is what we should expect for a large number.
Working with small numbers is similar. Take, for example, the radius of an electron, 0.00000000000047 m. Perform the same series of steps as above, except move the decimal point to the right.
Be careful not to include the leading 0 in your count. We move the decimal point 13 places to the right, so the exponent of 10 is 13. The exponent is negative because we moved the decimal point to the right. This is what we should expect for a small number.
A number is written in scientific notation if it is written in the form[latex]\,a\,×\,{10}^{n},[/latex]where[latex]\,1\le |a|<10\,[/latex]and[latex]\,n\,[/latex]is an integer.
Write each number in scientific notation.
Observe that, if the given number is greater than 1, as in examples a–c, the exponent of 10 is positive; and if the number is less than 1, as in examples d–e, the exponent is negative.
Write each number in scientific notation.
To convert a number in scientific notation to standard notation, simply reverse the process. Move the decimal[latex]\,n\,[/latex]places to the right if[latex]\,n\,[/latex]is positive or[latex]\,n\,[/latex]places to the left if[latex]\,n\,[/latex]is negative and add zeros as needed. Remember, if[latex]\,n\,[/latex]is positive, the value of the number is greater than 1, and if[latex]\,n\,[/latex]is negative, the value of the number is less than one.
Convert each number in scientific notation to standard notation.
Convert each number in scientific notation to standard notation.
Scientific notation, used with the rules of exponents, makes calculating with large or small numbers much easier than doing so using standard notation. For example, suppose we are asked to calculate the number of atoms in 1 L of water. Each water molecule contains 3 atoms (2 hydrogen and 1 oxygen). The average drop of water contains around[latex]\,1.32\,×\,{10}^{21}\,[/latex]molecules of water and 1 L of water holds about[latex]\,1.22\,×\,{10}^{4}\,[/latex]average drops. Therefore, there are approximately[latex]\,3\cdot \left(1.32\,×\,{10}^{21}\right)\cdot \left(1.22\,×\,{10}^{4}\right)\approx 4.83\,×\,{10}^{25}\,[/latex]atoms in 1 L of water. We simply multiply the decimal terms and add the exponents. Imagine having to perform the calculation without using scientific notation!
When performing calculations with scientific notation, be sure to write the answer in proper scientific notation. For example, consider the product[latex]\,\left(7\,×\,{10}^{4}\right)\cdot \left(5\,×\,{10}^{6}\right)=35\,×\,{10}^{10}.\,[/latex]The answer is not in proper scientific notation because 35 is greater than 10. Consider 35 as[latex]\,3.5\,×\,10.\,[/latex]That adds a ten to the exponent of the answer.
Perform the operations and write the answer in scientific notation.
Perform the operations and write the answer in scientific notation.
In April 2014, the population of the United States was about 308,000,000 people. The national debt was about $17,547,000,000,000. Write each number in scientific notation, rounding figures to two decimal places, and find the amount of the debt per U.S. citizen. Write the answer in both scientific and standard notations.
The population was[latex]\,308,000,000=3.08\,×\,{10}^{8}.[/latex]
The national debt was[latex]\,\text{\$}17,547,000,000,000\approx \text{\$}1.75\,×\,{10}^{13}.[/latex]
To find the amount of debt per citizen, divide the national debt by the number of citizens.
The debt per citizen at the time was about[latex]\,\text{\$}5.7\,×\,{10}^{4},[/latex]or $57,000.
[/hidden-answer]An average human body contains around 30,000,000,000,000 red blood cells. Each cell measures approximately 0.000008 m long. Write each number in scientific notation and find the total length if the cells were laid end-to-end. Write the answer in both scientific and standard notations.
Number of cells:[latex]\,3×{10}^{13};[/latex]length of a cell:[latex]\,8×{10}^{-6}\,[/latex]m; total length:[latex]\,2.4×{10}^{8}\,[/latex]m or[latex]\,240,000,000\,[/latex]m.
[/hidden-answer]Access these online resources for additional instruction and practice with exponents and scientific notation.
Rules of Exponents< For nonzero real numbers[latex]\,a\,[/latex]and[latex]\,b\,[/latex]and integers[latex]\,m\,[/latex]and[latex]\,n\,[/latex] | |
Product rule | [latex]{a}^{m}\cdot {a}^{n}={a}^{m+n}[/latex] |
Quotient rule | [latex]\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}[/latex] |
Power rule | [latex]{\left({a}^{m}\right)}^{n}={a}^{m\cdot n}[/latex] |
Zero exponent rule | [latex]{a}^{0}=1[/latex] |
Negative rule | [latex]{a}^{-n}=\frac{1}{{a}^{n}}[/latex] |
Power of a product rule | [latex]{\left(a\cdot b\right)}^{n}={a}^{n}\cdot {b}^{n}[/latex] |
Power of a quotient rule | [latex]{\left(\frac{a}{b}\right)}^{n}=\frac{{a}^{n}}{{b}^{n}}[/latex] |
Is[latex]\,{2}^{3}\,[/latex]the same as[latex]\,{3}^{2}?\,[/latex]Explain.
No, the two expressions are not the same. An exponent tells how many times you multiply the base. So[latex]\,{2}^{3}\,[/latex]is the same as[latex]\,2×2×2,[/latex]which is 8.[latex]\,{3}^{2}\,[/latex]is the same as[latex]\,3×3,[/latex]which is 9.
[/hidden-answer]When can you add two exponents?
What is the purpose of scientific notation?
It is a method of writing very small and very large numbers.
[/hidden-answer]Explain what a negative exponent does.
For the following exercises, simplify the given expression. Write answers with positive exponents.
[latex]\,{9}^{2}\,[/latex]
81
[/hidden-answer][latex]{15}^{-2}[/latex]
[latex]{3}^{2}\,×\,{3}^{3}[/latex]
243
[/hidden-answer][latex]{4}^{4}÷4[/latex]
[latex]{\left({2}^{2}\right)}^{-2}[/latex]
[latex]\frac{1}{16}[/latex]
[/hidden-answer][latex]{\left(5-8\right)}^{0}[/latex]
[latex]{11}^{3}÷{11}^{4}[/latex]
[latex]\frac{1}{11}[/latex]
[/hidden-answer][latex]{6}^{5}\,×\,{6}^{-7}[/latex]
[latex]{\left({8}^{0}\right)}^{2}[/latex]
1
[/hidden-answer][latex]{5}^{-2}÷{5}^{2}[/latex]
For the following exercises, write each expression with a single base. Do not simplify further. Write answers with positive exponents.
[latex]{4}^{2}\,×\,{4}^{3}÷{4}^{-4}[/latex]
[latex]{4}^{9}[/latex]
[/hidden-answer][latex]\frac{{6}^{12}}{{6}^{9}}[/latex]
[latex]{\left({12}^{3}\,×\,12\right)}^{10}[/latex]
[latex]{12}^{40}[/latex]
[/hidden-answer][latex]{10}^{6}÷{\left({10}^{10}\right)}^{-2}[/latex]
[latex]{7}^{-6}\,×\,{7}^{-3}[/latex]
[latex]\frac{1}{{7}^{9}}[/latex]
[/hidden-answer][latex]{\left({3}^{3}÷{3}^{4}\right)}^{5}[/latex]
For the following exercises, express the decimal in scientific notation.
0.0000314
[latex]3.14\,×{10}^{-5}[/latex]
[/hidden-answer]148,000,000
For the following exercises, convert each number in scientific notation to standard notation.
[latex]1.6\,×\,{10}^{10}[/latex]
16,000,000,000
[/hidden-answer][latex]9.8\,×\,{10}^{-9}[/latex]
For the following exercises, simplify the given expression. Write answers with positive exponents.
[latex]\frac{{a}^{3}{a}^{2}}{a}[/latex]
[latex]{a}^{4}[/latex]
[/hidden-answer][latex]\frac{m{n}^{2}}{{m}^{-2}}[/latex]
[latex]{\left({b}^{3}{c}^{4}\right)}^{2}[/latex]
[latex]{b}^{6}{c}^{8}[/latex]
[/hidden-answer][latex]{\left(\frac{{x}^{-3}}{{y}^{2}}\right)}^{-5}[/latex]
[latex]a{b}^{2}÷{d}^{-3}[/latex]
[latex]a{b}^{2}{d}^{3}[/latex]
[/hidden-answer][latex]{\left({w}^{0}{x}^{5}\right)}^{-1}[/latex]
[latex]\frac{{m}^{4}}{{n}^{0}}[/latex]
[latex]{m}^{4}[/latex]
[/hidden-answer][latex]{y}^{-4}{\left({y}^{2}\right)}^{2}[/latex]
[latex]\frac{{p}^{-4}{q}^{2}}{{p}^{2}{q}^{-3}}[/latex]
[latex]\frac{{q}^{5}}{{p}^{6}}[/latex]
[/hidden-answer][latex]{\left(l\,×\,w\right)}^{2}[/latex]
[latex]{\left({y}^{7}\right)}^{3}÷{x}^{14}[/latex]
[latex]\frac{{y}^{21}}{{x}^{14}}[/latex]
[/hidden-answer][latex]{\left(\frac{a}{{2}^{3}}\right)}^{2}[/latex]
[latex]{5}^{2}m÷{5}^{0}m[/latex]
[latex]25[/latex]
[/hidden-answer][latex]\frac{{\left(16\sqrt{x}\right)}^{2}}{{y}^{-1}}[/latex]
[latex]\frac{{2}^{3}}{{\left(3a\right)}^{-2}}[/latex]
[latex]72{a}^{2}[/latex]
[/hidden-answer][latex]{\left(m{a}^{6}\right)}^{2}\frac{1}{{m}^{3}{a}^{2}}[/latex]
[latex]{\left({b}^{-3}c\right)}^{3}[/latex]
[latex]\frac{{c}^{3}}{{b}^{9}}[/latex]
[/hidden-answer][latex]{\left({x}^{2}{y}^{13}÷{y}^{0}\right)}^{2}[/latex]
[latex]{\left(9{z}^{3}\right)}^{-2}y[/latex]
[latex]\frac{y}{81{z}^{6}}[/latex]
[/hidden-answer]To reach escape velocity, a rocket must travel at the rate of[latex]\,2.2\,×\,{10}^{6}\,[/latex]ft/min. Rewrite the rate in standard notation.
A dime is the thinnest coin in U.S. currency. A dime’s thickness measures[latex]\,1.35\,×\,{10}^{-3}\,[/latex]m. Rewrite the number in standard notation.
0.00135 m
[/hidden-answer]The average distance between Earth and the Sun is 92,960,000 mi. Rewrite the distance using scientific notation.
A terabyte is made of approximately 1,099,500,000,000 bytes. Rewrite in scientific notation.
[latex]1.0995×{10}^{12}[/latex]
[/hidden-answer]The Gross Domestic Product (GDP) for the United States in the first quarter of 2014 was[latex]\,\text{\$}1.71496\,×\,{10}^{13}.\,[/latex]Rewrite the GDP in standard notation.
One picometer is approximately[latex]\,3.397\,×\,{10}^{-11}\,[/latex]in. Rewrite this length using standard notation.
0.00000000003397 in.
[/hidden-answer]The value of the services sector of the U.S. economy in the first quarter of 2012 was $10,633.6 billion. Rewrite this amount in scientific notation.
For the following exercises, use a graphing calculator to simplify. Round the answers to the nearest hundredth.
[latex]{\left(\frac{{12}^{3}{m}^{33}}{{4}^{-3}}\right)}^{2}[/latex]
12,230,590,464[latex]\,{m}^{66}[/latex]
[/hidden-answer][latex]{17}^{3}÷{15}^{2}{x}^{3}[/latex]
For the following exercises, simplify the given expression. Write answers with positive exponents.
[latex]{\left(\frac{{3}^{2}}{{a}^{3}}\right)}^{-2}{\left(\frac{{a}^{4}}{{2}^{2}}\right)}^{2}[/latex]
[latex]\frac{{a}^{14}}{1296}[/latex]
[/hidden-answer][latex]{\left({6}^{2}-24\right)}^{2}÷{\left(\frac{x}{y}\right)}^{-5}[/latex]
[latex]\frac{{m}^{2}{n}^{3}}{{a}^{2}{c}^{-3}}\cdot \frac{{a}^{-7}{n}^{-2}}{{m}^{2}{c}^{4}}[/latex]
[latex]\frac{n}{{a}^{9}c}[/latex]
[/hidden-answer][latex]{\left(\frac{{x}^{6}{y}^{3}}{{x}^{3}{y}^{-3}}\cdot \frac{{y}^{-7}}{{x}^{-3}}\right)}^{10}[/latex]
[latex]{\left(\frac{{\left(a{b}^{2}c\right)}^{-3}}{{b}^{-3}}\right)}^{2}[/latex]
[latex]\frac{1}{{a}^{6}{b}^{6}{c}^{6}}[/latex]
[/hidden-answer]Avogadro’s constant is used to calculate the number of particles in a mole. A mole is a basic unit in chemistry to measure the amount of a substance. The constant is[latex]\,6.0221413\,×\,{10}^{23}.\,[/latex]Write Avogadro’s constant in standard notation.
Planck’s constant is an important unit of measure in quantum physics. It describes the relationship between energy and frequency. The constant is written as[latex]\,6.62606957\,×\,{10}^{-34}.\,[/latex]Write Planck’s constant in standard notation.
0.000000000000000000000000000000000662606957
[/hidden-answer]A hardware store sells 16-ft ladders and 24-ft ladders. A window is located 12 feet above the ground. A ladder needs to be purchased that will reach the window from a point on the ground 5 feet from the building. To find out the length of ladder needed, we can draw a right triangle as shown in (Figure), and use the Pythagorean Theorem.
When the square root of a number is squared, the result is the original number. Since[latex]\,{4}^{2}=16,[/latex]the square root of[latex]\,16\,[/latex]is[latex]\,4.\,[/latex]The square root function is the inverse of the squaring function just as subtraction is the inverse of addition. To undo squaring, we take the square root.
In general terms, if[latex]\,a\,[/latex]is a positive real number, then the square root of[latex]\,a\,[/latex]is a number that, when multiplied by itself, gives[latex]\,a.\,[/latex]The square root could be positive or negative because multiplying two negative numbers gives a positive number. The principal square root is the nonnegative number that when multiplied by itself equals[latex]\,a.\,[/latex]The square root obtained using a calculator is the principal square root.
The principal square root of[latex]\,a\,[/latex]is written as[latex]\,\sqrt{a}.\,[/latex]The symbol is called a radical, the term under the symbol is called the radicand, and the entire expression is called a radical expression.
The principal square root of[latex]\,a\,[/latex]is the nonnegative number that, when multiplied by itself, equals[latex]\,a.\,[/latex]It is written as a radical expression, with a symbol called a radical over the term called the radicand:[latex]\,\sqrt{a}.[/latex]
Does[latex]\,\sqrt{25}=±5?[/latex]
No. Although both[latex]\,{5}^{2}\,[/latex]and[latex]\,{\left(-5\right)}^{2}\,[/latex]are[latex]\,25,[/latex]the radical symbol implies only a nonnegative root, the principal square root. The principal square root of 25 is[latex]\,\sqrt{25}=5.[/latex]
Evaluate each expression.
For[latex]\,\sqrt{25+144},[/latex]can we find the square roots before adding?
No.[latex]\,\sqrt{25}+\sqrt{144}=5+12=17.\,[/latex]This is not equivalent to[latex]\,\sqrt{25+144}=13.\,[/latex]The order of operations requires us to add the terms in the radicand before finding the square root.
Evaluate each expression.
To simplify a square root, we rewrite it such that there are no perfect squares in the radicand. There are several properties of square roots that allow us to simplify complicated radical expressions. The first rule we will look at is the product rule for simplifying square roots, which allows us to separate the square root of a product of two numbers into the product of two separate rational expressions. For instance, we can rewrite[latex]\,\sqrt{15}\,[/latex]as[latex]\,\sqrt{3}\cdot \sqrt{5}.\,[/latex]We can also use the product rule to express the product of multiple radical expressions as a single radical expression.
If[latex]\,a\,[/latex]and[latex]\,b\,[/latex]are nonnegative, the square root of the product[latex]\,ab\,[/latex]is equal to the product of the square roots of[latex]\,a\,[/latex]and[latex]\,b.\,[/latex]
Given a square root radical expression, use the product rule to simplify it.
Simplify the radical expression.
Simplify[latex]\,\sqrt{50{x}^{2}{y}^{3}z}.[/latex]
[latex]5|x||y|\sqrt{2yz}.\,[/latex]Notice the absolute value signs around x and y? That’s because their value must be positive!
[/hidden-answer]Given the product of multiple radical expressions, use the product rule to combine them into one radical expression.
Simplify the radical expression.
[latex]\begin{array}{cc}\sqrt{12\cdot 3}\hfill & \phantom{\rule{2em}{0ex}}\text{Express the product as a single radical expression}.\hfill \\ \sqrt{36}\hfill & \phantom{\rule{2em}{0ex}}\text{Simplify}.\hfill \\ 6\hfill & \end{array}[/latex]
[/hidden-answer]Simplify[latex]\,\sqrt{50x}\cdot \sqrt{2x}\,[/latex]assuming[latex]\,x>0.[/latex]
[latex]10|x|[/latex]
[/hidden-answer]Just as we can rewrite the square root of a product as a product of square roots, so too can we rewrite the square root of a quotient as a quotient of square roots, using the quotient rule for simplifying square roots. It can be helpful to separate the numerator and denominator of a fraction under a radical so that we can take their square roots separately. We can rewrite[latex]\,\sqrt{\frac{5}{2}}\,[/latex]as[latex]\,\frac{\sqrt{5}}{\sqrt{2}}.[/latex]
The square root of the quotient[latex]\,\frac{a}{b}\,[/latex]is equal to the quotient of the square roots of[latex]\,a\,[/latex]and[latex]\,b,[/latex]where[latex]\,b\ne 0.[/latex]
Given a radical expression, use the quotient rule to simplify it.
Simplify the radical expression.
[latex]\sqrt{\frac{5}{36}}[/latex]
[latex]\begin{array}{cc}\frac{\sqrt{5}}{\sqrt{36}}\hfill & \phantom{\rule{2em}{0ex}}\text{Write as quotient of two radical expressions}.\hfill \\ \frac{\sqrt{5}}{6}\hfill & \phantom{\rule{2em}{0ex}}\text{Simplify denominator}.\hfill \end{array}[/latex]
[/hidden-answer]Simplify[latex]\,\sqrt{\frac{2{x}^{2}}{9{y}^{4}}}.[/latex]
[latex]\frac{x\sqrt{2}}{3{y}^{2}}.\,[/latex]We do not need the absolute value signs for[latex]\,{y}^{2}\,[/latex]because that term will always be nonnegative.
[/hidden-answer]Simplify the radical expression.
[latex]\frac{\sqrt{234{x}^{11}y}}{\sqrt{26{x}^{7}y}}[/latex]
[latex]\begin{array}{cc}\sqrt{\frac{234{x}^{11}y}{26{x}^{7}y}}\hfill & \phantom{\rule{2em}{0ex}}\text{Combine numerator and denominator into one radical expression}.\hfill \\ \sqrt{9{x}^{4}}\hfill & \phantom{\rule{2em}{0ex}}\text{Simplify fraction}.\hfill \\ 3{x}^{2}\text{ }\hfill & \phantom{\rule{2em}{0ex}}\text{Simplify square root}.\hfill \end{array}[/latex]
[/hidden-answer]Simplify[latex]\,\frac{\sqrt{9{a}^{5}{b}^{14}}}{\sqrt{3{a}^{4}{b}^{5}}}.[/latex]
[latex]{b}^{4}\sqrt{3ab}[/latex]
[/hidden-answer]We can add or subtract radical expressions only when they have the same radicand and when they have the same radical type such as square roots. For example, the sum of[latex]\,\sqrt{2}\,[/latex]and[latex]\,3\sqrt{2}\,[/latex]is[latex]\,4\sqrt{2}.\,[/latex]However, it is often possible to simplify radical expressions, and that may change the radicand. The radical expression[latex]\,\sqrt{18}\,[/latex]can be written with a[latex]\,2\,[/latex]in the radicand, as[latex]\,3\sqrt{2},[/latex]so[latex]\,\sqrt{2}+\sqrt{18}=\sqrt{2}+3\sqrt{2}=4\sqrt{2}.[/latex]
Given a radical expression requiring addition or subtraction of square roots, solve.
Add[latex]\,5\sqrt{12}+2\sqrt{3}.[/latex]
We can rewrite[latex]\,5\sqrt{12}\,[/latex]as[latex]\,5\sqrt{4·3}.\,[/latex]According the product rule, this becomes[latex]\,5\sqrt{4}\sqrt{3}.\,[/latex]The square root of[latex]\,\sqrt{4}\,[/latex]is 2, so the expression becomes[latex]\,5\left(2\right)\sqrt{3},[/latex]which is[latex]\,10\sqrt{3}.\,[/latex]Now we can the terms have the same radicand so we can add.
[latex]10\sqrt{3}+2\sqrt{3}=12\sqrt{3}[/latex]
[/hidden-answer]Add[latex]\,\sqrt{5}+6\sqrt{20}.[/latex]
[latex]13\sqrt{5}[/latex]
[/hidden-answer]Subtract[latex]\,20\sqrt{72{a}^{3}{b}^{4}c}\,-14\sqrt{8{a}^{3}{b}^{4}c}.[/latex]
Rewrite each term so they have equal radicands.
Now the terms have the same radicand so we can subtract.
Subtract[latex]\,3\sqrt{80x}\,-4\sqrt{45x}.[/latex]
[latex]0[/latex]
[/hidden-answer]When an expression involving square root radicals is written in simplest form, it will not contain a radical in the denominator. We can remove radicals from the denominators of fractions using a process called rationalizing the denominator.
We know that multiplying by 1 does not change the value of an expression. We use this property of multiplication to change expressions that contain radicals in the denominator. To remove radicals from the denominators of fractions, multiply by the form of 1 that will eliminate the radical.
For a denominator containing a single term, multiply by the radical in the denominator over itself. In other words, if the denominator is[latex]\,b\sqrt{c},[/latex]multiply by[latex]\,\frac{\sqrt{c}}{\sqrt{c}}.[/latex]
For a denominator containing the sum or difference of a rational and an irrational term, multiply the numerator and denominator by the conjugate of the denominator, which is found by changing the sign of the radical portion of the denominator. If the denominator is[latex]\,a+b\sqrt{c},[/latex]then the conjugate is[latex]\,a-b\sqrt{c}.[/latex]
Given an expression with a single square root radical term in the denominator, rationalize the denominator.
Write[latex]\,\frac{2\sqrt{3}}{3\sqrt{10}}\,[/latex]in simplest form.
The radical in the denominator is[latex]\,\sqrt{10}.\,[/latex]So multiply the fraction by[latex]\,\frac{\sqrt{10}}{\sqrt{10}}.\,[/latex]Then simplify.
Write[latex]\,\frac{12\sqrt{3}}{\sqrt{2}}\,[/latex]in simplest form.
[latex]6\sqrt{6}[/latex]
[/hidden-answer]Given an expression with a radical term and a constant in the denominator, rationalize the denominator.
Write[latex]\,\frac{4}{1+\sqrt{5}}\,[/latex]in simplest form.
Begin by finding the conjugate of the denominator by writing the denominator and changing the sign. So the conjugate of[latex]\,1+\sqrt{5}\,[/latex]is[latex]\,1-\sqrt{5}.\,[/latex]Then multiply the fraction by[latex]\,\frac{1-\sqrt{5}}{1-\sqrt{5}}.[/latex]
Write[latex]\,\frac{7}{2+\sqrt{3}}\,[/latex]in simplest form.
[latex]14-7\sqrt{3}[/latex]
[/hidden-answer]Although square roots are the most common rational roots, we can also find cube roots, 4th roots, 5th roots, and more. Just as the square root function is the inverse of the squaring function, these roots are the inverse of their respective power functions. These functions can be useful when we need to determine the number that, when raised to a certain power, gives a certain number.
Suppose we know that[latex]\,{a}^{3}=8.\,[/latex]We want to find what number raised to the 3rd power is equal to 8. Since[latex]\,{2}^{3}=8,[/latex]we say that 2 is the cube root of 8.
The nth root of[latex]\,a\,[/latex]is a number that, when raised to the nth power, gives[latex]\,a.\,[/latex]For example,[latex]\,-3\,[/latex]is the 5th root of[latex]\,-243\,[/latex]because[latex]\,{\left(-3\right)}^{5}=-243.\,[/latex]If[latex]\,a\,[/latex]is a real number with at least one nth root, then the principal nth root of[latex]\,a\,[/latex]is the number with the same sign as[latex]\,a\,[/latex]that, when raised to the nth power, equals[latex]\,a.[/latex]
The principal nth root of[latex]\,a\,[/latex]is written as[latex]\,\sqrt[n]{a},[/latex]where[latex]\,n\,[/latex]is a positive integer greater than or equal to 2. In the radical expression,[latex]\,n\,[/latex]is called the index of the radical.
If[latex]\,a\,[/latex]is a real number with at least one nth root, then the principal nth root of[latex]\,a,[/latex]written as[latex]\,\sqrt[n]{a},[/latex]is the number with the same sign as[latex]\,a\,[/latex]that, when raised to the nth power, equals[latex]\,a.\,[/latex]The index of the radical is[latex]\,n.[/latex]
Simplify each of the following:
Simplify.
Radical expressions can also be written without using the radical symbol. We can use rational (fractional) exponents. The index must be a positive integer. If the index[latex]\,n\,[/latex]is even, then[latex]\,a\,[/latex]cannot be negative.
We can also have rational exponents with numerators other than 1. In these cases, the exponent must be a fraction in lowest terms. We raise the base to a power and take an nth root. The numerator tells us the power and the denominator tells us the root.
All of the properties of exponents that we learned for integer exponents also hold for rational exponents.
Rational exponents are another way to express principal nth roots. The general form for converting between a radical expression with a radical symbol and one with a rational exponent is
Given an expression with a rational exponent, write the expression as a radical.
Write[latex]\,{343}^{\frac{2}{3}}\,[/latex]as a radical. Simplify.
The 2 tells us the power and the 3 tells us the root.
[latex]{343}^{\frac{2}{3}}={\left(\sqrt[3]{343}\right)}^{2}=\sqrt[3]{{343}^{2}}[/latex]
We know that[latex]\,\sqrt[3]{343}=7\,[/latex]because[latex]\,{7}^{3}=343.\,[/latex]Because the cube root is easy to find, it is easiest to find the cube root before squaring for this problem. In general, it is easier to find the root first and then raise it to a power.
[latex]{343}^{\frac{2}{3}}={\left(\sqrt[3]{343}\right)}^{2}={7}^{2}=49[/latex]
[/hidden-answer]Write[latex]\,{9}^{\frac{5}{2}}\,[/latex]as a radical. Simplify.
[latex]{\left(\sqrt{9}\right)}^{5}={3}^{5}=243[/latex]
[/hidden-answer]Write[latex]\,\frac{4}{\sqrt[7]{{a}^{2}}}\,[/latex]using a rational exponent.
Write[latex]\,x\sqrt{{\left(5y\right)}^{9}}\,[/latex]using a rational exponent.
[latex]x{\left(5y\right)}^{\frac{9}{2}}[/latex]
[/hidden-answer]Simplify:
Simplify[latex]\,\left[{\left(8x\right)}^{\frac{1}{3}}\right]\left(14{x}^{\frac{6}{5}}\right).[/latex]
[latex]28{x}^{\frac{23}{15}}[/latex]
[/hidden-answer]Access these online resources for additional instruction and practice with radicals and rational exponents.
What does it mean when a radical does not have an index? Is the expression equal to the radicand? Explain.
When there is no index, it is assumed to be 2 or the square root. The expression would only be equal to the radicand if the index were 1.
[/hidden-answer]Where would radicals come in the order of operations? Explain why.
Every number will have two square roots. What is the principal square root?
The principal square root is the nonnegative root of the number.
[/hidden-answer]Can a radical with a negative radicand have a real square root? Why or why not?
For the following exercises, simplify each expression.
[latex]\sqrt{256}[/latex]
16
[/hidden-answer][latex]\sqrt{\sqrt{256}}[/latex]
[latex]\sqrt{4\left(9+16\right)}[/latex]
10
[/hidden-answer][latex]\sqrt{289}-\sqrt{121}[/latex]
[latex]\sqrt{196}[/latex]
14
[/hidden-answer][latex]\sqrt{1}[/latex]
[latex]\sqrt{98}[/latex]
[latex]7\sqrt{2}[/latex]
[/hidden-answer][latex]\sqrt{\frac{27}{64}}[/latex]
[latex]\sqrt{\frac{81}{5}}[/latex]
[latex]\frac{9\sqrt{5}}{5}[/latex]
[/hidden-answer][latex]\sqrt{800}[/latex]
[latex]\sqrt{169}+\sqrt{144}[/latex]
25
[/hidden-answer][latex]\sqrt{\frac{8}{50}}[/latex]
[latex]\frac{18}{\sqrt{162}}[/latex]
[latex]\sqrt{2}[/latex]
[/hidden-answer][latex]\sqrt{192}[/latex]
[latex]14\sqrt{6}-6\sqrt{24}[/latex]
[latex]2\sqrt{6}[/latex]
[/hidden-answer][latex]15\sqrt{5}+7\sqrt{45}[/latex]
[latex]\sqrt{150}[/latex]
[latex]5\sqrt{6}[/latex]
[/hidden-answer][latex]\sqrt{\frac{96}{100}}[/latex]
[latex]\left(\sqrt{42}\right)\left(\sqrt{30}\right)[/latex]
[latex]6\sqrt{35}[/latex]
[/hidden-answer][latex]12\sqrt{3}-4\sqrt{75}[/latex]
[latex]\sqrt{\frac{4}{225}}[/latex]
[latex]\frac{2}{15}[/latex]
[/hidden-answer][latex]\sqrt{\frac{405}{324}}[/latex]
[latex]\sqrt{\frac{360}{361}}[/latex]
[latex]\frac{6\sqrt{10}}{19}[/latex]
[/hidden-answer][latex]\frac{5}{1+\sqrt{3}}[/latex]
[latex]\frac{8}{1-\sqrt{17}}[/latex]
[latex]-\frac{1+\sqrt{17}}{2}[/latex]
[/hidden-answer][latex]\sqrt[4]{16}[/latex]
[latex]\sqrt[3]{128}+3\sqrt[3]{2}[/latex]
[latex]7\sqrt[3]{2}[/latex]
[/hidden-answer][latex]\sqrt[5]{\frac{-32}{243}}[/latex]
[latex]\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}[/latex]
[latex]15\sqrt{5}[/latex]
[/hidden-answer][latex]3\sqrt[3]{-432}+\sqrt[3]{16}[/latex]
For the following exercises, simplify each expression.
[latex]\sqrt{400{x}^{4}}[/latex]
[latex]20{x}^{2}[/latex]
[/hidden-answer][latex]\sqrt{4{y}^{2}}[/latex]
[latex]\sqrt{49p}[/latex]
[latex]7\sqrt{p}[/latex]
[/hidden-answer][latex]{\left(144{p}^{2}{q}^{6}\right)}^{\frac{1}{2}}[/latex]
[latex]{m}^{\frac{5}{2}}\sqrt{289}[/latex]
[latex]17{m}^{2}\sqrt{m}[/latex]
[/hidden-answer][latex]9\sqrt{3{m}^{2}}+\sqrt{27}[/latex]
[latex]3\sqrt{a{b}^{2}}-b\sqrt{a}[/latex]
[latex]2b\sqrt{a}[/latex]
[/hidden-answer][latex]\frac{4\sqrt{2n}}{\sqrt{16{n}^{4}}}[/latex]
[latex]\sqrt{\frac{225{x}^{3}}{49x}}[/latex]
[latex]\frac{15x}{7}[/latex]
[/hidden-answer][latex]3\sqrt{44z}+\sqrt{99z}[/latex]
[latex]\sqrt{50{y}^{8}}[/latex]
[latex]5{y}^{4}\sqrt{2}[/latex]
[/hidden-answer][latex]\sqrt{490b{c}^{2}}[/latex]
[latex]\sqrt{\frac{32}{14d}}[/latex]
[latex]\frac{4\sqrt{7d}}{7d}[/latex]
[/hidden-answer][latex]{q}^{\frac{3}{2}}\sqrt{63p}[/latex]
[latex]\frac{\sqrt{8}}{1-\sqrt{3x}}[/latex]
[latex]\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}[/latex]
[/hidden-answer][latex]\sqrt{\frac{20}{121{d}^{4}}}[/latex]
[latex]{w}^{\frac{3}{2}}\sqrt{32}-{w}^{\frac{3}{2}}\sqrt{50}[/latex]
[latex]-w\sqrt{2w}[/latex]
[/hidden-answer][latex]\sqrt{108{x}^{4}}+\sqrt{27{x}^{4}}[/latex]
[latex]\frac{\sqrt{12x}}{2+2\sqrt{3}}[/latex]
[latex]\frac{3\sqrt{x}-\sqrt{3x}}{2}[/latex]
[/hidden-answer][latex]\sqrt{147{k}^{3}}[/latex]
[latex]\sqrt{125{n}^{10}}[/latex]
[latex]5{n}^{5}\sqrt{5}[/latex]
[/hidden-answer][latex]\sqrt{\frac{42q}{36{q}^{3}}}[/latex]
[latex]\sqrt{\frac{81m}{361{m}^{2}}}[/latex]
[latex]\frac{9\sqrt{m}}{19m}[/latex]
[/hidden-answer][latex]\sqrt{72c}-2\sqrt{2c}[/latex]
[latex]\sqrt{\frac{144}{324{d}^{2}}}[/latex]
[latex]\frac{2}{3d}[/latex]
[/hidden-answer][latex]\sqrt[3]{24{x}^{6}}+\sqrt[3]{81{x}^{6}}[/latex]
[latex]\sqrt[4]{\frac{162{x}^{6}}{16{x}^{4}}}[/latex]
[latex]\frac{3\sqrt[4]{2{x}^{2}}}{2}[/latex]
[/hidden-answer][latex]\sqrt[3]{64y}[/latex]
[latex]\sqrt[3]{128{z}^{3}}-\sqrt[3]{-16{z}^{3}}[/latex]
[latex]6z\sqrt[3]{2}[/latex]
[/hidden-answer][latex]\sqrt[5]{1,024{c}^{10}}[/latex]
A guy wire for a suspension bridge runs from the ground diagonally to the top of the closest pylon to make a triangle. We can use the Pythagorean Theorem to find the length of guy wire needed. The square of the distance between the wire on the ground and the pylon on the ground is 90,000 feet. The square of the height of the pylon is 160,000 feet. So the length of the guy wire can be found by evaluating[latex]\,\sqrt{90,000+160,000}.\,[/latex]What is the length of the guy wire?
500 feet
[/hidden-answer]A car accelerates at a rate of[latex]\,6-\frac{\sqrt{4}}{\sqrt{t}}{\text{ m/s}}^{2}\,[/latex]where t is the time in seconds after the car moves from rest. Simplify the expression.
For the following exercises, simplify each expression.
[latex]\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}}-{2}^{\frac{1}{2}}[/latex]
[latex]\frac{-5\sqrt{2}-6}{7}[/latex]
[/hidden-answer][latex]\frac{{4}^{\frac{3}{2}}-{16}^{\frac{3}{2}}}{{8}^{\frac{1}{3}}}[/latex]
[latex]\frac{\sqrt{m{n}^{3}}}{{a}^{2}\sqrt{{c}^{-3}}}\cdot \frac{{a}^{-7}{n}^{-2}}{\sqrt{{m}^{2}{c}^{4}}}[/latex]
[latex]\frac{\sqrt{mnc}}{{a}^{9}cmn}[/latex]
[/hidden-answer][latex]\frac{a}{a-\sqrt{c}}[/latex]
[latex]\frac{x\sqrt{64y}+4\sqrt{y}}{\sqrt{128y}}[/latex]
[latex]\frac{2\sqrt{2}x+\sqrt{2}}{4}[/latex]
[/hidden-answer][latex]\left(\frac{\sqrt{250{x}^{2}}}{\sqrt{100{b}^{3}}}\right)\left(\frac{7\sqrt{b}}{\sqrt{125x}}\right)[/latex]
[latex]\sqrt{\frac{\sqrt[3]{64}+\sqrt[4]{256}}{\sqrt{64}+\sqrt{256}}}[/latex]
[latex]\frac{\sqrt{3}}{3}[/latex]
[/hidden-answer]Earl is building a doghouse, whose front is in the shape of a square topped with a triangle. There will be a rectangular door through which the dog can enter and exit the house. Earl wants to find the area of the front of the doghouse so that he can purchase the correct amount of paint. Using the measurements of the front of the house, shown in (Figure), we can create an expression that combines several variable terms, allowing us to solve this problem and others like it.
First find the area of the square in square feet.
Then find the area of the triangle in square feet.
Next find the area of the rectangular door in square feet.
The area of the front of the doghouse can be found by adding the areas of the square and the triangle, and then subtracting the area of the rectangle. When we do this, we get[latex]\,4{x}^{2}+\frac{3}{2}x-x\,{\text{ft}}^{2},[/latex]or[latex]\,4{x}^{2}+\frac{1}{2}x\,[/latex]ft^{2}.
In this section, we will examine expressions such as this one, which combine several variable terms.
The formula just found is an example of a polynomial, which is a sum of or difference of terms, each consisting of a variable raised to a nonnegative integer power. A number multiplied by a variable raised to an exponent, such as[latex]\,384\pi ,[/latex]is known as a coefficient. Coefficients can be positive, negative, or zero, and can be whole numbers, decimals, or fractions. Each product[latex]\,{a}_{i}{x}^{i},[/latex]such as[latex]\,384\pi w,[/latex]is a term of a polynomial. If a term does not contain a variable, it is called a constant.
A polynomial containing only one term, such as[latex]\,5{x}^{4},[/latex]is called a monomial. A polynomial containing two terms, such as[latex]\,2x-9,[/latex]is called a binomial. A polynomial containing three terms, such as[latex]\,-3{x}^{2}+8x-7,[/latex]is called a trinomial.
We can find the degree of a polynomial by identifying the highest power of the variable that occurs in the polynomial. The term with the highest degree is called the leading term because it is usually written first. The coefficient of the leading term is called the leading coefficient. When a polynomial is written so that the powers are descending, we say that it is in standard form.
A polynomial is an expression that can be written in the form
Each real number a_{i}is called a coefficient. The number[latex]\,{a}_{0}\,[/latex]that is not multiplied by a variable is called a constant. Each product[latex]\,{a}_{i}{x}^{i}\,[/latex]is a term of a polynomial. The highest power of the variable that occurs in the polynomial is called the degree of a polynomial. The leading term is the term with the highest power, and its coefficient is called the leading coefficient.
Given a polynomial expression, identify the degree and leading coefficient.
For the following polynomials, identify the degree, the leading term, and the leading coefficient.
Identify the degree, leading term, and leading coefficient of the polynomial[latex]\,4{x}^{2}-{x}^{6}+2x-6.[/latex]
The degree is 6, the leading term is[latex]\,-{x}^{6},[/latex]and the leading coefficient is[latex]\,-1.[/latex]
[/hidden-answer]We can add and subtract polynomials by combining like terms, which are terms that contain the same variables raised to the same exponents. For example,[latex]\,5{x}^{2}\,[/latex]and[latex]\,-2{x}^{2}\,[/latex]are like terms, and can be added to get[latex]\,3{x}^{2},[/latex]but[latex]\,3x\,[/latex]and[latex]\,3{x}^{2}\,[/latex]are not like terms, and therefore cannot be added.
Given multiple polynomials, add or subtract them to simplify the expressions.
Find the sum.
[latex]\left(12{x}^{2}+9x-21\right)+\left(4{x}^{3}+8{x}^{2}-5x+20\right)[/latex]
[latex]\begin{array}{cc}4{x}^{3}+\left(12{x}^{2}+8{x}^{2}\right)+\left(9x-5x\right)+\left(-21+20\right) \hfill & \phantom{\rule{2em}{0ex}}\text{ }\text{Combine like terms}.\hfill \\ 4{x}^{3}+20{x}^{2}+4x-1\hfill & \phantom{\rule{2em}{0ex}}\text{ }\text{Simplify}.\hfill \end{array}[/latex]
[/hidden-answer]We can check our answers to these types of problems using a graphing calculator. To check, graph the problem as given along with the simplified answer. The two graphs should be equivalent. Be sure to use the same window to compare the graphs. Using different windows can make the expressions seem equivalent when they are not.
Find the sum.
[latex]\left(2{x}^{3}+5{x}^{2}-x+1\right)+\left(2{x}^{2}-3x-4\right)[/latex]
[latex]2{x}^{3}+7{x}^{2}-4x-3[/latex]
[/hidden-answer]Find the difference.
[latex]\left(7{x}^{4}-{x}^{2}+6x+1\right)-\left(5{x}^{3}-2{x}^{2}+3x+2\right)[/latex]
[latex]\begin{array}{cc}7{x}^{4}-5{x}^{3}+\left(-{x}^{2}+2{x}^{2}\right)+\left(6x-3x\right)+\left(1-2\right)\text{ }\hfill & \phantom{\rule{1em}{0ex}}\text{Combine like terms}.\hfill \\ 7{x}^{4}-5{x}^{3}+{x}^{2}+3x-1\hfill & \phantom{\rule{1em}{0ex}}\text{Simplify}.\hfill \end{array}[/latex]
[/hidden-answer]Note that finding the difference between two polynomials is the same as adding the opposite of the second polynomial to the first.
Find the difference.
[latex]\left(-7{x}^{3}-7{x}^{2}+6x-2\right)-\left(4{x}^{3}-6{x}^{2}-x+7\right)[/latex]
[latex]-11{x}^{3}-{x}^{2}+7x-9[/latex]
[/hidden-answer]Multiplying polynomials is a bit more challenging than adding and subtracting polynomials. We must use the distributive property to multiply each term in the first polynomial by each term in the second polynomial. We then combine like terms. We can also use a shortcut called the FOIL method when multiplying binomials. Certain special products follow patterns that we can memorize and use instead of multiplying the polynomials by hand each time. We will look at a variety of ways to multiply polynomials.
To multiply a number by a polynomial, we use the distributive property. The number must be distributed to each term of the polynomial. We can distribute the[latex]\,2\,[/latex]in[latex]\,2\left(x+7\right)\,[/latex]to obtain the equivalent expression[latex]\,2x+14.\,[/latex]When multiplying polynomials, the distributive property allows us to multiply each term of the first polynomial by each term of the second. We then add the products together and combine like terms to simplify.
Given the multiplication of two polynomials, use the distributive property to simplify the expression.
Find the product.
[latex]\left(2x+1\right)\left(3{x}^{2}-x+4\right)[/latex]
[latex]\begin{array}{cc}2x\left(3{x}^{2}-x+4\right)+1\left(3{x}^{2}-x+4\right) \hfill & \phantom{\rule{2em}{0ex}}\text{ }\text{Use the distributive property}.\hfill \\ \left(6{x}^{3}-2{x}^{2}+8x\right)+\left(3{x}^{2}-x+4\right)\hfill & \phantom{\rule{2em}{0ex}}\text{ }\text{Multiply}.\hfill \\ 6{x}^{3}+\left(-2{x}^{2}+3{x}^{2}\right)+\left(8x-x\right)+4\hfill & \phantom{\rule{2em}{0ex}}\text{ }\text{Combine like terms}.\hfill \\ 6{x}^{3}+{x}^{2}+7x+4 \hfill & \phantom{\rule{2em}{0ex}}\text{ }\text{Simplify}.\hfill \end{array}[/latex]
[/hidden-answer]We can use a table to keep track of our work, as shown in (Figure). Write one polynomial across the top and the other down the side. For each box in the table, multiply the term for that row by the term for that column. Then add all of the terms together, combine like terms, and simplify.
[latex]3{x}^{2}[/latex] | [latex]-x[/latex] | [latex]+4[/latex] | |
[latex]2x[/latex] | [latex]6{x}^{3}[/latex] | [latex]-2{x}^{2}[/latex] | [latex]8x[/latex] |
[latex]+1[/latex] | [latex]3{x}^{2}[/latex] | [latex]-x[/latex] | [latex]4[/latex] |
Find the product.
[latex]\left(3x+2\right)\left({x}^{3}-4{x}^{2}+7\right)[/latex]
[latex]3{x}^{4}-10{x}^{3}-8{x}^{2}+21x+14[/latex]
[/hidden-answer]A shortcut called FOIL is sometimes used to find the product of two binomials. It is called FOIL because we multiply the first terms, the outer terms, the inner terms, and then the last terms of each binomial.
The FOIL method arises out of the distributive property. We are simply multiplying each term of the first binomial by each term of the second binomial, and then combining like terms.
Given two binomials, use FOIL to simplify the expression.
[latex]\left(2x-18\right)\left(3x+3\right)[/latex]
Find the product of the first terms.
Find the product of the outer terms.
Find the product of the inner terms.
Find the product of the last terms.
[latex]\begin{array}{cc}6{x}^{2}+6x-54x-54\hfill & \phantom{\rule{2em}{0ex}}\text{Add the products}.\hfill \\ 6{x}^{2}+\left(6x-54x\right)-54\hfill & \phantom{\rule{2em}{0ex}}\text{Combine like terms}.\hfill \\ 6{x}^{2}-48x-54\hfill & \phantom{\rule{2em}{0ex}}\text{Simplify}.\hfill \end{array}[/latex]
[/hidden-answer]
Use FOIL to find the product.
[latex]\left(x+7\right)\left(3x-5\right)[/latex]
[latex]3{x}^{2}+16x-35[/latex]
[/hidden-answer]Certain binomial products have special forms. When a binomial is squared, the result is called a perfect square trinomial. We can find the square by multiplying the binomial by itself. However, there is a special form that each of these perfect square trinomials takes, and memorizing the form makes squaring binomials much easier and faster. Let’s look at a few perfect square trinomials to familiarize ourselves with the form.
Notice that the first term of each trinomial is the square of the first term of the binomial and, similarly, the last term of each trinomial is the square of the last term of the binomial. The middle term is double the product of the two terms. Lastly, we see that the first sign of the trinomial is the same as the sign of the binomial.
When a binomial is squared, the result is the first term squared added to double the product of both terms and the last term squared.
Given a binomial, square it using the formula for perfect square trinomials.
Expand[latex]\,{\left(3x-8\right)}^{2}.[/latex]
Begin by squaring the first term and the last term. For the middle term of the trinomial, double the product of the two terms.
Simplify
Expand [latex]\,{\left(4x-1\right)}^{2}.[/latex]
[latex]16{x}^{2}-8x+1[/latex]
[/hidden-answer]Another special product is called the difference of squares, which occurs when we multiply a binomial by another binomial with the same terms but the opposite sign. Let’s see what happens when we multiply[latex]\,\left(x+1\right)\left(x-1\right)\,[/latex]using the FOIL method.
The middle term drops out, resulting in a difference of squares. Just as we did with the perfect squares, let’s look at a few examples.
Because the sign changes in the second binomial, the outer and inner terms cancel each other out, and we are left only with the square of the first term minus the square of the last term.
Is there a special form for the sum of squares?
No. The difference of squares occurs because the opposite signs of the binomials cause the middle terms to disappear. There are no two binomials that multiply to equal a sum of squares.
When a binomial is multiplied by a binomial with the same terms separated by the opposite sign, the result is the square of the first term minus the square of the last term.
Given a binomial multiplied by a binomial with the same terms but the opposite sign, find the difference of squares.
Multiply[latex]\,\left(9x+4\right)\left(9x-4\right).[/latex]
Square the first term to get[latex]\,{\left(9x\right)}^{2}=81{x}^{2}.\,[/latex]Square the last term to get[latex]\,{4}^{2}=16.\,[/latex]Subtract the square of the last term from the square of the first term to find the product of[latex]\,81{x}^{2}-16.[/latex]
[/hidden-answer]Multiply[latex]\,\left(2x+7\right)\left(2x-7\right).[/latex]
[latex]4{x}^{2}-49[/latex]
[/hidden-answer]We have looked at polynomials containing only one variable. However, a polynomial can contain several variables. All of the same rules apply when working with polynomials containing several variables. Consider an example:
Multiply[latex]\,\left(x+4\right)\left(3x-2y+5\right).[/latex]
Follow the same steps that we used to multiply polynomials containing only one variable.
Multiply [latex]\left(3x-1\right)\left(2x+7y-9\right).[/latex]
[latex]\,6{x}^{2}+21xy-29x-7y+9[/latex]
[/hidden-answer]Access these online resources for additional instruction and practice with polynomials.
perfect square trinomial | [latex]{\left(x+a\right)}^{2}=\left(x+a\right)\left(x+a\right)={x}^{2}+2ax+{a}^{2}[/latex] |
difference of squares | [latex]\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}[/latex] |
Evaluate the following statement: The degree of a polynomial in standard form is the exponent of the leading term. Explain why the statement is true or false.
The statement is true. In standard form, the polynomial with the highest value exponent is placed first and is the leading term. The degree of a polynomial is the value of the highest exponent, which in standard form is also the exponent of the leading term.
[/hidden-answer]Many times, multiplying two binomials with two variables results in a trinomial. This is not the case when there is a difference of two squares. Explain why the product in this case is also a binomial.
You can multiply polynomials with any number of terms and any number of variables using four basic steps over and over until you reach the expanded polynomial. What are the four steps?
Use the distributive property, multiply, combine like terms, and simplify.
[/hidden-answer]State whether the following statement is true and explain why or why not: A trinomial is always a higher degree than a monomial.
For the following exercises, identify the degree of the polynomial.
[latex]7x-2{x}^{2}+13[/latex]
2
[/hidden-answer][latex]14{m}^{3}+{m}^{2}-16m+8[/latex]
[latex]-625{a}^{8}+16{b}^{4}[/latex]
8
[/hidden-answer][latex]200p-30{p}^{2}m+40{m}^{3}[/latex]
[latex]{x}^{2}+4x+4[/latex]
2
[/hidden-answer][latex]6{y}^{4}-{y}^{5}+3y-4[/latex]
For the following exercises, find the sum or difference.
[latex]\left(12{x}^{2}+3x\right)-\left(8{x}^{2}-19\right)[/latex]
[latex]4{x}^{2}+3x+19[/latex]
[/hidden-answer][latex]\left(4{z}^{3}+8{z}^{2}-z\right)+\left(-2{z}^{2}+z+6\right)[/latex]
[latex]\left(6{w}^{2}+24w+24\right)-\left(3w{}^{2}-6w+3\right)[/latex]
[latex]3{w}^{2}+30w+21[/latex]
[/hidden-answer][latex]\left(7{a}^{3}+6{a}^{2}-4a-13\right)+\left(-3{a}^{3}-4{a}^{2}+6a+17\right)[/latex]
[latex]\left(11{b}^{4}-6{b}^{3}+18{b}^{2}-4b+8\right)-\left(3{b}^{3}+6{b}^{2}+3b\right)[/latex]
[latex]11{b}^{4}-9{b}^{3}+12{b}^{2}-7b+8[/latex]
[/hidden-answer][latex]\left(49{p}^{2}-25\right)+\left(16{p}^{4}-32{p}^{2}+16\right)[/latex]
For the following exercises, find the product.
[latex]\left(4x+2\right)\left(6x-4\right)[/latex]
[latex]24{x}^{2}-4x-8[/latex]
[/hidden-answer][latex]\left(14{c}^{2}+4c\right)\left(2{c}^{2}-3c\right)[/latex]
[latex]\left(6{b}^{2}-6\right)\left(4{b}^{2}-4\right)[/latex]
[latex]24{b}^{4}-48{b}^{2}+24[/latex]
[/hidden-answer][latex]\left(3d-5\right)\left(2d+9\right)[/latex]
[latex]\left(9v-11\right)\left(11v-9\right)[/latex]
[latex]99{v}^{2}-202v+99[/latex]
[/hidden-answer][latex]\left(4{t}^{2}+7t\right)\left(-3{t}^{2}+4\right)[/latex]
[latex]\left(8n-4\right)\left({n}^{2}+9\right)[/latex]
[latex]8{n}^{3}-4{n}^{2}+72n-36[/latex]
[/hidden-answer]For the following exercises, expand the binomial.
[latex]{\left(4x+5\right)}^{2}[/latex]
[latex]{\left(3y-7\right)}^{2}[/latex]
[latex]9{y}^{2}-42y+49[/latex]
[/hidden-answer][latex]{\left(12-4x\right)}^{2}[/latex]
[latex]16{p}^{2}+72p+81[/latex]
[/hidden-answer][latex]{\left(2m-3\right)}^{2}[/latex]
[latex]{\left(3y-6\right)}^{2}[/latex]
[latex]9{y}^{2}-36y+36[/latex]
[/hidden-answer][latex]{\left(9b+1\right)}^{2}[/latex]
For the following exercises, multiply the binomials.
[latex]\left(4c+1\right)\left(4c-1\right)[/latex]
[latex]16{c}^{2}-1[/latex]
[/hidden-answer][latex]\left(9a-4\right)\left(9a+4\right)[/latex]
[latex]\left(15n-6\right)\left(15n+6\right)[/latex]
[latex]225{n}^{2}-36[/latex]
[/hidden-answer][latex]\left(25b+2\right)\left(25b-2\right)[/latex]
[latex]\left(4+4m\right)\left(4-4m\right)[/latex]
[latex]-16{m}^{2}+16[/latex]
[/hidden-answer][latex]\left(14p+7\right)\left(14p-7\right)[/latex]
[latex]\left(11q-10\right)\left(11q+10\right)[/latex]
[latex]121{q}^{2}-100[/latex]
[/hidden-answer]For the following exercises, multiply the polynomials.
[latex]\left(2{x}^{2}+2x+1\right)\left(4x-1\right)[/latex]
[latex]\left(4{t}^{2}+t-7\right)\left(4{t}^{2}-1\right)[/latex]
[latex]16{t}^{4}+4{t}^{3}-32{t}^{2}-t+7[/latex]
[/hidden-answer][latex]\left(x-1\right)\left({x}^{2}-2x+1\right)[/latex]
[latex]\left(y-2\right)\left({y}^{2}-4y-9\right)[/latex]
[latex]{y}^{3}-6{y}^{2}-y+18[/latex]
[/hidden-answer][latex]\left(6k-5\right)\left(6{k}^{2}+5k-1\right)[/latex]
[latex]\left(3{p}^{2}+2p-10\right)\left(p-1\right)[/latex]
[latex]3{p}^{3}-{p}^{2}-12p+10[/latex]
[/hidden-answer][latex]\left(4m-13\right)\left(2{m}^{2}-7m+9\right)[/latex]
[latex]\left(a+b\right)\left(a-b\right)[/latex]
[latex]{a}^{2}-{b}^{2}[/latex]
[/hidden-answer][latex]\left(4x-6y\right)\left(6x-4y\right)[/latex]
[latex]{\left(4t-5u\right)}^{2}[/latex]
[latex]16{t}^{2}-40tu+25{u}^{2}[/latex]
[/hidden-answer][latex]\left(9m+4n-1\right)\left(2m+8\right)[/latex]
[latex]\left(4t-x\right)\left(t-x+1\right)[/latex]
[latex]4{t}^{2}+{x}^{2}+4t-5tx-x[/latex]
[/hidden-answer][latex]\left({b}^{2}-1\right)\left({a}^{2}+2ab+{b}^{2}\right)[/latex]
[latex]\left(4r-d\right)\left(6r+7d\right)[/latex]
[latex]24{r}^{2}+22rd-7{d}^{2}[/latex]
[/hidden-answer][latex]\left(x+y\right)\left({x}^{2}-xy+{y}^{2}\right)[/latex]
A developer wants to purchase a plot of land to build a house. The area of the plot can be described by the following expression:[latex]\,\left(4x+1\right)\left(8x-3\right)\,[/latex]where x is measured in meters. Multiply the binomials to find the area of the plot in standard form.
[latex]32{x}^{2}-4x-3\,[/latex]m^{2}
[/hidden-answer]A prospective buyer wants to know how much grain a specific silo can hold. The area of the floor of the silo is[latex]\,{\left(2x+9\right)}^{2}.\,[/latex]The height of the silo is[latex]\,10x+10,[/latex]where x is measured in feet. Expand the square and multiply by the height to find the expression that shows how much grain the silo can hold.
For the following exercises, perform the given operations.
[latex]{\left(4t-7\right)}^{2}\left(2t+1\right)-\left(4{t}^{2}+2t+11\right)[/latex]
[latex]32{t}^{3}-100{t}^{2}+40t+38[/latex]
[/hidden-answer][latex]\left(3b+6\right)\left(3b-6\right)\left(9{b}^{2}-36\right)[/latex]
[latex]\left({a}^{2}+4ac+4{c}^{2}\right)\left({a}^{2}-4{c}^{2}\right)[/latex]
[latex]{a}^{4}+4{a}^{3}c-16a{c}^{3}-16{c}^{4}[/latex]
[/hidden-answer]Imagine that we are trying to find the area of a lawn so that we can determine how much grass seed to purchase. The lawn is the green portion in (Figure).
The area of the entire region can be found using the formula for the area of a rectangle.
The areas of the portions that do not require grass seed need to be subtracted from the area of the entire region. The two square regions each have an area of[latex]\,A={s}^{2}={4}^{2}=16\,[/latex]units^{2}. The other rectangular region has one side of length[latex]\,10x-8\,[/latex]and one side of length[latex]\,4,[/latex]giving an area of[latex]\,A=lw=4\left(10x-8\right)=40x-32\,[/latex]units^{2}. So the region that must be subtracted has an area of[latex]\,2\left(16\right)+40x-32=40x\,[/latex]units^{2}.
The area of the region that requires grass seed is found by subtracting[latex]\,60{x}^{2}-40x\,[/latex]units^{2}. This area can also be expressed in factored form as[latex]\,20x\left(3x-2\right)\,[/latex]units^{2}. We can confirm that this is an equivalent expression by multiplying.
Many polynomial expressions can be written in simpler forms by factoring. In this section, we will look at a variety of methods that can be used to factor polynomial expressions.
When we study fractions, we learn that the greatest common factor (GCF) of two numbers is the largest number that divides evenly into both numbers. For instance,[latex]\,4\,[/latex]is the GCF of[latex]\,16\,[/latex]and[latex]\,20\,[/latex]because it is the largest number that divides evenly into both[latex]\,16\,[/latex]and[latex]\,20\,[/latex]The GCF of polynomials works the same way:[latex]\,4x\,[/latex]is the GCF of[latex]\,16x\,[/latex]and[latex]\,20{x}^{2}\,[/latex]because it is the largest polynomial that divides evenly into both[latex]\,16x\,[/latex]and[latex]\,20{x}^{2}.[/latex]
When factoring a polynomial expression, our first step should be to check for a GCF. Look for the GCF of the coefficients, and then look for the GCF of the variables.
The greatest common factor (GCF) of polynomials is the largest polynomial that divides evenly into the polynomials.
Given a polynomial expression, factor out the greatest common factor.
Factor[latex]\,6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy.[/latex]
First, find the GCF of the expression. The GCF of[latex]\,6,45,[/latex]and[latex]\,21\,[/latex]is[latex]\,3.\,[/latex]The GCF of[latex]\,{x}^{3},{x}^{2},[/latex]and[latex]\,x\,[/latex]is[latex]\,x.\,[/latex](Note that the GCF of a set of expressions in the form[latex]\,{x}^{n}\,[/latex]will always be the exponent of lowest degree.) And the GCF of[latex]\,{y}^{3},{y}^{2},[/latex]and[latex]\,y\,[/latex]is[latex]\,y.\,[/latex]Combine these to find the GCF of the polynomial,[latex]\,3xy.[/latex]
Next, determine what the GCF needs to be multiplied by to obtain each term of the polynomial. We find that[latex]\,3xy\left(2{x}^{2}{y}^{2}\right)=6{x}^{3}{y}^{3},3xy\left(15xy\right)=45{x}^{2}{y}^{2},[/latex]and[latex]\,3xy\left(7\right)=21xy.[/latex]
Finally, write the factored expression as the product of the GCF and the sum of the terms we needed to multiply by.
After factoring, we can check our work by multiplying. Use the distributive property to confirm that[latex]\,\left(3xy\right)\left(2{x}^{2}{y}^{2}+15xy+7\right)=6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy.[/latex]
Factor[latex]\,x\left({b}^{2}-a\right)+6\left({b}^{2}-a\right)\,[/latex]by pulling out the GCF.
[latex]\left({b}^{2}-a\right)\left(x+6\right)[/latex]
[/hidden-answer]Although we should always begin by looking for a GCF, pulling out the GCF is not the only way that polynomial expressions can be factored. The polynomial[latex]\,{x}^{2}+5x+6\,[/latex]has a GCF of 1, but it can be written as the product of the factors[latex]\,\left(x+2\right)\,[/latex]and[latex]\,\left(x+3\right).[/latex]
Trinomials of the form[latex]\,{x}^{2}+bx+c\,[/latex]can be factored by finding two numbers with a product of[latex]c\,[/latex]and a sum of[latex]\,b.\,[/latex]The trinomial[latex]\,{x}^{2}+10x+16,[/latex]for example, can be factored using the numbers[latex]\,2\,[/latex]and[latex]\,8\,[/latex]because the product of those numbers is[latex]\,16\,[/latex]and their sum is[latex]\,10.\,[/latex]The trinomial can be rewritten as the product of[latex]\,\left(x+2\right)\,[/latex]and[latex]\,\left(x+8\right).[/latex]
A trinomial of the form[latex]\,{x}^{2}+bx+c\,[/latex]can be written in factored form as[latex]\,\left(x+p\right)\left(x+q\right)\,[/latex]where[latex]\,pq=c\,[/latex]and[latex]\,p+q=b.[/latex]
Can every trinomial be factored as a product of binomials?
No. Some polynomials cannot be factored. These polynomials are said to be prime.
Given a trinomial in the form[latex]\,{x}^{2}+bx+c,[/latex]factor it.
Factor[latex]\,{x}^{2}+2x-15.[/latex]
We have a trinomial with leading coefficient[latex]\,1,b=2,[/latex]and[latex]\,c=-15.\,[/latex]We need to find two numbers with a product of[latex]\,-15\,[/latex]and a sum of[latex]\,2.\,[/latex]In (Figure), we list factors until we find a pair with the desired sum.
Factors of[latex]\,-15[/latex] | Sum of Factors |
---|---|
[latex]1,-15[/latex] | [latex]-14[/latex] |
[latex]-1,15[/latex] | 14 |
[latex]3,-5[/latex] | [latex]-2[/latex] |
[latex]-3,5[/latex] | 2 |
Now that we have identified[latex]\,p\,[/latex]and[latex]\,q\,[/latex]as[latex]\,-3\,[/latex]and[latex]\,5,[/latex]write the factored form as[latex]\,\left(x-3\right)\left(x+5\right).[/latex][/hidden-answer]
We can check our work by multiplying. Use FOIL to confirm that[latex]\,\left(x-3\right)\left(x+5\right)={x}^{2}+2x-15.[/latex]
Does the order of the factors matter?
No. Multiplication is commutative, so the order of the factors does not matter.
Factor[latex]\,{x}^{2}-7x+6.[/latex]
[latex]\left(x-6\right)\left(x-1\right)[/latex]
[/hidden-answer]Trinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can factor by grouping by dividing the x term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial[latex]\,2{x}^{2}+5x+3\,[/latex]can be rewritten as[latex]\,\left(2x+3\right)\left(x+1\right)\,[/latex]using this process. We begin by rewriting the original expression as[latex]\,2{x}^{2}+2x+3x+3\,[/latex]and then factor each portion of the expression to obtain[latex]\,2x\left(x+1\right)+3\left(x+1\right).\,[/latex]We then pull out the GCF of[latex]\,\left(x+1\right)\,[/latex]to find the factored expression.
To factor a trinomial in the form[latex]\,a{x}^{2}+bx+c\,[/latex]by grouping, we find two numbers with a product of[latex]\,ac\,[/latex]and a sum of[latex]\,b.\,[/latex]We use these numbers to divide the[latex]\,x\,[/latex]term into the sum of two terms and factor each portion of the expression separately, then factor out the GCF of the entire expression.
Factor[latex]\,5{x}^{2}+7x-6\,[/latex]by grouping.
We have a trinomial with[latex]\,a=5,b=7,[/latex]and[latex]\,c=-6.\,[/latex]First, determine[latex]\,ac=-30.\,[/latex]We need to find two numbers with a product of[latex]\,-30\,[/latex]and a sum of[latex]\,7.\,[/latex]In (Figure), we list factors until we find a pair with the desired sum.
Factors of[latex]\,-30[/latex] | Sum of Factors |
---|---|
[latex]1,-30[/latex] | [latex]-29[/latex] |
[latex]-1,30[/latex] | 29 |
[latex]2,-15[/latex] | [latex]-13[/latex] |
[latex]-2,15[/latex] | 13 |
[latex]3,-10[/latex] | [latex]-7[/latex] |
[latex]-3,10[/latex] | 7 |
So[latex]\,p=-3\,[/latex]and[latex]\,q=10.[/latex]
We can check our work by multiplying. Use FOIL to confirm that[latex]\,\left(5x-3\right)\left(x+2\right)=5{x}^{2}+7x-6.[/latex]
Factor a.[latex]\,2{x}^{2}+9x+9\,[/latex]b.[latex]\,6{x}^{2}+x-1[/latex]
a.[latex]\,\left(2x+3\right)\left(x+3\right)\,[/latex]b.[latex]\,\left(3x-1\right)\left(2x+1\right)[/latex]
[/hidden-answer]A perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.
We can use this equation to factor any perfect square trinomial.
A perfect square trinomial can be written as the square of a binomial:
Given a perfect square trinomial, factor it into the square of a binomial.
Factor[latex]\,25{x}^{2}+20x+4.[/latex]
Notice that[latex]\,25{x}^{2}\,[/latex]and[latex]\,4\,[/latex]are perfect squares because[latex]\,25{x}^{2}={\left(5x\right)}^{2}\,[/latex]and[latex]\,4={2}^{2}.\,[/latex]Then check to see if the middle term is twice the product of[latex]\,5x\,[/latex]and[latex]\,2.\,[/latex]The middle term is, indeed, twice the product:[latex]\,2\left(5x\right)\left(2\right)=20x.\,[/latex]Therefore, the trinomial is a perfect square trinomial and can be written as[latex]\,{\left(5x+2\right)}^{2}.[/latex]
[/hidden-answer]Factor[latex]\,49{x}^{2}-14x+1.[/latex]
[latex]{\left(7x-1\right)}^{2}[/latex]
[/hidden-answer]A difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.
We can use this equation to factor any differences of squares.
A difference of squares can be rewritten as two factors containing the same terms but opposite signs.
Given a difference of squares, factor it into binomials.
Factor[latex]\,9{x}^{2}-25.[/latex]
Notice that[latex]\,9{x}^{2}\,[/latex]and[latex]\,25\,[/latex]are perfect squares because[latex]\,9{x}^{2}={\left(3x\right)}^{2}\,[/latex]and[latex]\,25={5}^{2}.\,[/latex]The polynomial represents a difference of squares and can be rewritten as[latex]\,\left(3x+5\right)\left(3x-5\right).[/latex]
[/hidden-answer]Factor[latex]\,81{y}^{2}-100.[/latex]
[latex]\left(9y+10\right)\left(9y-10\right)[/latex]
[/hidden-answer]Is there a formula to factor the sum of squares?
No. A sum of squares cannot be factored.
Now, we will look at two new special products: the sum and difference of cubes. Although the sum of squares cannot be factored, the sum of cubes can be factored into a binomial and a trinomial.
Similarly, the sum of cubes can be factored into a binomial and a trinomial, but with different signs.
We can use the acronym SOAP to remember the signs when factoring the sum or difference of cubes. The first letter of each word relates to the signs: Same Opposite Always Positive. For example, consider the following example.
The sign of the first 2 is the same as the sign between[latex]\,{x}^{3}-{2}^{3}.\,[/latex]The sign of the[latex]\,2x\,[/latex]term is opposite the sign between[latex]\,{x}^{3}-{2}^{3}.\,[/latex]And the sign of the last term, 4, is always positive.
We can factor the sum of two cubes as
We can factor the difference of two cubes as
Given a sum of cubes or difference of cubes, factor it.
Factor[latex]\,{x}^{3}+512.[/latex]
After writing the sum of cubes this way, we might think we should check to see if the trinomial portion can be factored further. However, the trinomial portion cannot be factored, so we do not need to check.
Factor the sum of cubes:[latex]\,216{a}^{3}+{b}^{3}.[/latex]
[latex]\left(6a+b\right)\left(36{a}^{2}-6ab+{b}^{2}\right)[/latex]
[/hidden-answer]Factor[latex]\,8{x}^{3}-125.[/latex]
Just as with the sum of cubes, we will not be able to further factor the trinomial portion.
Factor the difference of cubes:[latex]\,1,000{x}^{3}-1.[/latex]
[latex]\left(10x-1\right)\left(100{x}^{2}+10x+1\right)[/latex]
[/hidden-answer]Expressions with fractional or negative exponents can be factored by pulling out a GCF. Look for the variable or exponent that is common to each term of the expression and pull out that variable or exponent raised to the lowest power. These expressions follow the same factoring rules as those with integer exponents. For instance,[latex]\,2{x}^{\frac{1}{4}}+5{x}^{\frac{3}{4}}\,[/latex]can be factored by pulling out[latex]\,{x}^{\frac{1}{4}}\,[/latex]and being rewritten as[latex]\,{x}^{\frac{1}{4}}\left(2+5{x}^{\frac{1}{2}}\right).[/latex]
Factor[latex]\,3x{\left(x+2\right)}^{\frac{-1}{3}}+4{\left(x+2\right)}^{\frac{2}{3}}.[/latex]
Factor out the term with the lowest value of the exponent. In this case, that would be[latex]\,{\left(x+2\right)}^{-\frac{1}{3}}.[/latex]
Factor[latex]\,2{\left(5a-1\right)}^{\frac{3}{4}}+7a{\left(5a-1\right)}^{-\frac{1}{4}}.[/latex]
[latex]{\left(5a-1\right)}^{-\frac{1}{4}}\left(17a-2\right)[/latex]
[/hidden-answer]Access these online resources for additional instruction and practice with factoring polynomials.
difference of squares | [latex]{a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)[/latex] |
perfect square trinomial | [latex]{a}^{2}+2ab+{b}^{2}={\left(a+b\right)}^{2}[/latex] |
sum of cubes | [latex]{a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)[/latex] |
difference of cubes | [latex]{a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)[/latex] |
If the terms of a polynomial do not have a GCF, does that mean it is not factorable? Explain.
The terms of a polynomial do not have to have a common factor for the entire polynomial to be factorable. For example,[latex]\,4{x}^{2}\,[/latex]and[latex]\,-9{y}^{2}\,[/latex]don’t have a common factor, but the whole polynomial is still factorable:[latex]\,4{x}^{2}-9{y}^{2}=\left(2x+3y\right)\left(2x-3y\right).[/latex]
[/hidden-answer]A polynomial is factorable, but it is not a perfect square trinomial or a difference of two squares. Can you factor the polynomial without finding the GCF?
How do you factor by grouping?
Divide the[latex]\,x\,[/latex]term into the sum of two terms, factor each portion of the expression separately, and then factor out the GCF of the entire expression.
[/hidden-answer]For the following exercises, find the greatest common factor.
[latex]14x+4xy-18x{y}^{2}[/latex]
[latex]49m{b}^{2}-35{m}^{2}ba+77m{a}^{2}[/latex]
[latex]7m[/latex]
[/hidden-answer][latex]30{x}^{3}y-45{x}^{2}{y}^{2}+135x{y}^{3}[/latex]
[latex]200{p}^{3}{m}^{3}-30{p}^{2}{m}^{3}+40{m}^{3}[/latex]
[latex]10{m}^{3}[/latex]
[/hidden-answer][latex]36{j}^{4}{k}^{2}-18{j}^{3}{k}^{3}+54{j}^{2}{k}^{4}[/latex]
[latex]6{y}^{4}-2{y}^{3}+3{y}^{2}-y[/latex]
[latex]y[/latex]
[/hidden-answer]For the following exercises, factor by grouping.
[latex]6{x}^{2}+5x-4[/latex]
[latex]2{a}^{2}+9a-18[/latex]
[latex]\left(2a-3\right)\left(a+6\right)[/latex]
[/hidden-answer][latex]6{c}^{2}+41c+63[/latex]
[latex]6{n}^{2}-19n-11[/latex]
[latex]\left(3n-11\right)\left(2n+1\right)[/latex]
[/hidden-answer][latex]20{w}^{2}-47w+24[/latex]
[latex]2{p}^{2}-5p-7[/latex]
[latex]\left(p+1\right)\left(2p-7\right)[/latex]
[/hidden-answer]For the following exercises, factor the polynomial.
[latex]7{x}^{2}+48x-7[/latex]
[latex]10{h}^{2}-9h-9[/latex]
[latex]\left(5h+3\right)\left(2h-3\right)[/latex]
[/hidden-answer][latex]2{b}^{2}-25b-247[/latex]
[latex]9{d}^{2}-73d+8[/latex]
[latex]\left(9d-1\right)\left(d-8\right)[/latex]
[/hidden-answer][latex]90{v}^{2}-181v+90[/latex]
[latex]12{t}^{2}+t-13[/latex]
[latex]\left(12t+13\right)\left(t-1\right)[/latex]
[/hidden-answer][latex]2{n}^{2}-n-15[/latex]
[latex]16{x}^{2}-100[/latex]
[latex]25{y}^{2}-196[/latex]
[latex]121{p}^{2}-169[/latex]
[latex]\left(11p+13\right)\left(11p-13\right)[/latex]
[/hidden-answer][latex]4{m}^{2}-9[/latex]
[latex]361{d}^{2}-81[/latex]
[latex]\left(19d+9\right)\left(19d-9\right)[/latex]
[/hidden-answer][latex]324{x}^{2}-121[/latex]
[latex]144{b}^{2}-25{c}^{2}[/latex]
[latex]\left(12b+5c\right)\left(12b-5c\right)[/latex]
[/hidden-answer][latex]16{a}^{2}-8a+1[/latex]
[latex]49{n}^{2}+168n+144[/latex]
[latex]{\left(7n+12\right)}^{2}[/latex]
[/hidden-answer][latex]121{x}^{2}-88x+16[/latex]
[latex]225{y}^{2}+120y+16[/latex]
[latex]{\left(15y+4\right)}^{2}[/latex]
[/hidden-answer][latex]{m}^{2}-20m+100[/latex]
[latex]25{p}^{2}-120m+144[/latex]
[latex]{\left(5p-12\right)}^{2}[/latex]
[/hidden-answer][latex]36{q}^{2}+60q+25[/latex]
For the following exercises, factor the polynomials.
[latex]{x}^{3}+216[/latex]
[latex]\left(x+6\right)\left({x}^{2}-6x+36\right)[/latex]
[/hidden-answer][latex]27{y}^{3}-8[/latex]
[latex]125{a}^{3}+343[/latex]
[latex]\left(5a+7\right)\left(25{a}^{2}-35a+49\right)[/latex]
[/hidden-answer][latex]{b}^{3}-8{d}^{3}[/latex]
[latex]64{x}^{3}-125[/latex]
[latex]\left(4x-5\right)\left(16{x}^{2}+20x+25\right)[/latex]
[/hidden-answer][latex]729{q}^{3}+1331[/latex]
[latex]125{r}^{3}+1,728{s}^{3}[/latex]
[latex]\left(5r+12s\right)\left(25{r}^{2}-60rs+144{s}^{2}\right)[/latex]
[/hidden-answer][latex]4x{\left(x-1\right)}^{-\frac{2}{3}}+3{\left(x-1\right)}^{\frac{1}{3}}[/latex]
[latex]3c{\left(2c+3\right)}^{-\frac{1}{4}}-5{\left(2c+3\right)}^{\frac{3}{4}}[/latex]
[latex]{\left(2c+3\right)}^{-\frac{1}{4}}\left(-7c-15\right)[/latex]
[/hidden-answer][latex]3t{\left(10t+3\right)}^{\frac{1}{3}}+7{\left(10t+3\right)}^{\frac{4}{3}}[/latex]
[latex]14x{\left(x+2\right)}^{-\frac{2}{5}}+5{\left(x+2\right)}^{\frac{3}{5}}[/latex]
[latex]{\left(x+2\right)}^{-\frac{2}{5}}\left(19x+10\right)[/latex]
[/hidden-answer][latex]9y{\left(3y-13\right)}^{\frac{1}{5}}-2{\left(3y-13\right)}^{\frac{6}{5}}[/latex]
[latex]5z{\left(2z-9\right)}^{-\frac{3}{2}}+11{\left(2z-9\right)}^{-\frac{1}{2}}[/latex]
[latex]{\left(2z-9\right)}^{-\frac{3}{2}}\left(27z-99\right)[/latex]
[/hidden-answer][latex]6d{\left(2d+3\right)}^{-\frac{1}{6}}+5{\left(2d+3\right)}^{\frac{5}{6}}[/latex]
For the following exercises, consider this scenario:
Charlotte has appointed a chairperson to lead a city beautification project. The first act is to install statues and fountains in one of the city’s parks. The park is a rectangle with an area of[latex]\,98{x}^{2}+105x-27\,[/latex]m^{2}, as shown in the figure below. The length and width of the park are perfect factors of the area.
Factor by grouping to find the length and width of the park.
[latex]\left(14x-3\right)\left(7x+9\right)[/latex]
[/hidden-answer]A statue is to be placed in the center of the park. The area of the base of the statue is[latex]\,4{x}^{2}+12x+9{\text{m}}^{2}.\,[/latex]Factor the area to find the lengths of the sides of the statue.
At the northwest corner of the park, the city is going to install a fountain. The area of the base of the fountain is[latex]\,9{x}^{2}-25{\text{m}}^{2}.\,[/latex]Factor the area to find the lengths of the sides of the fountain.
[latex]\left(3x+5\right)\left(3x-5\right)[/latex]
[/hidden-answer]For the following exercise, consider the following scenario:
A school is installing a flagpole in the central plaza. The plaza is a square with side length 100 yd. as shown in the figure below. The flagpole will take up a square plot with area[latex]\,{x}^{2}-6x+9[/latex]yd^{2}.
Find the length of the base of the flagpole by factoring.
For the following exercises, factor the polynomials completely.
[latex]16{x}^{4}-200{x}^{2}+625[/latex]
[latex]{\left(2x+5\right)}^{2}{\left(2x-5\right)}^{2}[/latex]
[/hidden-answer][latex]81{y}^{4}-256[/latex]
[latex]16{z}^{4}-2,401{a}^{4}[/latex]
[latex]\left(4{z}^{2}+49{a}^{2}\right)\left(2z+7a\right)\left(2z-7a\right)[/latex]
[/hidden-answer][latex]5x{\left(3x+2\right)}^{-\frac{2}{4}}+{\left(12x+8\right)}^{\frac{3}{2}}[/latex]
[latex]{\left(32{x}^{3}+48{x}^{2}-162x-243\right)}^{-1}[/latex]
[latex]\frac{1}{\left(4x+9\right)\left(4x-9\right)\left(2x+3\right)}[/latex]
[/hidden-answer]A pastry shop has fixed costs of[latex]\,\text{\$}280\,[/latex]per week and variable costs of[latex]\,\text{\$}9\,[/latex]per box of pastries. The shop’s costs per week in terms of[latex]\,x,[/latex]the number of boxes made, is[latex]\,280+9x.\,[/latex]We can divide the costs per week by the number of boxes made to determine the cost per box of pastries.
Notice that the result is a polynomial expression divided by a second polynomial expression. In this section, we will explore quotients of polynomial expressions.
The quotient of two polynomial expressions is called a rational expression. We can apply the properties of fractions to rational expressions, such as simplifying the expressions by canceling common factors from the numerator and the denominator. To do this, we first need to factor both the numerator and denominator. Let’s start with the rational expression shown.
We can factor the numerator and denominator to rewrite the expression.
Then we can simplify that expression by canceling the common factor[latex]\,\left(x+4\right).[/latex]
Given a rational expression, simplify it.
Simplify[latex]\,\frac{{x}^{2}-9}{{x}^{2}+4x+3}.[/latex]
We can cancel the common factor because any expression divided by itself is equal to 1.
Can the[latex]\,{x}^{2}\,[/latex]term be cancelled in (Figure)?
No. A factor is an expression that is multiplied by another expression. The[latex]\,{x}^{2}\,[/latex]term is not a factor of the numerator or the denominator.
Simplify[latex]\,\frac{x-6}{{x}^{2}-36}.[/latex]
[latex]\frac{1}{x+6}[/latex]
[/hidden-answer]Multiplication of rational expressions works the same way as multiplication of any other fractions. We multiply the numerators to find the numerator of the product, and then multiply the denominators to find the denominator of the product. Before multiplying, it is helpful to factor the numerators and denominators just as we did when simplifying rational expressions. We are often able to simplify the product of rational expressions.
Given two rational expressions, multiply them.
Multiply the rational expressions and show the product in simplest form:
Multiply the rational expressions and show the product in simplest form:
[latex]\frac{\left(x+5\right)\left(x+6\right)}{\left(x+2\right)\left(x+4\right)}[/latex]
[/hidden-answer]Division of rational expressions works the same way as division of other fractions. To divide a rational expression by another rational expression, multiply the first expression by the reciprocal of the second. Using this approach, we would rewrite[latex]\,\frac{1}{x}÷\frac{{x}^{2}}{3}\,[/latex]as the product[latex]\,\frac{1}{x}\cdot \frac{3}{{x}^{2}}.\,[/latex]Once the division expression has been rewritten as a multiplication expression, we can multiply as we did before.
Given two rational expressions, divide them.
Divide the rational expressions and express the quotient in simplest form:
[latex]1[/latex]
[/hidden-answer]Adding and subtracting rational expressions works just like adding and subtracting numerical fractions. To add fractions, we need to find a common denominator. Let’s look at an example of fraction addition.
We have to rewrite the fractions so they share a common denominator before we are able to add. We must do the same thing when adding or subtracting rational expressions.
The easiest common denominator to use will be the least common denominator, or LCD. The LCD is the smallest multiple that the denominators have in common. To find the LCD of two rational expressions, we factor the expressions and multiply all of the distinct factors. For instance, if the factored denominators were[latex]\,\left(x+3\right)\left(x+4\right)\,[/latex]and[latex]\,\left(x+4\right)\left(x+5\right),[/latex]then the LCD would be[latex]\,\left(x+3\right)\left(x+4\right)\left(x+5\right).[/latex]
Once we find the LCD, we need to multiply each expression by the form of 1 that will change the denominator to the LCD. We would need to multiply the expression with a denominator of[latex]\,\left(x+3\right)\left(x+4\right)\,[/latex]by[latex]\,\frac{x+5}{x+5}\,[/latex]and the expression with a denominator of[latex]\,\left(x+4\right)\left(x+5\right)\,[/latex]by[latex]\,\frac{x+3}{x+3}.[/latex]
Given two rational expressions, add or subtract them.
Add the rational expressions:
First, we have to find the LCD. In this case, the LCD will be[latex]\,xy.\,[/latex]We then multiply each expression by the appropriate form of 1 to obtain[latex]\,xy\,[/latex]as the denominator for each fraction.
Now that the expressions have the same denominator, we simply add the numerators to find the sum.
Multiplying by[latex]\,\frac{y}{y}\,[/latex]or[latex]\,\frac{x}{x}\,[/latex]does not change the value of the original expression because any number divided by itself is 1, and multiplying an expression by 1 gives the original expression.
Subtract the rational expressions:
Do we have to use the LCD to add or subtract rational expressions?
No. Any common denominator will work, but it is easiest to use the LCD.
Subtract the rational expressions:[latex]\,\frac{3}{x+5}-\frac{1}{x-3}.[/latex]
[latex]\frac{2\left(x-7\right)}{\left(x+5\right)\left(x-3\right)}[/latex]
[/hidden-answer]Given a complex rational expression, simplify it.
Simplify:[latex]\frac{y+\frac{1}{x}}{\frac{x}{y}}[/latex].
Begin by combining the expressions in the numerator into one expression.
Now the numerator is a single rational expression and the denominator is a single rational expression.
We can rewrite this as division, and then multiplication.
Simplify:[latex]\frac{\frac{x}{y}-\frac{y}{x}}{y}[/latex]
[latex]\frac{{x}^{2}-{y}^{2}}{x{y}^{2}}[/latex]
[/hidden-answer]Can a complex rational expression always be simplified?
Yes. We can always rewrite a complex rational expression as a simplified rational expression.
Access these online resources for additional instruction and practice with rational expressions.
How can you use factoring to simplify rational expressions?
You can factor the numerator and denominator to see if any of the terms can cancel one another out.
[/hidden-answer]How do you use the LCD to combine two rational expressions?
Tell whether the following statement is true or false and explain why: You only need to find the LCD when adding or subtracting rational expressions.
True. Multiplication and division do not require finding the LCD because the denominators can be combined through those operations, whereas addition and subtraction require like terms.
[/hidden-answer]For the following exercises, simplify the rational expressions.
[latex]\frac{{x}^{2}-16}{{x}^{2}-5x+4}[/latex]
[latex]\frac{{y}^{2}+10y+25}{{y}^{2}+11y+30}[/latex]
[latex]\frac{y+5}{y+6}[/latex]
[/hidden-answer][latex]\frac{6{a}^{2}-24a+24}{6{a}^{2}-24}[/latex]
[latex]\frac{9{b}^{2}+18b+9}{3b+3}[/latex]
[latex]3b+3[/latex]
[/hidden-answer][latex]\frac{m-12}{{m}^{2}-144}[/latex]
[latex]\frac{2{x}^{2}+7x-4}{4{x}^{2}+2x-2}[/latex]
[latex]\frac{x+4}{2x+2}[/latex]
[/hidden-answer][latex]\frac{6{x}^{2}+5x-4}{3{x}^{2}+19x+20}[/latex]
[latex]\frac{{a}^{2}+9a+18}{{a}^{2}+3a-18}[/latex]
[latex]\frac{a+3}{a-3}[/latex]
[/hidden-answer][latex]\frac{3{c}^{2}+25c-18}{3{c}^{2}-23c+14}[/latex]
[latex]\frac{12{n}^{2}-29n-8}{28{n}^{2}-5n-3}[/latex]
[latex]\frac{3n-8}{7n-3}[/latex]
[/hidden-answer]For the following exercises, multiply the rational expressions and express the product in simplest form.
[latex]\frac{{x}^{2}-x-6}{2{x}^{2}+x-6}\cdot \frac{2{x}^{2}+7x-15}{{x}^{2}-9}[/latex]
[latex]\frac{{c}^{2}+2c-24}{{c}^{2}+12c+36}\cdot \frac{{c}^{2}-10c+24}{{c}^{2}-8c+16}[/latex]
[latex]\frac{c-6}{c+6}[/latex]
[/hidden-answer][latex]\frac{2{d}^{2}+9d-35}{{d}^{2}+10d+21}\cdot \frac{3{d}^{2}+2d-21}{3{d}^{2}+14d-49}[/latex]
[latex]\frac{10{h}^{2}-9h-9}{2{h}^{2}-19h+24}\cdot \frac{{h}^{2}-16h+64}{5{h}^{2}-37h-24}[/latex]
[latex]1[/latex]
[/hidden-answer][latex]\frac{6{b}^{2}+13b+6}{4{b}^{2}-9}\cdot \frac{6{b}^{2}+31b-30}{18{b}^{2}-3b-10}[/latex]
[latex]\frac{2{d}^{2}+15d+25}{4{d}^{2}-25}\cdot \frac{2{d}^{2}-15d+25}{25{d}^{2}-1}[/latex]
[latex]\frac{{d}^{2}-25}{25{d}^{2}-1}[/latex]
[/hidden-answer][latex]\frac{6{x}^{2}-5x-50}{15{x}^{2}-44x-20}\cdot \frac{20{x}^{2}-7x-6}{2{x}^{2}+9x+10}[/latex]
[latex]\frac{{t}^{2}-1}{{t}^{2}+4t+3}\cdot \frac{{t}^{2}+2t-15}{{t}^{2}-4t+3}[/latex]
[latex]\frac{t+5}{t+3}[/latex]
[/hidden-answer][latex]\frac{2{n}^{2}-n-15}{6{n}^{2}+13n-5}\cdot \frac{12{n}^{2}-13n+3}{4{n}^{2}-15n+9}[/latex]
[latex]\frac{36{x}^{2}-25}{6{x}^{2}+65x+50}\cdot \frac{3{x}^{2}+32x+20}{18{x}^{2}+27x+10}[/latex]
[latex]\frac{6x-5}{6x+5}[/latex]
[/hidden-answer]For the following exercises, divide the rational expressions.
[latex]\frac{3{y}^{2}-7y-6}{2{y}^{2}-3y-9}÷\frac{{y}^{2}+y-2}{2{y}^{2}+y-3}[/latex]
[latex]\frac{6{p}^{2}+p-12}{8{p}^{2}+18p+9}÷\frac{6{p}^{2}-11p+4}{2{p}^{2}+11p-6}[/latex]
[latex]\frac{p+6}{4p+3}[/latex]
[/hidden-answer][latex]\frac{{q}^{2}-9}{{q}^{2}+6q+9}÷\frac{{q}^{2}-2q-3}{{q}^{2}+2q-3}[/latex]
[latex]\frac{18{d}^{2}+77d-18}{27{d}^{2}-15d+2}÷\frac{3{d}^{2}+29d-44}{9{d}^{2}-15d+4}[/latex]
[latex]\frac{2d+9}{d+11}[/latex]
[/hidden-answer][latex]\frac{16{x}^{2}+18x-55}{32{x}^{2}-36x-11}÷\frac{2{x}^{2}+17x+30}{4{x}^{2}+25x+6}[/latex]
[latex]\frac{144{b}^{2}-25}{72{b}^{2}-6b-10}÷\frac{18{b}^{2}-21b+5}{36{b}^{2}-18b-10}[/latex]
[latex]\frac{16{a}^{2}-24a+9}{4{a}^{2}+17a-15}÷\frac{16{a}^{2}-9}{4{a}^{2}+11a+6}[/latex]
[latex]\frac{22{y}^{2}+59y+10}{12{y}^{2}+28y-5}÷\frac{11{y}^{2}+46y+8}{24{y}^{2}-10y+1}[/latex]
[latex]\frac{4y-1}{y+4}[/latex]
[/hidden-answer][latex]\frac{9{x}^{2}+3x-20}{3{x}^{2}-7x+4}÷\frac{6{x}^{2}+4x-10}{{x}^{2}-2x+1}[/latex]
For the following exercises, add and subtract the rational expressions, and then simplify.
[latex]\frac{4}{x}+\frac{10}{y}[/latex]
[latex]\frac{10x+4y}{xy}[/latex]
[/hidden-answer][latex]\frac{12}{2q}-\frac{6}{3p}[/latex]
[latex]\frac{4}{a+1}+\frac{5}{a-3}[/latex]
[latex]\frac{9a-7}{{a}^{2}-2a-3}[/latex]
[/hidden-answer][latex]\frac{c+2}{3}-\frac{c-4}{4}[/latex]
[latex]\frac{y+3}{y-2}+\frac{y-3}{y+1}[/latex]
[latex]\frac{2{y}^{2}-y+9}{{y}^{2}-y-2}[/latex]
[/hidden-answer][latex]\frac{x-1}{x+1}-\frac{2x+3}{2x+1}[/latex]
[latex]\frac{3z}{z+1}+\frac{2z+5}{z-2}[/latex]
[latex]\frac{5{z}^{2}+z+5}{{z}^{2}-z-2}[/latex]
[/hidden-answer][latex]\frac{x}{x+1}+\frac{y}{y+1}[/latex]
[latex]\frac{x+2xy+y}{x+xy+y+1}[/latex]
[/hidden-answer]For the following exercises, simplify the rational expression.
[latex]\frac{\frac{6}{y}-\frac{4}{x}}{y}[/latex]
[latex]\frac{\frac{2}{a}+\frac{7}{b}}{b}[/latex]
[latex]\frac{\frac{x}{4}-\frac{p}{8}}{p}[/latex]
[latex]\frac{\frac{3}{a}+\frac{b}{6}}{\frac{2b}{3a}}[/latex]
[latex]\frac{18+ab}{4b}[/latex]
[/hidden-answer][latex]\frac{\frac{3}{x+1}+\frac{2}{x-1}}{\frac{x-1}{x+1}}[/latex]
[latex]\frac{\frac{a}{b}-\frac{b}{a}}{\frac{a+b}{ab}}[/latex]
[latex]a-b[/latex]
[/hidden-answer][latex]\frac{\frac{2x}{3}+\frac{4x}{7}}{\frac{x}{2}}[/latex]
[latex]\frac{\frac{2c}{c+2}+\frac{c-1}{c+1}}{\frac{2c+1}{c+1}}[/latex]
[latex]\frac{3{c}^{2}+3c-2}{2{c}^{2}+5c+2}[/latex]
[/hidden-answer][latex]\frac{\frac{x}{y}-\frac{y}{x}}{\frac{x}{y}+\frac{y}{x}}[/latex]
Brenda is placing tile on her bathroom floor. The area of the floor is[latex]\,15{x}^{2}-8x-7\,[/latex]ft^{2}. The area of one tile is[latex]\,{x}^{2}-2x+1{\text{ft}}^{2}.\,[/latex]To find the number of tiles needed, simplify the rational expression:[latex]\,\frac{15{x}^{2}-8x-7}{{x}^{2}-2x+1}.[/latex]
[latex]\frac{15x+7}{x-1}[/latex]
[/hidden-answer]The area of Sandy’s yard is[latex]\,25{x}^{2}-625\,[/latex]ft^{2}. A patch of sod has an area of[latex]\,{x}^{2}-10x+25\,[/latex]ft^{2}. Divide the two areas and simplify to find how many pieces of sod Sandy needs to cover her yard.
Aaron wants to mulch his garden. His garden is[latex]\,{x}^{2}+18x+81\,[/latex]ft^{2}. One bag of mulch covers[latex]\,{x}^{2}-81\,[/latex]ft^{2}. Divide the expressions and simplify to find how many bags of mulch Aaron needs to mulch his garden.
[latex]\frac{x+9}{x-9}[/latex]
[/hidden-answer]For the following exercises, perform the given operations and simplify.
[latex]\frac{{x}^{2}+x-6}{{x}^{2}-2x-3}\cdot \frac{2{x}^{2}-3x-9}{{x}^{2}-x-2}÷\frac{10{x}^{2}+27x+18}{{x}^{2}+2x+1}[/latex]
[latex]\frac{\frac{3{y}^{2}-10y+3}{3{y}^{2}+5y-2}\cdot \frac{2{y}^{2}-3y-20}{2{y}^{2}-y-15}}{y-4}[/latex]
[latex]\frac{1}{y+2}[/latex]
[/hidden-answer][latex]\frac{\frac{4a+1}{2a-3}+\frac{2a-3}{2a+3}}{\frac{4{a}^{2}+9}{a}}[/latex]
[latex]\frac{{x}^{2}+7x+12}{{x}^{2}+x-6}÷\frac{3{x}^{2}+19x+28}{8{x}^{2}-4x-24}÷\frac{2{x}^{2}+x-3}{3{x}^{2}+4x-7}[/latex]
[latex]4[/latex]
[/hidden-answer]For the following exercises, perform the given operations.
[latex]{\left(5-3\cdot 2\right)}^{2}-6[/latex]
[latex]-5[/latex]
[/hidden-answer][latex]64÷\left(2\cdot 8\right)+14÷7[/latex]
[latex]2\cdot {5}^{2}+6÷2[/latex]
53
[/hidden-answer]For the following exercises, solve the equation.
[latex]5x+9=-11[/latex]
[latex]2y+{4}^{2}=64[/latex]
[latex]y=24[/latex]
[/hidden-answer]For the following exercises, simplify the expression.
[latex]9\left(y+2\right)÷3\cdot 2+1[/latex]
[latex]3m\left(4+7\right)-m[/latex]
[latex]32m[/latex]
[/hidden-answer]For the following exercises, identify the number as rational, irrational, whole, or natural. Choose the most descriptive answer.
11
0
whole
[/hidden-answer][latex]\frac{5}{6}[/latex]
[latex]\sqrt{11}[/latex]
irrational
[/hidden-answer]For the following exercises, simplify the expression.
[latex]{2}^{2}\cdot {2}^{4}[/latex]
[latex]\frac{{4}^{5}}{{4}^{3}}[/latex]
[latex]16[/latex]
[/hidden-answer][latex]{\left(\frac{{a}^{2}}{{b}^{3}}\right)}^{4}[/latex]
[latex]\frac{6{a}^{2}\cdot {a}^{0}}{2{a}^{-4}}[/latex]
[latex]{a}^{6}[/latex]
[/hidden-answer][latex]\frac{{\left(xy\right)}^{4}}{{y}^{3}}\cdot \frac{2}{{x}^{5}}[/latex]
[latex]\frac{{4}^{-2}{x}^{3}{y}^{-3}}{2{x}^{0}}[/latex]
[latex]\frac{{x}^{3}}{32{y}^{3}}[/latex]
[/hidden-answer][latex]{\left(\frac{2{x}^{2}}{y}\right)}^{-2}[/latex]
[latex]\left(\frac{16{a}^{3}}{{b}^{2}}\right){\left(4a{b}^{-1}\right)}^{-2}[/latex]
[latex]a[/latex]
[/hidden-answer]Write the number in standard notation:[latex]\,2.1314\,×\,{10}^{-6}[/latex]
Write the number in scientific notation: 16,340,000
[latex]1.634\,×\,{10}^{7}[/latex]
[/hidden-answer]For the following exercises, find the principal square root.
[latex]\sqrt{121}[/latex]
[latex]\sqrt{196}[/latex]
14
[/hidden-answer][latex]\sqrt{361}[/latex]
[latex]\sqrt{75}[/latex]
[latex]5\sqrt{3}[/latex]
[/hidden-answer][latex]\sqrt{162}[/latex]
[latex]\sqrt{\frac{32}{25}}[/latex]
[latex]\frac{4\sqrt{2}}{5}[/latex]
[/hidden-answer][latex]\sqrt{\frac{80}{81}}[/latex]
[latex]\sqrt{\frac{49}{1250}}[/latex]
[latex]\frac{7\sqrt{2}}{50}[/latex]
[/hidden-answer][latex]\frac{2}{4+\sqrt{2}}[/latex]
[latex]4\sqrt{3}+6\sqrt{3}[/latex]
[latex]10\sqrt{3}[/latex]
[/hidden-answer][latex]12\sqrt{5}-13\sqrt{5}[/latex]
[latex]\sqrt[5]{-243}[/latex]
[latex]-3[/latex]
[/hidden-answer][latex]\frac{\sqrt[3]{250}}{\sqrt[3]{-8}}[/latex]
For the following exercises, perform the given operations and simplify.
[latex]\left(3{x}^{3}+2x-1\right)+\left(4{x}^{2}-2x+7\right)[/latex]
[latex]3{x}^{3}+4{x}^{2}+6[/latex]
[/hidden-answer][latex]\left(2y+1\right)-\left(2{y}^{2}-2y-5\right)[/latex]
[latex]\left(2{x}^{2}+3x-6\right)+\left(3{x}^{2}-4x+9\right)[/latex]
[latex]5{x}^{2}-x+3[/latex]
[/hidden-answer][latex]\left(6{a}^{2}+3a+10\right)-\left(6{a}^{2}-3a+5\right)[/latex]
[latex]\left(k+3\right)\left(k-6\right)[/latex]
[latex]{k}^{2}-3k-18[/latex]
[/hidden-answer][latex]\left(2h+1\right)\left(3h-2\right)[/latex]
[latex]\left(x+1\right)\left({x}^{2}+1\right)[/latex]
[latex]{x}^{3}+{x}^{2}+x+1[/latex]
[/hidden-answer][latex]\left(m-2\right)\left({m}^{2}+2m-3\right)[/latex]
[latex]\left(a+2b\right)\left(3a-b\right)[/latex]
[latex]3{a}^{2}+5ab-2{b}^{2}[/latex]
[/hidden-answer][latex]\left(x+y\right)\left(x-y\right)[/latex]
For the following exercises, find the greatest common factor.
[latex]81p+9pq-27{p}^{2}{q}^{2}[/latex]
[latex]9p[/latex]
[/hidden-answer][latex]12{x}^{2}y+4x{y}^{2}-18xy[/latex]
[latex]88{a}^{3}b+4{a}^{2}b-144{a}^{2}[/latex]
[latex]4{a}^{2}[/latex]
[/hidden-answer]For the following exercises, factor the polynomial.
[latex]2{x}^{2}-9x-18[/latex]
[latex]8{a}^{2}+30a-27[/latex]
[latex]\left(4a-3\right)\left(2a+9\right)[/latex]
[/hidden-answer][latex]{d}^{2}-5d-66[/latex]
[latex]{x}^{2}+10x+25[/latex]
[latex]{\left(x+5\right)}^{2}[/latex]
[/hidden-answer][latex]{y}^{2}-6y+9[/latex]
[latex]4{h}^{2}-12hk+9{k}^{2}[/latex]
[latex]{\left(2h-3k\right)}^{2}[/latex]
[/hidden-answer][latex]361{x}^{2}-121[/latex]
[latex]{p}^{3}+216[/latex]
[latex]\left(p+6\right)\left({p}^{2}-6p+36\right)[/latex]
[/hidden-answer][latex]8{x}^{3}-125[/latex]
[latex]64{q}^{3}-27{p}^{3}[/latex]
[latex]\left(4q-3p\right)\left(16{q}^{2}+12pq+9{p}^{2}\right)[/latex]
[/hidden-answer][latex]4x{\left(x-1\right)}^{-\frac{1}{4}}+3{\left(x-1\right)}^{\frac{3}{4}}[/latex]
[latex]3p{\left(p+3\right)}^{\frac{1}{3}}-8{\left(p+3\right)}^{\frac{4}{3}}[/latex]
[latex]{\left(p+3\right)}^{\frac{1}{3}}\left(-5p-24\right)[/latex]
[/hidden-answer][latex]4r{\left(2r-1\right)}^{-\frac{2}{3}}-5{\left(2r-1\right)}^{\frac{1}{3}}[/latex]
For the following exercises, simplify the expression.
[latex]\frac{{x}^{2}-x-12}{{x}^{2}-8x+16}[/latex]
[latex]\frac{x+3}{x-4}[/latex]
[/hidden-answer][latex]\frac{4{y}^{2}-25}{4{y}^{2}-20y+25}[/latex]
[latex]\frac{2{a}^{2}-a-3}{2{a}^{2}-6a-8}\cdot \frac{5{a}^{2}-19a-4}{10{a}^{2}-13a-3}[/latex]
[latex]\frac{1}{2}[/latex]
[/hidden-answer][latex]\frac{d-4}{{d}^{2}-9}\cdot \frac{d-3}{{d}^{2}-16}[/latex]
[latex]\frac{{m}^{2}+5m+6}{2{m}^{2}-5m-3}÷\frac{2{m}^{2}+3m-9}{4{m}^{2}-4m-3}[/latex]
[latex]\frac{m+2}{m-3}[/latex]
[/hidden-answer][latex]\frac{4{d}^{2}-7d-2}{6{d}^{2}-17d+10}÷\frac{8{d}^{2}+6d+1}{6{d}^{2}+7d-10}[/latex]
[latex]\frac{10}{x}+\frac{6}{y}[/latex]
[latex]\frac{6x+10y}{xy}[/latex]
[/hidden-answer][latex]\frac{12}{{a}^{2}+2a+1}-\frac{3}{{a}^{2}-1}[/latex]
[latex]\frac{\frac{1}{d}+\frac{2}{c}}{\frac{6c+12d}{dc}}[/latex]
[latex]\frac{1}{6}[/latex]
[/hidden-answer][latex]\frac{\frac{3}{x}-\frac{7}{y}}{\frac{2}{x}}[/latex]
For the following exercises, identify the number as rational, irrational, whole, or natural. Choose the most descriptive answer.
[latex]-13[/latex]
rational
[/hidden-answer][latex]\sqrt{2}[/latex]
For the following exercises, evaluate the equations.
[latex]2\left(x+3\right)-12=18[/latex]
[latex]x=12[/latex]
[/hidden-answer][latex]y{\left(3+3\right)}^{2}-26=10[/latex]
Write the number in standard notation:[latex]3.1415\,×\,{10}^{6}[/latex]
3,141,500
[/hidden-answer]Write the number in scientific notation: 0.0000000212.
For the following exercises, simplify the expression.
[latex]-2\cdot {\left(2+3\cdot 2\right)}^{2}+144[/latex]
[latex]16[/latex]
[/hidden-answer][latex]4\left(x+3\right)-\left(6x+2\right)[/latex]
[latex]{3}^{5}\cdot {3}^{-3}[/latex]
9
[/hidden-answer][latex]{\left(\frac{2}{3}\right)}^{3}[/latex]
[latex]\frac{8{x}^{3}}{{\left(2x\right)}^{2}}[/latex]
[latex]2x[/latex]
[/hidden-answer][latex]\left(16{y}^{0}\right)2{y}^{-2}[/latex]
[latex]\sqrt{441}[/latex]
21
[/hidden-answer][latex]\sqrt{490}[/latex]
[latex]\sqrt{\frac{9x}{16}}[/latex]
[latex]\frac{3\sqrt{x}}{4}[/latex]
[/hidden-answer][latex]\frac{\sqrt{121{b}^{2}}}{1+\sqrt{b}}[/latex]
[latex]6\sqrt{24}+7\sqrt{54}-12\sqrt{6}[/latex]
[latex]21\sqrt{6}[/latex]
[/hidden-answer][latex]\frac{\sqrt[3]{-8}}{\sqrt[4]{625}}[/latex]
[latex]\left(13{q}^{3}+2{q}^{2}-3\right)-\left(6{q}^{2}+5q-3\right)[/latex]
[latex]13{q}^{3}-4{q}^{2}-5q[/latex]
[/hidden-answer][latex]\left(6{p}^{2}+2p+1\right)+\left(9{p}^{2}-1\right)[/latex]
[latex]\left(n-2\right)\left({n}^{2}-4n+4\right)[/latex]
[latex]{n}^{3}-6{n}^{2}+12n-8[/latex]
[/hidden-answer][latex]\left(a-2b\right)\left(2a+b\right)[/latex]
For the following exercises, factor the polynomial.
[latex]16{x}^{2}-81[/latex]
[latex]\left(4x+9\right)\left(4x-9\right)[/latex]
[/hidden-answer][latex]{y}^{2}+12y+36[/latex]
[latex]27{c}^{3}-1331[/latex]
[latex]\left(3c-11\right)\left(9{c}^{2}+33c+121\right)[/latex]
[/hidden-answer][latex]3x{\left(x-6\right)}^{-\frac{1}{4}}+2{\left(x-6\right)}^{\frac{3}{4}}[/latex]
For the following exercises, simplify the expression.
[latex]\frac{2{z}^{2}+7z+3}{{z}^{2}-9}\cdot \frac{4{z}^{2}-15z+9}{4{z}^{2}-1}[/latex]
[latex]\frac{4z-3}{2z-1}[/latex]
[/hidden-answer][latex]\frac{x}{y}+\frac{2}{x}[/latex]
[latex]\frac{\frac{a}{2b}-\frac{2b}{9a}}{\frac{3a-2b}{6a}}[/latex]
[latex]\frac{3a+2b}{3b}[/latex]
[/hidden-answer]For most people, the term territorial possession indicates restrictions, usually dealing with trespassing or rite of passage and takes place in some foreign location. What most Americans do not realize is that from September through December, territorial possession dominates our lifestyles while watching the NFL. In this area, territorial possession is governed by the referees who make their decisions based on what the chains reveal. If the ball is at point A[latex]\,\left({x}_{1},{y}_{1}\right),[/latex]then it is up to the quarterback to decide which route to point B[latex]\,\left({x}_{2},{y}_{2}\right),[/latex]the end zone, is most feasible.
]]>Tracie set out from Elmhurst, IL, to go to Franklin Park. On the way, she made a few stops to do errands. Each stop is indicated by a red dot in (Figure). Laying a rectangular coordinate grid over the map, we can see that each stop aligns with an intersection of grid lines. In this section, we will learn how to use grid lines to describe locations and changes in locations.
An old story describes how seventeenth-century philosopher/mathematician René Descartes invented the system that has become the foundation of algebra while sick in bed. According to the story, Descartes was staring at a fly crawling on the ceiling when he realized that he could describe the fly’s location in relation to the perpendicular lines formed by the adjacent walls of his room. He viewed the perpendicular lines as horizontal and vertical axes. Further, by dividing each axis into equal unit lengths, Descartes saw that it was possible to locate any object in a two-dimensional plane using just two numbers—the displacement from the horizontal axis and the displacement from the vertical axis.
While there is evidence that ideas similar to Descartes’ grid system existed centuries earlier, it was Descartes who introduced the components that comprise the Cartesian coordinate system, a grid system having perpendicular axes. Descartes named the horizontal axis the x-axis and the vertical axis the y-axis.
The Cartesian coordinate system, also called the rectangular coordinate system, is based on a two-dimensional plane consisting of the x-axis and the y-axis. Perpendicular to each other, the axes divide the plane into four sections. Each section is called a quadrant; the quadrants are numbered counterclockwise as shown in (Figure)
The center of the plane is the point at which the two axes cross. It is known as the origin, or point[latex]\left(0,0\right).[/latex]From the origin, each axis is further divided into equal units: increasing, positive numbers to the right on the x-axis and up the y-axis; decreasing, negative numbers to the left on the x-axis and down the y-axis. The axes extend to positive and negative infinity as shown by the arrowheads in (Figure).
Each point in the plane is identified by its x-coordinate, or horizontal displacement from the origin, and its y-coordinate, or vertical displacement from the origin. Together, we write them as an ordered pair indicating the combined distance from the origin in the form[latex]\,\left(x,y\right).\,[/latex]An ordered pair is also known as a coordinate pair because it consists of x- and y-coordinates. For example, we can represent the point[latex]\,\left(3,-1\right)\,[/latex]in the plane by moving three units to the right of the origin in the horizontal direction, and one unit down in the vertical direction. See (Figure).
When dividing the axes into equally spaced increments, note that the x-axis may be considered separately from the y-axis. In other words, while the x-axis may be divided and labeled according to consecutive integers, the y-axis may be divided and labeled by increments of 2, or 10, or 100. In fact, the axes may represent other units, such as years against the balance in a savings account, or quantity against cost, and so on. Consider the rectangular coordinate system primarily as a method for showing the relationship between two quantities.
A two-dimensional plane where the
A point in the plane is defined as an ordered pair,[latex]\,\left(x,y\right),[/latex]such that x is determined by its horizontal distance from the origin and y is determined by its vertical distance from the origin.
Plot the points[latex]\,\left(-2,4\right),[/latex][latex]\left(3,3\right),[/latex]and[latex]\,\left(0,-3\right)\,[/latex]in the plane.
To plot the point[latex]\,\left(-2,4\right),[/latex]begin at the origin. The x-coordinate is –2, so move two units to the left. The y-coordinate is 4, so then move four units up in the positive y direction.
To plot the point[latex]\,\left(3,3\right),[/latex]begin again at the origin. The x-coordinate is 3, so move three units to the right. The y-coordinate is also 3, so move three units up in the positive y direction.
To plot the point[latex]\,\left(0,-3\right),[/latex]begin again at the origin. The x-coordinate is 0. This tells us not to move in either direction along the x-axis. The y-coordinate is –3, so move three units down in the negative y direction. See the graph in (Figure).
Note that when either coordinate is zero, the point must be on an axis. If the x-coordinate is zero, the point is on the y-axis. If the y-coordinate is zero, the point is on the x-axis.
We can plot a set of points to represent an equation. When such an equation contains both an x variable and a y variable, it is called an equation in two variables. Its graph is called a graph in two variables. Any graph on a two-dimensional plane is a graph in two variables.
Suppose we want to graph the equation[latex]\,y=2x-1.\,[/latex]We can begin by substituting a value for x into the equation and determining the resulting value of y. Each pair of x- and y-values is an ordered pair that can be plotted. (Figure) lists values of x from –3 to 3 and the resulting values for y.
[latex]x[/latex] | [latex]y=2x-1[/latex] | [latex]\left(x,y\right)[/latex] |
[latex]-3[/latex] | [latex]y=2\left(-3\right)-1=-7[/latex] | [latex]\left(-3,-7\right)[/latex] |
[latex]-2[/latex] | [latex]y=2\left(-2\right)-1=-5[/latex] | [latex]\left(-2,-5\right)[/latex] |
[latex]-1[/latex] | [latex]y=2\left(-1\right)-1=-3[/latex] | [latex]\left(-1,-3\right)[/latex] |
[latex]0[/latex] | [latex]y=2\left(0\right)-1=-1[/latex] | [latex]\left(0,-1\right)[/latex] |
[latex]1[/latex] | [latex]y=2\left(1\right)-1=1[/latex] | [latex]\left(1,1\right)[/latex] |
[latex]2[/latex] | [latex]y=2\left(2\right)-1=3[/latex] | [latex]\left(2,3\right)[/latex] |
[latex]3[/latex] | [latex]y=2\left(3\right)-1=5[/latex] | [latex]\left(3,5\right)[/latex] |
We can plot the points in the table. The points for this particular equation form a line, so we can connect them. See (Figure). This is not true for all equations.
Note that the x-values chosen are arbitrary, regardless of the type of equation we are graphing. Of course, some situations may require particular values of x to be plotted in order to see a particular result. Otherwise, it is logical to choose values that can be calculated easily, and it is always a good idea to choose values that are both negative and positive. There is no rule dictating how many points to plot, although we need at least two to graph a line. Keep in mind, however, that the more points we plot, the more accurately we can sketch the graph.
Given an equation, graph by plotting points.
Graph the equation[latex]\,y=-x+2\,[/latex]by plotting points.
First, we construct a table similar to (Figure). Choose x values and calculate y.
[latex]x[/latex] | [latex]y=-x+2[/latex] | [latex]\left(x,y\right)[/latex] |
[latex]-5[/latex] | [latex]y=-\left(-5\right)+2=7[/latex] | [latex]\left(-5,7\right)[/latex] |
[latex]-3[/latex] | [latex]y=-\left(-3\right)+2=5[/latex] | [latex]\left(-3,5\right)[/latex] |
[latex]-1[/latex] | [latex]y=-\left(-1\right)+2=3[/latex] | [latex]\left(-1,3\right)[/latex] |
[latex]0[/latex] | [latex]y=-\left(0\right)+2=2[/latex] | [latex]\left(0,2\right)[/latex] |
[latex]1[/latex] | [latex]y=-\left(1\right)+2=1[/latex] | [latex]\left(1,1\right)[/latex] |
[latex]3[/latex] | [latex]y=-\left(3\right)+2=-1[/latex] | [latex]\left(3,-1\right)[/latex] |
[latex]5[/latex] | [latex]y=-\left(5\right)+2=-3[/latex] | [latex]\left(5,-3\right)[/latex] |
Now, plot the points. Connect them if they form a line. See (Figure)
Construct a table and graph the equation by plotting points:[latex]\,y=\frac{1}{2}x+2.[/latex]
[latex]x[/latex] | [latex]y=\frac{1}{2}x+2[/latex] | [latex]\left(x,y\right)[/latex] |
[latex]-2[/latex] | [latex]y=\frac{1}{2}\left(-2\right)+2=1[/latex] | [latex]\left(-2,1\right)[/latex] |
[latex]-1[/latex] | [latex]y=\frac{1}{2}\left(-1\right)+2=\frac{3}{2}[/latex] | [latex]\left(-1,\frac{3}{2}\right)[/latex] |
[latex]0[/latex] | [latex]y=\frac{1}{2}\left(0\right)+2=2[/latex] | [latex]\left(0,2\right)[/latex] |
[latex]1[/latex] | [latex]y=\frac{1}{2}\left(1\right)+2=\frac{5}{2}[/latex] | [latex]\left(1,\frac{5}{2}\right)[/latex] |
[latex]2[/latex] | [latex]y=\frac{1}{2}\left(2\right)+2=3[/latex] | [latex]\left(2,3\right)[/latex] |
Most graphing calculators require similar techniques to graph an equation. The equations sometimes have to be manipulated so they are written in the style[latex]\,y\,[/latex]=_____. The TI-84 Plus, and many other calculator makes and models, have a mode function, which allows the window (the screen for viewing the graph) to be altered so the pertinent parts of a graph can be seen.
For example, the equation[latex]\,y=2x-20\,[/latex]has been entered in the TI-84 Plus shown in (Figure)a. In (Figure)b, the resulting graph is shown. Notice that we cannot see on the screen where the graph crosses the axes. The standard window screen on the TI-84 Plus shows[latex]\,-10\le x\le 10,[/latex]and[latex]\,-10\le y\le 10.\,[/latex]See (Figure)c.
By changing the window to show more of the positive x-axis and more of the negative y-axis, we have a much better view of the graph and the x- and y-intercepts. See (Figure)a and (Figure)b.
The intercepts of a graph are points at which the graph crosses the axes. The x-intercept is the point at which the graph crosses the x-axis. At this point, the y-coordinate is zero. The y-intercept is the point at which the graph crosses the y-axis. At this point, the x-coordinate is zero.
To determine the x-intercept, we set y equal to zero and solve for x. Similarly, to determine the y-intercept, we set x equal to zero and solve for y. For example, lets find the intercepts of the equation[latex]\,y=3x-1.[/latex]
To find the x-intercept, set[latex]\,y=0.[/latex]
To find the y-intercept, set[latex]\,x=0.[/latex]
We can confirm that our results make sense by observing a graph of the equation as in (Figure). Notice that the graph crosses the axes where we predicted it would.
Set[latex]\,y=0\,[/latex]to find the x-intercept.
Set[latex]\,x=0\,[/latex]to find the y-intercept.
[/hidden-answer]
Find the intercepts of the equation and sketch the graph:[latex]\,y=-\frac{3}{4}x+3.[/latex]
x-intercept is[latex]\,\left(4,0\right);[/latex]y-intercept is[latex]\,\left(0,3\right).[/latex]
[/hidden-answer]Derived from the Pythagorean Theorem, the distance formula is used to find the distance between two points in the plane. The Pythagorean Theorem,[latex]\,{a}^{2}+{b}^{2}={c}^{2},[/latex]is based on a right triangle where a and b are the lengths of the legs adjacent to the right angle, and c is the length of the hypotenuse. See (Figure).
The relationship of sides[latex]\,|{x}_{2}-{x}_{1}|\,[/latex]and[latex]\,|{y}_{2}-{y}_{1}|\,[/latex]to side d is the same as that of sides a and b to side c. We use the absolute value symbol to indicate that the length is a positive number because the absolute value of any number is positive. (For example,[latex]\,|-3|=3.\,[/latex]) The symbols[latex]\,|{x}_{2}-{x}_{1}|\,[/latex]and[latex]\,|{y}_{2}-{y}_{1}|\,[/latex]indicate that the lengths of the sides of the triangle are positive. To find the length c, take the square root of both sides of the Pythagorean Theorem.
It follows that the distance formula is given as
We do not have to use the absolute value symbols in this definition because any number squared is positive.
Given endpoints[latex]\,\left({x}_{1},{y}_{1}\right)\,[/latex]and[latex]\,\left({x}_{2},{y}_{2}\right),[/latex]the distance between two points is given by
Find the distance between the points[latex]\,\left(-3,-1\right)\,[/latex]and[latex]\,\left(2,3\right).[/latex]
Let us first look at the graph of the two points. Connect the points to form a right triangle as in (Figure).
Then, calculate the length of d using the distance formula.
Find the distance between two points:[latex]\,\left(1,4\right)\,[/latex]and[latex]\,\left(11,9\right).[/latex]
[latex]\sqrt{125}=5\sqrt{5}[/latex]
[/hidden-answer]Let’s return to the situation introduced at the beginning of this section.
Tracie set out from Elmhurst, IL, to go to Franklin Park. On the way, she made a few stops to do errands. Each stop is indicated by a red dot in (Figure). Find the total distance that Tracie traveled. Compare this with the distance between her starting and final positions.
The first thing we should do is identify ordered pairs to describe each position. If we set the starting position at the origin, we can identify each of the other points by counting units east (right) and north (up) on the grid. For example, the first stop is 1 block east and 1 block north, so it is at[latex]\,\left(1,1\right).\,[/latex]The next stop is 5 blocks to the east, so it is at[latex]\,\left(5,1\right).\,[/latex]After that, she traveled 3 blocks east and 2 blocks north to[latex]\,\left(8,3\right).\,[/latex]Lastly, she traveled 4 blocks north to[latex]\,\left(8,7\right).\,[/latex]We can label these points on the grid as in (Figure).
Next, we can calculate the distance. Note that each grid unit represents 1,000 feet.
Next, we will add the distances listed in (Figure).
From/To | Number of Feet Driven |
---|---|
[latex]\left(0,0\right)\,[/latex]to[latex]\,\left(1,1\right)[/latex] | 2,000 |
[latex]\left(1,1\right)\,[/latex]to[latex]\left(5,1\right)\,[/latex] | 4,000 |
[latex]\left(5,1\right)\,[/latex]to[latex]\,\left(8,3\right)[/latex] | 5,000 |
[latex]\left(8,3\right)\,[/latex]to[latex]\,\left(8,7\right)[/latex] | 4,000 |
Total | 15,000 |
The total distance Tracie drove is 15,000 feet, or 2.84 miles. This is not, however, the actual distance between her starting and ending positions. To find this distance, we can use the distance formula between the points[latex]\,\left(0,0\right)\,[/latex]and[latex]\,\left(8,7\right).[/latex]
At 1,000 feet per grid unit, the distance between Elmhurst, IL, to Franklin Park is 10,630.14 feet, or 2.01 miles. The distance formula results in a shorter calculation because it is based on the hypotenuse of a right triangle, a straight diagonal from the origin to the point[latex]\,\left(8,7\right).\,[/latex]Perhaps you have heard the saying “as the crow flies,” which means the shortest distance between two points because a crow can fly in a straight line even though a person on the ground has to travel a longer distance on existing roadways.[/hidden-answer]
When the endpoints of a line segment are known, we can find the point midway between them. This point is known as the midpoint and the formula is known as the midpoint formula. Given the endpoints of a line segment,[latex]\,\left({x}_{1},{y}_{1}\right)\,[/latex]and[latex]\,\left({x}_{2},{y}_{2}\right),[/latex]the midpoint formula states how to find the coordinates of the midpoint[latex]\,M.[/latex]
A graphical view of a midpoint is shown in (Figure). Notice that the line segments on either side of the midpoint are congruent.
Find the midpoint of the line segment with the endpoints[latex]\,\left(7,-2\right)\,[/latex]and[latex]\,\left(9,5\right).[/latex]
Use the formula to find the midpoint of the line segment.
Find the midpoint of the line segment with endpoints[latex]\,\left(-2,-1\right)\,[/latex]and[latex]\,\left(-8,6\right).[/latex]
[latex]\left(-5,\frac{5}{2}\right)[/latex]
[/hidden-answer]The diameter of a circle has endpoints[latex]\,\left(-1,-4\right)\,[/latex]and[latex]\,\left(5,-4\right).\,[/latex]Find the center of the circle.
The center of a circle is the center, or midpoint, of its diameter. Thus, the midpoint formula will yield the center point.
Access these online resources for additional instruction and practice with the Cartesian coordinate system.
Is it possible for a point plotted in the Cartesian coordinate system to not lie in one of the four quadrants? Explain.
Answers may vary. Yes. It is possible for a point to be on the x-axis or on the y-axis and therefore is considered to NOT be in one of the quadrants.
[/hidden-answer]Describe the process for finding the x-intercept and the y-intercept of a graph algebraically.
Describe in your own words what the y-intercept of a graph is.
The y-intercept is the point where the graph crosses the y-axis.
[/hidden-answer]When using the distance formula[latex]\,d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}},[/latex]explain the correct order of operations that are to be performed to obtain the correct answer.
For each of the following exercises, find the x-intercept and the y-intercept without graphing. Write the coordinates of each intercept.
[latex]y=-3x+6[/latex]
The x-intercept is[latex]\,\left(2,0\right)\,[/latex]and the y-intercept is[latex]\,\left(0,6\right).[/latex]
[/hidden-answer][latex]4y=2x-1[/latex]
[latex]3x-2y=6[/latex]
The x-intercept is[latex]\,\left(2,0\right)\,[/latex]and the y-intercept is[latex]\,\left(0,-3\right).[/latex]
[/hidden-answer][latex]4x-3=2y[/latex]
[latex]3x+8y=9[/latex]
The x-intercept is[latex]\,\left(3,0\right)\,[/latex]and the y-intercept is[latex]\,\left(0,\frac{9}{8}\right).[/latex]
[/hidden-answer][latex]2x-\frac{2}{3}=\frac{3}{4}y+3[/latex]
For each of the following exercises, solve the equation for y in terms of x.
[latex]4x+2y=8[/latex]
[latex]y=4-2x[/latex]
[/hidden-answer][latex]3x-2y=6[/latex]
[latex]2x=5-3y[/latex]
[latex]y=\frac{5-2x}{3}[/latex]
[/hidden-answer][latex]x-2y=7[/latex]
[latex]5y+4=10x[/latex]
[latex]y=2x-\frac{4}{5}[/latex]
[/hidden-answer][latex]5x+2y=0[/latex]
For each of the following exercises, find the distance between the two points. Simplify your answers, and write the exact answer in simplest radical form for irrational answers.
[latex]\left(-4,1\right)\,[/latex]and[latex]\,\left(3,-4\right)[/latex]
[latex]d=\sqrt{74}[/latex]
[/hidden-answer][latex]\left(2,-5\right)\,[/latex]and[latex]\,\left(7,4\right)[/latex]
[latex]\left(5,0\right)\,[/latex]and[latex]\,\left(5,6\right)[/latex]
[latex]d=\sqrt{36}=6[/latex]
[/hidden-answer][latex]\left(-4,3\right)\,[/latex]and[latex]\,\left(10,3\right)[/latex]
Find the distance between the two points given using your calculator, and round your answer to the nearest hundredth.
[latex]\left(19,12\right)\,[/latex]and[latex]\,\left(41,71\right)[/latex]
[latex]d\approx 62.97[/latex]
[/hidden-answer]For each of the following exercises, find the coordinates of the midpoint of the line segment that joins the two given points.
[latex]\left(-5,-6\right)\,[/latex]and[latex]\,\left(4,2\right)[/latex]
[latex]\left(-1,1\right)\,[/latex]and[latex]\,\left(7,-4\right)[/latex]
[latex]\left(3,\frac{-3}{2}\right)[/latex]
[/hidden-answer][latex]\left(-5,-3\right)\,[/latex]and[latex]\,\left(-2,-8\right)[/latex]
[latex]\left(0,7\right)\,[/latex]and[latex]\,\left(4,-9\right)[/latex]
[latex]\left(2,-1\right)[/latex]
[/hidden-answer][latex]\left(-43,17\right)\,[/latex]and[latex]\,\left(23,-34\right)[/latex]
For each of the following exercises, identify the information requested.
What are the coordinates of the origin?
[latex]\left(0,0\right)[/latex]
[/hidden-answer]If a point is located on the y-axis, what is the x-coordinate?
If a point is located on the x-axis, what is the y-coordinate?
[latex]y=0[/latex]
[/hidden-answer]For each of the following exercises, plot the three points on the given coordinate plane. State whether the three points you plotted appear to be collinear (on the same line).
[latex]\left(4,1\right)\left(-2,-3\right)\left(5,0\right)[/latex]
[latex]\left(-1,2\right)\left(0,4\right)\left(2,1\right)[/latex]
[latex]\left(-3,0\right)\left(-3,4\right)\left(-3,-3\right)[/latex]
Name the coordinates of the points graphed.
[latex]\left(-3,2\right),\left(1,3\right),\left(4,0\right)[/latex]
[/hidden-answer]Name the quadrant in which the following points would be located. If the point is on an axis, name the axis.
[latex]\begin{array}{l}a.\left(-3,-4\right)\\ b.\left(-5,0\right)\\ c.\left(1,-4\right)\\ d.\left(-2,7\right)\\ e.\left(0,-3\right)\end{array}[/latex]
For each of the following exercises, construct a table and graph the equation by plotting at least three points.
[latex]y=\frac{1}{3}x+2[/latex]
[latex]x[/latex] | [latex]y[/latex] |
[latex]-3[/latex] | 1 |
0 | 2 |
3 | 3 |
6 | 4 |
[latex]y=-3x+1[/latex]
[latex]2y=x+3[/latex]
x | y |
–3 | 0 |
0 | 1.5 |
3 | 3 |
For each of the following exercises, find and plot the x- and y-intercepts, and graph the straight line based on those two points.
[latex]4x-3y=12[/latex]
[latex]x-2y=8[/latex]
[latex]y-5=5x[/latex]
[latex]3y=-2x+6[/latex]
[latex]y=\frac{x-3}{2}[/latex]
For each of the following exercises, use the graph in the figure below.
Find the distance between the two endpoints using the distance formula. Round to three decimal places.
[latex]d=8.246[/latex]
[/hidden-answer]Find the coordinates of the midpoint of the line segment connecting the two points.
Find the distance that[latex]\,\left(-3,4\right)\,[/latex]is from the origin.
[latex]d=5[/latex]
[/hidden-answer]Find the distance that[latex]\,\left(5,2\right)\,[/latex]is from the origin. Round to three decimal places.
Which point is closer to the origin?
[latex]\left(-3,4\right)[/latex]
[/hidden-answer]For the following exercises, use your graphing calculator to input the linear graphs in the Y= graph menu.
After graphing it, use the 2^{nd} CALC button and 1:value button, hit enter. At the lower part of the screen you will see “x=” and a blinking cursor. You may enter any number for x and it will display the y value for any x value you input. Use this and plug in x = 0, thus finding the y-intercept, for each of the following graphs.
[latex]{\text{Y}}_{1}=-2x+5[/latex]
[latex]{\text{Y}}_{1}=\frac{3x-8}{4}[/latex]
[latex]x=0\text{ }y=-2[/latex]
[/hidden-answer][latex]{\text{Y}}_{1}=\frac{x+5}{2}[/latex]
For the following exercises, use your graphing calculator to input the linear graphs in the Y= graph menu.
After graphing it, use the 2^{nd} CALC button and 2:zero button, hit enter. At the lower part of the screen you will see “left bound?” and a blinking cursor on the graph of the line. Move this cursor to the left of the x-intercept, hit ENTER. Now it says “right bound?” Move the cursor to the right of the x-intercept, hit enter. Now it says “guess?” Move your cursor to the left somewhere in between the left and right bound near the x-intercept. Hit enter. At the bottom of your screen it will display the coordinates of the x-intercept or the “zero” to the y-value. Use this to find the x-intercept.
Note: With linear/straight line functions the zero is not really a “guess,” but it is necessary to enter a “guess” so it will search and find the exact x-intercept between your right and left boundaries. With other types of functions (more than one x-intercept), they may be irrational numbers so “guess” is more appropriate to give it the correct limits to find a very close approximation between the left and right boundaries.
[latex]{\text{Y}}_{1}=-8x+6[/latex]
[latex]x=0.75\text{ }y=0[/latex]
[/hidden-answer][latex]{\text{Y}}_{1}=4x-7[/latex]
[latex]{\text{Y}}_{1}=\frac{3x+5}{4}\,[/latex]Round your answer to the nearest thousandth.
[latex]x=-1.667\text{ }y=0[/latex]
[/hidden-answer]A man drove 10 mi directly east from his home, made a left turn at an intersection, and then traveled 5 mi north to his place of work. If a road was made directly from his home to his place of work, what would its distance be to the nearest tenth of a mile?
If the road was made in the previous exercise, how much shorter would the man’s one-way trip be every day?
[latex]\text{15}\text{−11}.\text{2 }=\text{ 3}.8\,[/latex]mi shorter
[/hidden-answer]Given these four points:[latex]\,A\left(1,3\right),\text{}B\left(-3,5\right),\text{}C\left(4,7\right),\text{ and }D\left(5,-4\right),[/latex]find the coordinates of the midpoint of line segments[latex]\,\overline{\text{AB}}\,[/latex]and[latex]\,\overline{\text{CD}}.[/latex]
After finding the two midpoints in the previous exercise, find the distance between the two midpoints to the nearest thousandth.
[latex]\text{6}.0\text{42}[/latex]
[/hidden-answer]Given the graph of the rectangle shown and the coordinates of its vertices, prove that the diagonals of the rectangle are of equal length.
In the previous exercise, find the coordinates of the midpoint for each diagonal.
Midpoint of each diagonal is the same point[latex]\,\left(2,2\right).\,[/latex]Note this is a characteristic of rectangles, but not other quadrilaterals.
[/hidden-answer]The coordinates on a map for San Francisco are[latex]\,\left(53,17\right)\,[/latex]and those for Sacramento are[latex]\,\left(123,78\right).\,[/latex]Note that coordinates represent miles. Find the distance between the cities to the nearest mile.
If San Jose’s coordinates are[latex]\,\left(76,-12\right),[/latex]where the coordinates represent miles, find the distance between San Jose and San Francisco to the nearest mile.
[latex]\text{37}\,[/latex]mi
[/hidden-answer]A small craft in Lake Ontario sends out a distress signal. The coordinates of the boat in trouble were[latex]\,\left(49,64\right).\,[/latex]One rescue boat is at the coordinates[latex]\,\left(60,82\right)\,[/latex]and a second Coast Guard craft is at coordinates[latex]\,\left(58,47\right).\,[/latex]Assuming both rescue craft travel at the same rate, which one would get to the distressed boat the fastest?
A man on the top of a building wants to have a guy wire extend to a point on the ground 20 ft from the building. To the nearest foot, how long will the wire have to be if the building is 50 ft tall?
54 ft
[/hidden-answer]If we rent a truck and pay a $75/day fee plus $.20 for every mile we travel, write a linear equation that would express the total cost[latex]\,y,[/latex]using[latex]\,x\,[/latex]to represent the number of miles we travel. Graph this function on your graphing calculator and find the total cost for one day if we travel 70 mi.
Caroline is a full-time college student planning a spring break vacation. To earn enough money for the trip, she has taken a part-time job at the local bank that pays $15.00/hr, and she opened a savings account with an initial deposit of $400 on January 15. She arranged for direct deposit of her payroll checks. If spring break begins March 20 and the trip will cost approximately $2,500, how many hours will she have to work to earn enough to pay for her vacation? If she can only work 4 hours per day, how many days per week will she have to work? How many weeks will it take? In this section, we will investigate problems like this and others, which generate graphs like the line in (Figure).
A linear equation is an equation of a straight line, written in one variable. The only power of the variable is 1. Linear equations in one variable may take the form[latex]\,ax+b=0\,[/latex]and are solved using basic algebraic operations.
We begin by classifying linear equations in one variable as one of three types: identity, conditional, or inconsistent. An identity equation is true for all values of the variable. Here is an example of an identity equation.
The solution set consists of all values that make the equation true. For this equation, the solution set is all real numbers because any real number substituted for[latex]\,x\,[/latex]will make the equation true.
A conditional equation is true for only some values of the variable. For example, if we are to solve the equation[latex]\,5x+2=3x-6,[/latex]we have the following:
The solution set consists of one number:[latex]\,\left\{-4\right\}.\,[/latex]It is the only solution and, therefore, we have solved a conditional equation.
An inconsistent equation results in a false statement. For example, if we are to solve[latex]\,5x-15=5\left(x-4\right),[/latex]we have the following:
Indeed,[latex]\,-15\ne \,-20.\,[/latex]There is no solution because this is an inconsistent equation.
Solving linear equations in one variable involves the fundamental properties of equality and basic algebraic operations. A brief review of those operations follows.
A linear equation in one variable can be written in the form
where a and b are real numbers,[latex]\,a\ne 0.[/latex]
Given a linear equation in one variable, use algebra to solve it.
The following steps are used to manipulate an equation and isolate the unknown variable, so that the last line reads[latex]\,x=\text{_________,}[/latex]if x is the unknown. There is no set order, as the steps used depend on what is given:
Solve the following equation:[latex]\,2x+7=19.[/latex]
The solution is 6
[/hidden-answer]Solve the linear equation in one variable:[latex]\,2x+1=-9.[/latex]
[latex]x=-5[/latex]
[/hidden-answer]Solve the following equation:[latex]\,4\left(x-3\right)+12=15-5\left(x+6\right).[/latex]
Apply standard algebraic properties.
This problem requires the distributive property to be applied twice, and then the properties of algebra are used to reach the final line,[latex]\,x=-\frac{5}{3}.[/latex]
Solve the equation in one variable:[latex]\,-2\left(3x-1\right)+x=14-x.[/latex]
[latex]x=-3[/latex]
[/hidden-answer]In this section, we look at rational equations that, after some manipulation, result in a linear equation. If an equation contains at least one rational expression, it is a considered a rational equation.
Recall that a rational number is the ratio of two numbers, such as[latex]\,\frac{2}{3}\,[/latex]or[latex]\,\frac{7}{2}.\,[/latex]A rational expression is the ratio, or quotient, of two polynomials. Here are three examples.
Rational equations have a variable in the denominator in at least one of the terms. Our goal is to perform algebraic operations so that the variables appear in the numerator. In fact, we will eliminate all denominators by multiplying both sides of the equation by the least common denominator (LCD).
Finding the LCD is identifying an expression that contains the highest power of all of the factors in all of the denominators. We do this because when the equation is multiplied by the LCD, the common factors in the LCD and in each denominator will equal one and will cancel out.
Solve the rational equation:[latex]\,\frac{7}{2x}-\frac{5}{3x}=\frac{22}{3}.[/latex]
We have three denominators;[latex]\,2x,3x,[/latex]and 3. The LCD must contain[latex]\,2x,3x,[/latex]and 3. An LCD of[latex]\,6x\,[/latex]contains all three denominators. In other words, each denominator can be divided evenly into the LCD. Next, multiply both sides of the equation by the LCD[latex]\,6x.[/latex]
A common mistake made when solving rational equations involves finding the LCD when one of the denominators is a binomial—two terms added or subtracted—such as[latex]\,\left(x+1\right).\,[/latex]Always consider a binomial as an individual factor—the terms cannot be separated. For example, suppose a problem has three terms and the denominators are[latex]\,x,[/latex][latex]x-1,[/latex]and[latex]\,3x-3.\,[/latex]First, factor all denominators. We then have[latex]\,x,[/latex][latex]\left(x-1\right),[/latex]and[latex]\,3\left(x-1\right)\,[/latex]as the denominators. (Note the parentheses placed around the second denominator.) Only the last two denominators have a common factor of[latex]\,\left(x-1\right).\,[/latex]The[latex]\,x\,[/latex]in the first denominator is separate from the[latex]\,x\,[/latex]in the[latex]\,\left(x-1\right)\,[/latex]denominators. An effective way to remember this is to write factored and binomial denominators in parentheses, and consider each parentheses as a separate unit or a separate factor. The LCD in this instance is found by multiplying together the[latex]\,x,[/latex]one factor of[latex]\,\left(x-1\right),[/latex]and the 3. Thus, the LCD is the following:
So, both sides of the equation would be multiplied by[latex]\,3x\left(x-1\right).\,[/latex]Leave the LCD in factored form, as this makes it easier to see how each denominator in the problem cancels out.
Another example is a problem with two denominators, such as[latex]\,x\,[/latex]and[latex]\,{x}^{2}+2x.\,[/latex]Once the second denominator is factored as[latex]\,{x}^{2}+2x=x\left(x+2\right),[/latex]there is a common factor of x in both denominators and the LCD is[latex]\,x\left(x+2\right).[/latex]Sometimes we have a rational equation in the form of a proportion; that is, when one fraction equals another fraction and there are no other terms in the equation.
We can use another method of solving the equation without finding the LCD: cross-multiplication. We multiply terms by crossing over the equal sign.
Multiply[latex]\,a\left(d\right)\,[/latex]and[latex]\,b\left(c\right),[/latex]which results in[latex]\,ad=bc.[/latex]
Any solution that makes a denominator in the original expression equal zero must be excluded from the possibilities.
A rational equation contains at least one rational expression where the variable appears in at least one of the denominators.
Given a rational equation, solve it.
Solve the following rational equation:
We have three denominators:[latex]\,x,[/latex][latex]2,[/latex]and[latex]\,2x.\,[/latex]No factoring is required. The product of the first two denominators is equal to the third denominator, so, the LCD is[latex]\,2x.\,[/latex]Only one value is excluded from a solution set, 0.[latex][/latex] Next, multiply the whole equation (both sides of the equal sign) by[latex]\,2x.[/latex]
The proposed solution is −1,[latex][/latex] which is not an excluded value, so the solution set contains one number,[latex]\,x=-1,[/latex]or[latex]\,\left\{-1\right\}\,[/latex]written in set notation.[/hidden-answer]
Solve the rational equation:[latex]\,\frac{2}{3x}=\frac{1}{4}-\frac{1}{6x}.[/latex]
[latex]x=\frac{10}{3}[/latex]
[/hidden-answer]Solve the following rational equation:[latex]\,\frac{1}{x}=\frac{1}{10}-\frac{3}{4x}.[/latex]
First find the common denominator. The three denominators in factored form are[latex]\,x,10=2\cdot 5,[/latex]and[latex]\,4x=2\cdot 2\cdot x.\,[/latex]The smallest expression that is divisible by each one of the denominators is[latex]\,20x.\,[/latex]Only[latex]\,x=0\,[/latex]is an excluded value. Multiply the whole equation by[latex]\,20x.[/latex]
The solution is[latex]\,\frac{35}{2}.[/latex][/hidden-answer]
Solve the rational equation:[latex]\,-\frac{5}{2x}+\frac{3}{4x}=-\frac{7}{4}.[/latex]
[latex]x=1[/latex]
[/hidden-answer]Solve the following rational equations and state the excluded values:
The denominators[latex]\,x\,[/latex]and[latex]\,x-6\,[/latex]have nothing in common. Therefore, the LCD is the product[latex]\,x\left(x-6\right).\,[/latex]However, for this problem, we can cross-multiply.
The solution is 15.[latex][/latex] The excluded values are [latex]6[/latex] and [latex]0.[/latex]
The LCD is[latex]\,2\left(x-3\right).\,[/latex]Multiply both sides of the equation by[latex]\,2\left(x-3\right).[/latex]
The solution is[latex]\,\frac{13}{3}.\,[/latex]The excluded value is [latex]3.[/latex]
The least common denominator is[latex]\,2\left(x-2\right).\,[/latex]Multiply both sides of the equation by[latex]\,x\left(x-2\right).[/latex]
The solution is 4. The excluded value is [latex]2.[/latex]
Solve[latex]\,\frac{-3}{2x+1}=\frac{4}{3x+1}.\,[/latex]State the excluded values.
[latex]x=-\frac{7}{17}.\,[/latex]Excluded values are[latex]\,x=-\frac{1}{2}\,[/latex]and[latex]\,x=-\frac{1}{3}.[/latex]
[/hidden-answer]Solve the rational equation after factoring the denominators:[latex]\,\frac{2}{x+1}-\frac{1}{x-1}=\frac{2x}{{x}^{2}-1}.\,[/latex]State the excluded values.
We must factor the denominator[latex]\,{x}^{2}-1.\,[/latex]We recognize this as the difference of squares, and factor it as[latex]\,\left(x-1\right)\left(x+1\right).\,[/latex]Thus, the LCD that contains each denominator is[latex]\,\left(x-1\right)\left(x+1\right).\,[/latex]Multiply the whole equation by the LCD, cancel out the denominators, and solve the remaining equation.
The solution is [latex]\,-3.\,[/latex]The excluded values are [latex]\,1\,[/latex] and [latex]\,-1.[/latex][/hidden-answer]
Solve the rational equation:[latex]\,\frac{2}{x-2}+\frac{1}{x+1}=\frac{1}{{x}^{2}-x-2}.[/latex]
[latex]x=\frac{1}{3}[/latex]
[/hidden-answer]Perhaps the most familiar form of a linear equation is the slope-intercept form, written as[latex]\,y=mx+b,[/latex]where[latex]\,m=\text{slope}\,[/latex]and[latex]\,b=y\text{−intercept}\text{.}\,[/latex]Let us begin with the slope.
The slope of a line refers to the ratio of the vertical change in y over the horizontal change in x between any two points on a line. It indicates the direction in which a line slants as well as its steepness. Slope is sometimes described as rise over run.
If the slope is positive, the line slants to the right. If the slope is negative, the line slants to the left. As the slope increases, the line becomes steeper. Some examples are shown in (Figure). The lines indicate the following slopes:[latex]\,m=-3,[/latex][latex]m=2,[/latex]and[latex]\,m=\frac{1}{3}.[/latex]
The slope of a line, m, represents the change in y over the change in x. Given two points,[latex]\,\left({x}_{1},{y}_{1}\right)\,[/latex]and[latex]\,\left({x}_{2},{y}_{2}\right),[/latex]the following formula determines the slope of a line containing these points:
We substitute the y-values and the x-values into the formula.
It does not matter which point is called[latex]\,\left({x}_{1},{y}_{1}\right)\,[/latex]or[latex]\,\left({x}_{2},{y}_{2}\right).\,[/latex]As long as we are consistent with the order of the y terms and the order of the x terms in the numerator and denominator, the calculation will yield the same result.
Find the slope of the line that passes through the points[latex]\,\left(-2,6\right)\,[/latex]and[latex]\,\left(1,4\right).[/latex]
[latex]m=-\frac{2}{3}[/latex]
[/hidden-answer]As the line is in[latex]\,y=mx+b\,[/latex]form, the given line has a slope of[latex]\,m=-\frac{3}{4}.\,[/latex]The y-intercept is[latex]\,b=-4.[/latex]
[/hidden-answer]The y-intercept is the point at which the line crosses the y-axis. On the y-axis,[latex]\,x=0.\,[/latex]We can always identify the y-intercept when the line is in slope-intercept form, as it will always equal b. Or, just substitute[latex]\,x=0\,[/latex]and solve for y.
Given the slope and one point on a line, we can find the equation of the line using the point-slope formula.
This is an important formula, as it will be used in other areas of college algebra and often in calculus to find the equation of a tangent line. We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form.
Write the equation of the line with slope[latex]\,m=-3\,[/latex]and passing through the point[latex]\,\left(4,8\right).\,[/latex]Write the final equation in slope-intercept form.
Using the point-slope formula, substitute[latex]\,-3\,[/latex]for m and the point[latex]\,\left(4,8\right)\,[/latex]for[latex]\,\left({x}_{1},{y}_{1}\right).[/latex]
Note that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained.
Given[latex]\,m=4,[/latex]find the equation of the line in slope-intercept form passing through the point[latex]\,\left(2,5\right).[/latex]
[latex]y=4x-3[/latex]
[/hidden-answer]Find the equation of the line passing through the points[latex]\,\left(3,4\right)\,[/latex]and[latex]\,\left(0,-3\right).\,[/latex]Write the final equation in slope-intercept form.
First, we calculate the slope using the slope formula and two points.
To prove that either point can be used, let us use the second point[latex]\,\left(0,-3\right)\,[/latex]and see if we get the same equation.
We see that the same line will be obtained using either point. This makes sense because we used both points to calculate the slope.
Another way that we can represent the equation of a line is in standard form. Standard form is given as
where[latex]\,A,[/latex][latex]B,[/latex]and[latex]\,C[/latex]are integers. The x- and y-terms are on one side of the equal sign and the constant term is on the other side.
Find the equation of the line with[latex]\,m=-6\,[/latex]and passing through the point[latex]\,\left(\frac{1}{4},-2\right).\,[/latex]Write the equation in standard form.
We begin using the point-slope formula.
Find the equation of the line in standard form with slope[latex]\,m=-\frac{1}{3}\,[/latex]and passing through the point[latex]\,\left(1,\frac{1}{3}\right).[/latex]
[latex]x+3y=2[/latex]
[/hidden-answer]The equations of vertical and horizontal lines do not require any of the preceding formulas, although we can use the formulas to prove that the equations are correct. The equation of a vertical line is given as
where c is a constant. The slope of a vertical line is undefined, and regardless of the y-value of any point on the line, the x-coordinate of the point will be c.
Suppose that we want to find the equation of a line containing the following points:[latex]\,\left(-3,-5\right),\left(-3,1\right),\left(-3,3\right),[/latex]and[latex]\,\left(-3,5\right).\,[/latex]First, we will find the slope.
Zero in the denominator means that the slope is undefined and, therefore, we cannot use the point-slope formula. However, we can plot the points. Notice that all of the x-coordinates are the same and we find a vertical line through[latex]\,x=-3.\,[/latex]See (Figure).
The equation of a horizontal line is given as
where c is a constant. The slope of a horizontal line is zero, and for any x-value of a point on the line, the y-coordinate will be c.
Suppose we want to find the equation of a line that contains the following set of points:[latex]\,\left(-2,-2\right),\left(0,-2\right),\left(3,-2\right),[/latex]and[latex]\,\left(5,-2\right).[/latex]We can use the point-slope formula. First, we find the slope using any two points on the line.
The graph is a horizontal line through[latex]\,y=-2.\,[/latex]Notice that all of the y-coordinates are the same. See (Figure).
Find the equation of the line passing through the given points:[latex]\,\left(1,-3\right)\,[/latex]and[latex]\,\left(1,4\right).[/latex]
The x-coordinate of both points is 1. Therefore, we have a vertical line,[latex]\,x=1.[/latex]
[/hidden-answer]Find the equation of the line passing through[latex]\,\left(-5,2\right)\,[/latex]and[latex]\,\left(2,2\right).[/latex]
Horizontal line:[latex]\,y=2[/latex]
[/hidden-answer]Parallel lines have the same slope and different y-intercepts. Lines that are parallel to each other will never intersect. For example, (Figure) shows the graphs of various lines with the same slope,[latex]\,m=2.[/latex]
All of the lines shown in the graph are parallel because they have the same slope and different y-intercepts.
Lines that are perpendicular intersect to form a[latex]\,90°[/latex]-angle. The slope of one line is the negative reciprocal of the other. We can show that two lines are perpendicular if the product of the two slopes is[latex]\,-1:{m}_{1}\cdot {m}_{2}=-1.\,[/latex]For example, (Figure) shows the graph of two perpendicular lines. One line has a slope of 3; the other line has a slope of[latex]\,-\frac{1}{3}.[/latex]
Graph the equations of the given lines, and state whether they are parallel, perpendicular, or neither:[latex]\,3y=-4x+3\,[/latex]and[latex]\,3x-4y=8.[/latex]
The first thing we want to do is rewrite the equations so that both equations are in slope-intercept form.
First equation:
Second equation:
See the graph of both lines in (Figure)
From the graph, we can see that the lines appear perpendicular, but we must compare the slopes.
The slopes are negative reciprocals of each other, confirming that the lines are perpendicular.[/hidden-answer]
Graph the two lines and determine whether they are parallel, perpendicular, or neither:[latex]\,2y-x=10\,[/latex]and[latex]\,2y=x+4.[/latex]
Parallel lines: equations are written in slope-intercept form.
[/hidden-answer]As we have learned, determining whether two lines are parallel or perpendicular is a matter of finding the slopes. To write the equation of a line parallel or perpendicular to another line, we follow the same principles as we do for finding the equation of any line. After finding the slope, use the point-slope formula to write the equation of the new line.
Given an equation for a line, write the equation of a line parallel or perpendicular to it.
First, we will write the equation in slope-intercept form to find the slope.
The slope is[latex]\,m=-\frac{5}{3}.\,[/latex]The y-intercept is[latex]\,\frac{1}{3},[/latex]but that really does not enter into our problem, as the only thing we need for two lines to be parallel is the same slope. The one exception is that if the y-intercepts are the same, then the two lines are the same line. The next step is to use this slope and the given point with the point-slope formula.
Find the equation of the line parallel to[latex]\,5x=7+y\,[/latex]and passing through the point[latex]\,\left(-1,-2\right).[/latex]
[latex]y=5x+3[/latex]
[/hidden-answer]Find the equation of the line perpendicular to[latex]\,5x-3y+4=0\,\,\left(-4,1\right).[/latex]
The first step is to write the equation in slope-intercept form.
We see that the slope is[latex]\,m=\frac{5}{3}.\,[/latex]This means that the slope of the line perpendicular to the given line is the negative reciprocal, or [latex]-\frac{3}{5}.\,[/latex]Next, we use the point-slope formula with this new slope and the given point.
Access these online resources for additional instruction and practice with linear equations.
What does it mean when we say that two lines are parallel?
It means they have the same slope.
[/hidden-answer]What is the relationship between the slopes of perpendicular lines (assuming neither is horizontal nor vertical)?
How do we recognize when an equation, for example[latex]\,y=4x+3,[/latex]will be a straight line (linear) when graphed?
The exponent of the[latex]\,x\,[/latex]variable is 1. It is called a first-degree equation.
[/hidden-answer]What does it mean when we say that a linear equation is inconsistent?
When solving the following equation:
[latex]\frac{2}{x-5}=\frac{4}{x+1}[/latex]
explain why we must exclude[latex]\,x=5\,[/latex]and[latex]\,x=-1\,[/latex]as possible solutions from the solution set.
If we insert either value into the equation, they make an expression in the equation undefined (zero in the denominator).
[/hidden-answer]For the following exercises, solve the equation for[latex]\,x.[/latex]
[latex]7x+2=3x-9[/latex]
[latex]4x-3=5[/latex]
[latex]x=2[/latex]
[/hidden-answer][latex]3\left(x+2\right)-12=5\left(x+1\right)[/latex]
[latex]12-5\left(x+3\right)=2x-5[/latex]
[latex]x=\frac{2}{7}[/latex]
[/hidden-answer][latex]\frac{1}{2}-\frac{1}{3}x=\frac{4}{3}[/latex]
[latex]\frac{x}{3}-\frac{3}{4}=\frac{2x+3}{12}[/latex]
[latex]x=6[/latex]
[/hidden-answer][latex]\frac{2}{3}x+\frac{1}{2}=\frac{31}{6}[/latex]
[latex]3\left(2x-1\right)+x=5x+3[/latex]
[latex]x=3[/latex]
[/hidden-answer][latex]\frac{2x}{3}-\frac{3}{4}=\frac{x}{6}+\frac{21}{4}[/latex]
[latex]\frac{x+2}{4}-\frac{x-1}{3}=2[/latex]
[latex]x=-14[/latex]
[/hidden-answer]For the following exercises, solve each rational equation for[latex]\,x.\,[/latex]State all x-values that are excluded from the solution set.
[latex]\frac{3}{x}-\frac{1}{3}=\frac{1}{6}[/latex]
[latex]2-\frac{3}{x+4}=\frac{x+2}{x+4}[/latex]
[latex]x\ne -4;[/latex][latex]x=-3[/latex]
[/hidden-answer][latex]\frac{3}{x-2}=\frac{1}{x-1}+\frac{7}{\left(x-1\right)\left(x-2\right)}[/latex]
[latex]\frac{3x}{x-1}+2=\frac{3}{x-1}[/latex]
[latex]x\ne 1;[/latex]when we solve this we get[latex]\,x=1,[/latex]which is excluded, therefore NO solution
[/hidden-answer][latex]\frac{5}{x+1}+\frac{1}{x-3}=\frac{-6}{{x}^{2}-2x-3}[/latex]
[latex]\frac{1}{x}=\frac{1}{5}+\frac{3}{2x}[/latex]
[latex]x\ne 0;[/latex][latex]x=\frac{-5}{2}[/latex]
[/hidden-answer]For the following exercises, find the equation of the line using the point-slope formula.
Write all the final equations using the slope-intercept form.
[latex]\left(0,3\right)\,[/latex]with a slope of[latex]\,\frac{2}{3}[/latex]
[latex]\left(1,2\right)\,[/latex]with a slope of[latex]\,\frac{-4}{5}[/latex]
[latex]y=\frac{-4}{5}x+\frac{14}{5}[/latex]
[/hidden-answer]x-intercept is 1, and[latex]\,\left(-2,6\right)[/latex]
y-intercept is 2, and[latex]\,\left(4,-1\right)[/latex]
[latex]y=\frac{-3}{4}x+2[/latex]
[/hidden-answer][latex]\left(-3,10\right)\,[/latex]and[latex]\,\left(5,-6\right)[/latex]
[latex]\left(1,3\right)\text{ and }\left(5,5\right)[/latex]
[latex]y=\frac{1}{2}x+\frac{5}{2}[/latex]
[/hidden-answer]parallel to[latex]\,y=2x+5\,[/latex]and passes through the point[latex]\,\left(4,3\right)[/latex]
perpendicular to[latex]\,\text{3}y=x-4\,[/latex]and passes through the point[latex]\,\left(-2,1\right)[/latex].
[latex]y=-3x-5[/latex]
[/hidden-answer]For the following exercises, find the equation of the line using the given information.
[latex]\left(-2,0\right)\,[/latex]and[latex]\,\left(-2,5\right)[/latex]
[latex]\left(1,7\right)\,[/latex]and[latex]\,\left(3,7\right)[/latex]
[latex]y=7[/latex]
[/hidden-answer]The slope is undefined and it passes through the point[latex]\,\left(2,3\right).[/latex]
The slope equals zero and it passes through the point[latex]\,\left(1,-4\right).[/latex]
[latex]y=-4[/latex]
[/hidden-answer]The slope is[latex]\,\frac{3}{4}\,[/latex]and it passes through the point[latex]\,\text{(1,4)}\text{.}[/latex]
[latex]\left(-1,3\right)\,[/latex]and[latex]\,\left(4,-5\right)[/latex]
[latex]8x+5y=7[/latex]
[/hidden-answer]For the following exercises, graph the pair of equations on the same axes, and state whether they are parallel, perpendicular, or neither.
[latex]\begin{array}{l}\\ \begin{array}{l}y=2x+7\hfill \\ y=\frac{-1}{2}x-4\hfill \end{array}\end{array}[/latex]
[latex]\begin{array}{l}3x-2y=5\hfill \\ 6y-9x=6\hfill \end{array}[/latex]
Parallel
[/hidden-answer][latex]\begin{array}{l}y=\frac{3x+1}{4}\hfill \\ y=3x+2\hfill \end{array}[/latex]
[latex]\begin{array}{l}x=4\\ y=-3\end{array}[/latex]
Perpendicular
[/hidden-answer]For the following exercises, find the slope of the line that passes through the given points.
[latex]\left(5,4\right)\,[/latex]and[latex]\,\left(7,9\right)[/latex]
[latex]\left(-3,2\right)\,[/latex]and[latex]\,\left(4,-7\right)[/latex]
[latex]m=\frac{-9}{7}[/latex]
[/hidden-answer][latex]\left(-5,4\right)\,[/latex]and[latex]\,\left(2,4\right)[/latex]
[latex]\left(-1,-2\right)\,[/latex]and[latex]\,\left(3,4\right)[/latex]
[latex]m=\frac{3}{2}[/latex]
[/hidden-answer][latex]\,\left(3,-2\right)[/latex]and[latex]\,\left(3,-2\right)[/latex]
For the following exercises, find the slope of the lines that pass through each pair of points and determine whether the lines are parallel or perpendicular.
[latex]\begin{array}{l}\left(-1,3\right)\text{ and }\left(5,1\right)\\ \left(-2,3\right)\text{ and }\left(0,9\right)\end{array}[/latex]
[latex]{m}_{1}=\frac{-1}{3},\text{ }{m}_{2}=3;\text{ }\text{Perpendicular}\text{.}[/latex]
[/hidden-answer][latex]\begin{array}{l}\left(2,5\right)\text{ and }\left(5,9\right)\\ \left(-1,-1\right)\text{ and }\left(2,3\right)\end{array}[/latex]
For the following exercises, express the equations in slope intercept form (rounding each number to the thousandths place). Enter this into a graphing calculator as Y1, then adjust the ymin and ymax values for your window to include where the y-intercept occurs. State your ymin and ymax values.
[latex]0.537x-2.19y=100[/latex]
[latex]y=0.245x-45.662.\,[/latex]Answers may vary.[latex]\,{y}_{\text{min}}=-50,\text{ }{y}_{\text{max}}=-40[/latex]
[/hidden-answer][latex]4,500x-200y=9,528[/latex]
[latex]\frac{200-30y}{x}=70[/latex]
[latex]y=-2.333x+6.667.\,[/latex]Answers may vary.[latex]\,{y}_{\mathrm{min}}=-10, {y}_{\mathrm{max}}=10[/latex]
[/hidden-answer]Starting with the point-slope formula[latex]\,y-{y}_{1}=m\left(x-{x}_{1}\right),[/latex]solve this expression for[latex]\,x\,[/latex]in terms of[latex]\,{x}_{1},y,{y}_{1},[/latex]and[latex]\,m.[/latex]
Starting with the standard form of an equation[latex]\,\text{A}x\text{ + B}y\text{ = C,}[/latex]solve this expression for y in terms of[latex]\,A,B,C,\,[/latex]and[latex]\,x.\,[/latex]Then put the expression in slope-intercept form.
[latex]y=\frac{-A}{B}x+\frac{C}{B}[/latex]
[/hidden-answer]Use the above derived formula to put the following standard equation in slope intercept form:[latex]\,7x-5y=25.[/latex]
Given that the following coordinates are the vertices of a rectangle, prove that this truly is a rectangle by showing the slopes of the sides that meet are perpendicular.
[latex]\left(-1,1\right),\left(2,0\right),\left(3,3\right)\text{,}[/latex]and[latex]\,\left(0,4\right)[/latex]
[latex]\begin{array}{l}\text{The slope for }\left(-1,1\right)\text{ to }\left(0,4\right)\text{ is }3.\\ \text{The slope for }\left(-1,1\right)\text{ to }\left(2,0\right)\text{ is }\frac{-1}{3}.\\ \text{The slope for }\left(2,0\right)\text{ to }\left(3,3\right)\text{ is }3.\\ \text{The slope for }\left(0,4\right)\text{ to }\left(3,3\right)\text{ is }\frac{-1}{3}.\end{array}[/latex]
Yes they are perpendicular.
[/hidden-answer]Find the slopes of the diagonals in the previous exercise. Are they perpendicular?
The slope for a wheelchair ramp for a home has to be[latex]\,\frac{1}{12}.\,[/latex]If the vertical distance from the ground to the door bottom is 2.5 ft, find the distance the ramp has to extend from the home in order to comply with the needed slope.
30 ft
[/hidden-answer]If the profit equation for a small business selling[latex]\,x\,[/latex]number of item one and[latex]\,y\,[/latex]number of item two is[latex]\,p=3x+4y,[/latex]find the[latex]\,y\,[/latex]value when[latex]\,p=\text{\$}453\text{ and }x=75.[/latex]
For the following exercises, use this scenario: The cost of renting a car is $45/wk plus $0.25/mi traveled during that week. An equation to represent the cost would be[latex]\,y=45+.25x,[/latex]where[latex]\,x\,[/latex]is the number of miles traveled.
What is your cost if you travel 50 mi?
$57.50
[/hidden-answer]If your cost were[latex]\,\text{\$}63.75,[/latex]how many miles were you charged for traveling?
Suppose you have a maximum of $100 to spend for the car rental. What would be the maximum number of miles you could travel?
220 mi
[/hidden-answer]Josh is hoping to get an A in his college algebra class. He has scores of 75, 82, 95, 91, and 94 on his first five tests. Only the final exam remains, and the maximum of points that can be earned is 100. Is it possible for Josh to end the course with an A? A simple linear equation will give Josh his answer.
Many real-world applications can be modeled by linear equations. For example, a cell phone package may include a monthly service fee plus a charge per minute of talk-time; it costs a widget manufacturer a certain amount to produce x widgets per month plus monthly operating charges; a car rental company charges a daily fee plus an amount per mile driven. These are examples of applications we come across every day that are modeled by linear equations. In this section, we will set up and use linear equations to solve such problems.
To set up or model a linear equation to fit a real-world application, we must first determine the known quantities and define the unknown quantity as a variable. Then, we begin to interpret the words as mathematical expressions using mathematical symbols. Let us use the car rental example above. In this case, a known cost, such as $0.10/mi, is multiplied by an unknown quantity, the number of miles driven. Therefore, we can write[latex]\,0.10x.\,[/latex]This expression represents a variable cost because it changes according to the number of miles driven.
If a quantity is independent of a variable, we usually just add or subtract it, according to the problem. As these amounts do not change, we call them fixed costs. Consider a car rental agency that charges $0.10/mi plus a daily fee of $50. We can use these quantities to model an equation that can be used to find the daily car rental cost[latex]\,C.[/latex]
When dealing with real-world applications, there are certain expressions that we can translate directly into math. (Figure) lists some common verbal expressions and their equivalent mathematical expressions.
Verbal | Translation to Math Operations |
---|---|
One number exceeds another by a | [latex]x,\text{}\,x+a[/latex] |
Twice a number | [latex]2x[/latex] |
One number is a more than another number | [latex]x,\text{}\,x+a[/latex] |
One number is a less than twice another number | [latex]x,\,2x-a[/latex] |
The product of a number and a, decreased by b | [latex]ax-b[/latex] |
The quotient of a number and the number plus a is three times the number | [latex]\frac{x}{x+a}=3x[/latex] |
The product of three times a number and the number decreased by b is c | [latex]3x\left(x-b\right)=c[/latex] |
Given a real-world problem, model a linear equation to fit it.
Find a linear equation to solve for the following unknown quantities: One number exceeds another number by[latex]\,17\,[/latex]and their sum is[latex]\,31.\,[/latex]Find the two numbers.
Let[latex]\,x\,[/latex]equal the first number. Then, as the second number exceeds the first by 17, we can write the second number as[latex]\,x+17.\,[/latex]The sum of the two numbers is 31. We usually interpret the word is as an equal sign.
The two numbers are[latex]\,7\,[/latex]and[latex]\,24.[/latex][/hidden-answer]
Find a linear equation to solve for the following unknown quantities: One number is three more than twice another number. If the sum of the two numbers is[latex]\,36,[/latex]find the numbers.
There are two cell phone companies that offer different packages. Company A charges a monthly service fee of $34 plus $.05/min talk-time. Company B charges a monthly service fee of $40 plus $.04/min talk-time.
If the average number of minutes used each month is 1,160, we have the following:
So, Company B offers the lower monthly cost of $86.40 as compared with the $92 monthly cost offered by Company A when the average number of minutes used each month is 1,160.
If the average number of minutes used each month is 420, we have the following:
If the average number of minutes used each month is 420, then Company A offers a lower monthly cost of $55 compared to Company B’s monthly cost of $56.80.
To answer the question of how many talk-time minutes would yield the same bill from both companies, we should think about the problem in terms of[latex]\,\left(x,y\right)\,[/latex]coordinates: At what point are both the x-value and the y-value equal? We can find this point by setting the equations equal to each other and solving for x.
Check the x-value in each equation.
Therefore, a monthly average of 600 talk-time minutes renders the plans equal. See (Figure)
Find a linear equation to model this real-world application: It costs ABC electronics company $2.50 per unit to produce a part used in a popular brand of desktop computers. The company has monthly operating expenses of $350 for utilities and $3,300 for salaries. What are the company’s monthly expenses?
[latex]C=2.5x+3,650[/latex]
[/hidden-answer]Many applications are solved using known formulas. The problem is stated, a formula is identified, the known quantities are substituted into the formula, the equation is solved for the unknown, and the problem’s question is answered. Typically, these problems involve two equations representing two trips, two investments, two areas, and so on. Examples of formulas include the area of a rectangular region,[latex]\,A=LW;[/latex]the perimeter of a rectangle,[latex]\,P=2L+2W;[/latex]and the volume of a rectangular solid,[latex]\,V=LWH.\,[/latex]When there are two unknowns, we find a way to write one in terms of the other because we can solve for only one variable at a time.
It takes Andrew 30 min to drive to work in the morning. He drives home using the same route, but it takes 10 min longer, and he averages 10 mi/h less than in the morning. How far does Andrew drive to work?
This is a distance problem, so we can use the formula[latex]\,d=rt,[/latex]where distance equals rate multiplied by time. Note that when rate is given in mi/h, time must be expressed in hours. Consistent units of measurement are key to obtaining a correct solution.
First, we identify the known and unknown quantities. Andrew’s morning drive to work takes 30 min, or[latex]\,\frac{1}{2}\,[/latex]h at rate[latex]\,r.\,[/latex]His drive home takes 40 min, or[latex]\,\frac{2}{3}\,[/latex]h, and his speed averages 10 mi/h less than the morning drive. Both trips cover distance[latex]\,d.\,[/latex]A table, such as (Figure), is often helpful for keeping track of information in these types of problems.
[latex]d[/latex] | [latex]r[/latex] | [latex]t[/latex] | |
---|---|---|---|
To Work | [latex]d[/latex] | [latex]r[/latex] | [latex]\frac{1}{2}[/latex] |
To Home | [latex]d[/latex] | [latex]r-10[/latex] | [latex]\frac{2}{3}[/latex] |
Write two equations, one for each trip.
As both equations equal the same distance, we set them equal to each other and solve for r.
We have solved for the rate of speed to work, 40 mph. Substituting 40 into the rate on the return trip yields 30 mi/h. Now we can answer the question. Substitute the rate back into either equation and solve for d.
Note that we could have cleared the fractions in the equation by multiplying both sides of the equation by the LCD to solve for[latex]\,r.[/latex]
On Saturday morning, it took Jennifer 3.6 h to drive to her mother’s house for the weekend. On Sunday evening, due to heavy traffic, it took Jennifer 4 h to return home. Her speed was 5 mi/h slower on Sunday than on Saturday. What was her speed on Sunday?
45 mi/h
[/hidden-answer]The perimeter of a rectangular outdoor patio is[latex]\,54\,[/latex]ft. The length is[latex]\,3\,[/latex]ft greater than the width. What are the dimensions of the patio?
The perimeter formula is standard:[latex]\,P=2L+2W.\,[/latex]We have two unknown quantities, length and width. However, we can write the length in terms of the width as[latex]\,L=W+3.\,[/latex]Substitute the perimeter value and the expression for length into the formula. It is often helpful to make a sketch and label the sides as in (Figure).
Now we can solve for the width and then calculate the length.
The dimensions are[latex]\,L=15\,[/latex]ft and[latex]\,W=12\,[/latex]ft.[/hidden-answer]
Find the dimensions of a rectangle given that the perimeter is[latex]\,110\,[/latex]cm and the length is 1 cm more than twice the width.
[latex]L=37\,[/latex]cm,[latex]\,W=18\,[/latex]cm
[/hidden-answer]The perimeter of a tablet of graph paper is 48 in. The length is[latex]\,6\,[/latex]in. more than the width. Find the area of the graph paper.
The standard formula for area is[latex]\,A=LW;[/latex]however, we will solve the problem using the perimeter formula. The reason we use the perimeter formula is because we know enough information about the perimeter that the formula will allow us to solve for one of the unknowns. As both perimeter and area use length and width as dimensions, they are often used together to solve a problem such as this one.
We know that the length is 6 in. more than the width, so we can write length as[latex]\,L=W+6.\,[/latex]Substitute the value of the perimeter and the expression for length into the perimeter formula and find the length.
Now, we find the area given the dimensions of[latex]\,L=15\,[/latex]in. and[latex]\,W=9\,[/latex]in.
The area is[latex]\,135\,[/latex]in.^{2}.[/hidden-answer]
A game room has a perimeter of 70 ft. The length is five more than twice the width. How many ft^{2} of new carpeting should be ordered?
250 ft2
[/hidden-answer]Find the dimensions of a shipping box given that the length is twice the width, the height is[latex]\,8\,[/latex]inches, and the volume is 1,600 in.3.
The formula for the volume of a box is given as[latex]\,V=LWH,[/latex]the product of length, width, and height. We are given that[latex]\,L=2W,[/latex]and[latex]\,H=8.\,[/latex]The volume is[latex]\,1,600\,[/latex]cubic inches.
Note that the square root of[latex]\,{W}^{2}\,[/latex]would result in a positive and a negative value. However, because we are describing width, we can use only the positive result.
Access these online resources for additional instruction and practice with models and applications of linear equations.
To set up a model linear equation to fit real-world applications, what should always be the first step?
Use your own words to describe this equation where n is a number:
[latex]5\left(n+3\right)=2n[/latex]
If the total amount of money you had to invest was $2,000 and you deposit[latex]\,x\,[/latex]amount in one investment, how can you represent the remaining amount?
[latex]2,000-x[/latex]
[/hidden-answer]If a man sawed a 10-ft board into two sections and one section was[latex]\,n\,[/latex]ft long, how long would the other section be in terms of[latex]\,n[/latex]?
If Bill was traveling[latex]\,v\,[/latex]mi/h, how would you represent Daemon’s speed if he was traveling 10 mi/h faster?
[latex]v+10[/latex]
[/hidden-answer]For the following exercises, use the information to find a linear algebraic equation model to use to answer the question being asked.
Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?
Beth and Ann are joking that their combined ages equal Sam’s age. If Beth is twice Ann’s age and Sam is 69 yr old, what are Beth and Ann’s ages?
Ann:[latex]\,23;[/latex]Beth:[latex]\,46[/latex]
[/hidden-answer]Ben originally filled out 8 more applications than Henry. Then each boy filled out 3 additional applications, bringing the total to 28. How many applications did each boy originally fill out?
For the following exercises, use this scenario: Two different telephone carriers offer the following plans that a person is considering. Company A has a monthly fee of $20 and charges of $.05/min for calls. Company B has a monthly fee of $5 and charges $.10/min for calls.
Find the model of the total cost of Company A’s plan, using[latex]\,m\,[/latex]for the minutes.
[latex]20+0.05m[/latex]
[/hidden-answer]Find the model of the total cost of Company B’s plan, using[latex]\,m\,[/latex]for the minutes.
Find out how many minutes of calling would make the two plans equal.
300 min
[/hidden-answer]If the person makes a monthly average of 200 min of calls, which plan should for the person choose?
For the following exercises, use this scenario: A wireless carrier offers the following plans that a person is considering. The Family Plan: $90 monthly fee, unlimited talk and text on up to 8 lines, and data charges of $40 for each device for up to 2 GB of data per device. The Mobile Share Plan: $120 monthly fee for up to 10 devices, unlimited talk and text for all the lines, and data charges of $35 for each device up to a shared total of 10 GB of data. Use[latex]\,P\,[/latex]for the number of devices that need data plans as part of their cost.
Find the model of the total cost of the Family Plan.
[latex]90+40P[/latex]
[/hidden-answer]Find the model of the total cost of the Mobile Share Plan.
Assuming they stay under their data limit, find the number of devices that would make the two plans equal in cost.
6 devices
[/hidden-answer]If a family has 3 smart phones, which plan should they choose?
For exercises 17 and 18, use this scenario: A retired woman has $50,000 to invest but needs to make $6,000 a year from the interest to meet certain living expenses. One bond investment pays 15% annual interest. The rest of it she wants to put in a CD that pays 7%.
If we let[latex]\,x\,[/latex]be the amount the woman invests in the 15% bond, how much will she be able to invest in the CD?
[latex]50,000-x[/latex]
[/hidden-answer]Set up and solve the equation for how much the woman should invest in each option to sustain a $6,000 annual return.
Two planes fly in opposite directions. One travels 450 mi/h and the other 550 mi/h. How long will it take before they are 4,000 mi apart?
4 h
[/hidden-answer]Ben starts walking along a path at 4 mi/h. One and a half hours after Ben leaves, his sister Amanda begins jogging along the same path at 6 mi/h. How long will it be before Amanda catches up to Ben?
Fiora starts riding her bike at 20 mi/h. After a while, she slows down to 12 mi/h, and maintains that speed for the rest of the trip. The whole trip of 70 mi takes her 4.5 h. For what distance did she travel at 20 mi/h?
She traveled for 2 h at 20 mi/h, or 40 miles.
[/hidden-answer]A chemistry teacher needs to mix a 30% salt solution with a 70% salt solution to make 20 qt of a 40% salt solution. How many quarts of each solution should the teacher mix to get the desired result?
Paul has $20,000 to invest. His intent is to earn 11% interest on his investment. He can invest part of his money at 8% interest and part at 12% interest. How much does Paul need to invest in each option to make get a total 11% return on his $20,000?
$5,000 at 8% and $15,000 at 12%
[/hidden-answer]For the following exercises, use this scenario: A truck rental agency offers two kinds of plans. Plan A charges $75/wk plus $.10/mi driven. Plan B charges $100/wk plus $.05/mi driven.
Write the model equation for the cost of renting a truck with plan A.
Write the model equation for the cost of renting a truck with plan B.
[latex]B=100+.05x[/latex]
[/hidden-answer]Find the number of miles that would generate the same cost for both plans.
If Tim knows he has to travel 300 mi, which plan should he choose?
Plan A
[/hidden-answer]For the following exercises, use the given formulas to answer the questions.
[latex]A=P\left(1+rt\right)\,[/latex]is used to find the principal amount Pdeposited, earning r% interest, for t years. Use this to find what principal amount P David invested at a 3% rate for 20 yr if[latex]\,A=\text{\$}8,000.[/latex]
The formula[latex]\,F=\frac{m{v}^{2}}{R}\,[/latex]relates force (F), velocity (v), mass (m), and resistance (R). Find[latex]\,R\,[/latex]when[latex]\,m=45,[/latex][latex]v=7,[/latex]and[latex]\,F=245.[/latex]
[latex]R=9[/latex]
[/hidden-answer][latex]F=ma\,[/latex]indicates that force (F) equals mass (m) times acceleration (a). Find the acceleration of a mass of 50 kg if a force of 12 N is exerted on it.
[latex]Sum=\frac{1}{1-r}\,[/latex]is the formula for an infinite series sum. If the sum is 5, find[latex]\,r.[/latex]
[latex]r=\frac{4}{5}\,[/latex]or 0.8
[/hidden-answer]For the following exercises, solve for the given variable in the formula. After obtaining a new version of the formula, you will use it to solve a question.
Solve for W:[latex]\,P=2L+2W[/latex]
Use the formula from the previous question to find the width,[latex]\,W,[/latex]of a rectangle whose length is 15 and whose perimeter is 58.
[latex]W=\frac{P-2L}{2}=\frac{58-2\left(15\right)}{2}=14[/latex]
[/hidden-answer]Solve for[latex]\,f:\frac{1}{p}+\frac{1}{q}=\frac{1}{f}[/latex]
Use the formula from the previous question to find[latex]\,f\,[/latex]when[latex]\,p=8\,\text{and }q=13.[/latex]
[latex]f=\frac{pq}{p+q}=\frac{8\left(13\right)}{8+13}=\frac{104}{21}[/latex]
[/hidden-answer]Solve for[latex]\,m\,[/latex]in the slope-intercept formula:[latex]\,y=mx+b[/latex]
Use the formula from the previous question to find[latex]\,m\,[/latex]when the coordinates of the point are[latex]\,\left(4,7\right)\,[/latex]and[latex]\,b=12.[/latex]
[latex]m=\frac{-5}{4}[/latex]
[/hidden-answer]The area of a trapezoid is given by[latex]\,A=\frac{1}{2}h\left({b}_{1}+{b}_{2}\right).\,[/latex]Use the formula to find the area of a trapezoid with[latex]\,h=6,\text{ }{b}_{1}=14,\text{ and }{b}_{2}=8.[/latex]
Solve for h:[latex]\,A=\frac{1}{2}h\left({b}_{1}+{b}_{2}\right)[/latex]
[latex]h=\frac{2A}{{b}_{1}+{b}_{2}}[/latex]
[/hidden-answer]Use the formula from the previous question to find the height of a trapezoid with[latex]\,A=150,\text{ }{b}_{1}=19,\text{ and }{b}_{2}=11.[/latex]
Find the dimensions of an American football field. The length is 200 ft more than the width, and the perimeter is 1,040 ft. Find the length and width. Use the perimeter formula[latex]\,P=2L+2W.[/latex]
length = 360 ft; width = 160 ft
[/hidden-answer]Distance equals rate times time,[latex]\,d=rt.\,[/latex]Find the distance Tom travels if he is moving at a rate of 55 mi/h for 3.5 h.
405 mi
[/hidden-answer]What is the total distance that two people travel in 3 h if one of them is riding a bike at 15 mi/h and the other is walking at 3 mi/h?
If the area model for a triangle is[latex]\,A=\frac{1}{2}bh,[/latex]find the area of a triangle with a height of 16 in. and a base of 11 in.
[latex]A=88\text{ in}{.}^{2}[/latex]
[/hidden-answer]Solve for h:[latex]\,A=\frac{1}{2}bh[/latex]
Use the formula from the previous question to find the height to the nearest tenth of a triangle with a base of 15 and an area of 215.
28.7
[/hidden-answer]The volume formula for a cylinder is[latex]\,V=\pi {r}^{2}h.\,[/latex]Using the symbol[latex]\,\pi \,[/latex]in your answer, find the volume of a cylinder with a radius,[latex]\,r,[/latex]of 4 cm and a height of 14 cm.
Solve for h:[latex]\,V=\pi {r}^{2}h[/latex]
[latex]h=\frac{V}{\pi {r}^{2}}[/latex]
[/hidden-answer]Use the formula from the previous question to find the height of a cylinder with a radius of 8 and a volume of[latex]\,16\pi [/latex]
Solve for r:[latex]\,V=\pi {r}^{2}h[/latex]
[latex]r=\sqrt{\frac{V}{\pi h}}[/latex]
[/hidden-answer]Use the formula from the previous question to find the radius of a cylinder with a height of 36 and a volume of[latex]\,324\pi .[/latex]
The formula for the circumference of a circle is[latex]\,C=2\pi r.\,[/latex]Find the circumference of a circle with a diameter of 12 in. (diameter = 2r). Use the symbol[latex]\,\pi \,[/latex]in your final answer.
[latex]C=12\pi [/latex]
[/hidden-answer]Solve the formula from the previous question for[latex]\,\pi .\,[/latex]Notice why[latex]\,\pi \,[/latex]is sometimes defined as the ratio of the circumference to its diameter.
Discovered by Benoit Mandelbrot around 1980, the Mandelbrot Set is one of the most recognizable fractal images. The image is built on the theory of self-similarity and the operation of iteration. Zooming in on a fractal image brings many surprises, particularly in the high level of repetition of detail that appears as magnification increases. The equation that generates this image turns out to be rather simple.
In order to better understand it, we need to become familiar with a new set of numbers. Keep in mind that the study of mathematics continuously builds upon itself. Negative integers, for example, fill a void left by the set of positive integers. The set of rational numbers, in turn, fills a void left by the set of integers. The set of real numbers fills a void left by the set of rational numbers. Not surprisingly, the set of real numbers has voids as well. In this section, we will explore a set of numbers that fills voids in the set of real numbers and find out how to work within it.
We know how to find the square root of any positive real number. In a similar way, we can find the square root of any negative number. The difference is that the root is not real. If the value in the radicand is negative, the root is said to be an imaginary number. The imaginary number[latex]\,i\,[/latex]is defined as the square root of[latex]\,-1.[/latex]
So, using properties of radicals,
We can write the square root of any negative number as a multiple of[latex]\,i.\,[/latex]Consider the square root of [latex]\,-49.[/latex]
We use[latex]\,7i\,[/latex]and not[latex]\,-7i\,[/latex]because the principal root of[latex]\,49\,[/latex]is the positive root.
A complex number is the sum of a real number and an imaginary number. A complex number is expressed in standard form when written[latex]\,a+bi\,[/latex]where[latex]\,a\,[/latex]is the real part and[latex]\,b\,[/latex]is the imaginary part. For example,[latex]\,5+2i\,[/latex]is a complex number. So, too, is[latex]\,3+4i\sqrt{3}.[/latex]
Imaginary numbers differ from real numbers in that a squared imaginary number produces a negative real number. Recall that when a positive real number is squared, the result is a positive real number and when a negative real number is squared, the result is also a positive real number. Complex numbers consist of real and imaginary numbers.A complex number is a number of the form[latex]\,a+bi\,[/latex]where
If[latex]\,b=0,[/latex]then[latex]\,a+bi\,[/latex]is a real number. If[latex]\,a=0\,[/latex]and[latex]\,b\,[/latex]is not equal to 0, the complex number is called a pure imaginary number. An imaginary number is an even root of a negative number.
Given an imaginary number, express it in the standard form of a complex number.
Express[latex]\,\sqrt{-9}\,[/latex]in standard form.
In standard form, this is[latex]\,0+3i.[/latex][/hidden-answer]
Express[latex]\,\sqrt{-24}\,[/latex]in standard form.
[latex]\sqrt{-24}=0+2i\sqrt{6}[/latex]
[/hidden-answer]We cannot plot complex numbers on a number line as we might real numbers. However, we can still represent them graphically. To represent a complex number, we need to address the two components of the number. We use the complex plane, which is a coordinate system in which the horizontal axis represents the real component and the vertical axis represents the imaginary component. Complex numbers are the points on the plane, expressed as ordered pairs[latex]\,\left(a,b\right),[/latex]where[latex]\,a\,[/latex]represents the coordinate for the horizontal axis and[latex]\,b\,[/latex]represents the coordinate for the vertical axis.
Let’s consider the number[latex]\,-2+3i.\,[/latex]The real part of the complex number is[latex]\,-2\,[/latex]and the imaginary part is 3. We plot the ordered pair[latex]\,\left(-2,3\right)\,[/latex]to represent the complex number[latex]\,-2+3i,[/latex]as shown in (Figure).
In the complex plane, the horizontal axis is the real axis, and the vertical axis is the imaginary axis, as shown in (Figure).
Given a complex number, represent its components on the complex plane.
Plot the complex number[latex]\,3-4i\,[/latex]on the complex plane.
The real part of the complex number is[latex]\,3,[/latex]and the imaginary part is –4. We plot the ordered pair[latex]\,\left(3,-4\right)\,[/latex]as shown in (Figure).
Plot the complex number[latex]\,-4-i\,[/latex]on the complex plane.
Just as with real numbers, we can perform arithmetic operations on complex numbers. To add or subtract complex numbers, we combine the real parts and then combine the imaginary parts.
Adding complex numbers:
Subtracting complex numbers:
Given two complex numbers, find the sum or difference.
Add or subtract as indicated.
We add the real parts and add the imaginary parts.
Subtract[latex]\,2+5i\,[/latex]from[latex]\,3–4i.[/latex]
[latex]\left(3-4i\right)-\left(2+5i\right)=1-9i[/latex]
[/hidden-answer]Multiplying complex numbers is much like multiplying binomials. The major difference is that we work with the real and imaginary parts separately.
Lets begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. Consider, for example,[latex]\,3\left(6+2i\right)[/latex]:
Given a complex number and a real number, multiply to find the product.
Find the product[latex]\,4\left(2+5i\right).[/latex]
Distribute the 4.
Find the product:[latex]\,\frac{1}{2}\left(5-2i\right).[/latex]
[latex]\frac{5}{2}-i[/latex]
[/hidden-answer]Now, let’s multiply two complex numbers. We can use either the distributive property or more specifically the FOIL method because we are dealing with binomials. Recall that FOIL is an acronym for multiplying First, Inner, Outer, and Last terms together. The difference with complex numbers is that when we get a squared term,[latex]\,{i}^{2},[/latex]it equals[latex]\,-1.[/latex]
Given two complex numbers, multiply to find the product.
Multiply:[latex]\,\left(3-4i\right)\left(2+3i\right).[/latex]
[latex]18+i[/latex]
[/hidden-answer]Dividing two complex numbers is more complicated than adding, subtracting, or multiplying because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator to write the answer in standard form[latex]\,a+bi.\,[/latex]We need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator so that we end up with a real number as the denominator. This term is called the complex conjugate of the denominator, which is found by changing the sign of the imaginary part of the complex number. In other words, the complex conjugate of[latex]\,a+bi\,[/latex]is[latex]\,a-bi.\,[/latex]For example, the product of[latex]\,a+bi\,[/latex]and[latex]\,a-bi\,[/latex]is
The result is a real number.
Note that complex conjugates have an opposite relationship: The complex conjugate of[latex]\,a+bi\,[/latex]is[latex]\,a-bi,[/latex]and the complex conjugate of[latex]\,a-bi\,[/latex]is[latex]\,a+bi.\,[/latex]Further, when a quadratic equation with real coefficients has complex solutions, the solutions are always complex conjugates of one another.
Suppose we want to divide[latex]\,c+di\,[/latex]by[latex]\,a+bi,[/latex]where neither[latex]\,a\,[/latex]nor[latex]\,b\,[/latex]equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply.
Multiply the numerator and denominator by the complex conjugate of the denominator.
Simplify, remembering that[latex]\,{i}^{2}=-1.[/latex]
The complex conjugate of a complex number[latex]\,a+bi\,[/latex]is[latex]\,a-bi.\,[/latex]It is found by changing the sign of the imaginary part of the complex number. The real part of the number is left unchanged.
Find the complex conjugate of each number.
Find the complex conjugate of[latex]\,-3+4i.[/latex]
[latex]-3-4i[/latex]
[/hidden-answer]Given two complex numbers, divide one by the other.
Divide:[latex]\,\left(2+5i\right)\,[/latex]by[latex]\,\left(4-i\right).[/latex]
We begin by writing the problem as a fraction.
Then we multiply the numerator and denominator by the complex conjugate of the denominator.
To multiply two complex numbers, we expand the product as we would with polynomials (using FOIL).
Note that this expresses the quotient in standard form.[/hidden-answer]
The powers of[latex]\,i\,[/latex]are cyclic. Let’s look at what happens when we raise[latex]\,i\,[/latex]to increasing powers.
We can see that when we get to the fifth power of[latex]\,i,[/latex]it is equal to the first power. As we continue to multiply[latex]\,i\,[/latex] by increasing powers, we will see a cycle of four. Let’s examine the next four powers of[latex]\,i.[/latex]
The cycle is repeated continuously:[latex]\,i,-1,-i,1,[/latex]every four powers.
Evaluate:[latex]\,{i}^{35}.[/latex]
Since[latex]\,{i}^{4}=1,[/latex]we can simplify the problem by factoring out as many factors of[latex]\,{i}^{4}\,[/latex]as possible. To do so, first determine how many times 4 goes into 35:[latex]\,35=4\cdot 8+3.[/latex]
Evaluate:[latex]\,{i}^{18}[/latex]
[latex]-1[/latex]
[/hidden-answer]Can we write[latex]\,{i}^{35}\,[/latex]in other helpful ways?
As we saw in (Figure), we reduced[latex]\,{i}^{35}\,[/latex]to[latex]\,{i}^{3}\,[/latex]by dividing the exponent by 4 and using the remainder to find the simplified form. But perhaps another factorization of[latex]\,{i}^{35}\,[/latex]may be more useful. (Figure) shows some other possible factorizations.
Factorization of[latex]\,{i}^{35}[/latex] | [latex]{i}^{34}\cdot i[/latex] | [latex]{i}^{33}\cdot {i}^{2}[/latex] | [latex]{i}^{31}\cdot {i}^{4}[/latex] | [latex]{i}^{19}\cdot {i}^{16}[/latex] |
Reduced form | [latex]{\left({i}^{2}\right)}^{17}\cdot i[/latex] | [latex]{i}^{33}\cdot \left(-1\right)[/latex] | [latex]{i}^{31}\cdot 1[/latex] | [latex]{i}^{19}\cdot {\left({i}^{4}\right)}^{4}[/latex] |
Simplified form | [latex]{\left(-1\right)}^{17}\cdot i[/latex] | [latex]-{i}^{33}[/latex] | [latex]{i}^{31}[/latex] | [latex]{i}^{19}[/latex] |
Each of these will eventually result in the answer we obtained above but may require several more steps than our earlier method.
Access these online resources for additional instruction and practice with complex numbers.
Explain how to add complex numbers.
What is the basic principle in multiplication of complex numbers?
Give an example to show that the product of two imaginary numbers is not always imaginary.
Possible answer:[latex]\,i\,[/latex]times[latex]\,i\,[/latex]equals -1, which is not imaginary.
[/hidden-answer]What is a characteristic of the plot of a real number in the complex plane?
For the following exercises, evaluate the algebraic expressions.
If[latex]\,y={x}^{2}+x-4,[/latex]evaluate[latex]\,y\,[/latex]given[latex]\,x=2i.[/latex]
[latex]-8+2i[/latex]
[/hidden-answer]If[latex]\,y={x}^{3}-2,[/latex]evaluate[latex]\,y\,[/latex]given[latex]\,x=i.[/latex]
If[latex]\,y={x}^{2}+3x+5,[/latex]evaluate[latex]\,y\,[/latex]given[latex]\,x=2+i.[/latex]
[latex]14+7i[/latex]
[/hidden-answer]If[latex]\,y=2{x}^{2}+x-3,[/latex]evaluate[latex]\,y\,[/latex]given[latex]\,x=2-3i.[/latex]
If[latex]\,y=\frac{x+1}{2-x},[/latex]evaluate[latex]\,y\,[/latex]given[latex]\,x=5i.[/latex]
[latex]-\frac{23}{29}+\frac{15}{29}i[/latex]
[/hidden-answer]If[latex]\,y=\frac{1+2x}{x+3},[/latex]evaluate[latex]\,y\,[/latex]given[latex]\,x=4i.[/latex]
For the following exercises, plot the complex numbers on the complex plane.
[latex]1-2i[/latex]
[latex]-2+3i[/latex]
[latex]i[/latex]
[latex]-3-4i[/latex]
For the following exercises, perform the indicated operation and express the result as a simplified complex number.
[latex]\left(3+2i\right)+\left(5-3i\right)[/latex]
[latex]8-i[/latex]
[/hidden-answer][latex]\left(-2-4i\right)+\left(1+6i\right)[/latex]
[latex]\left(-5+3i\right)-\left(6-i\right)[/latex]
[latex]-11+4i[/latex]
[/hidden-answer][latex]\left(2-3i\right)-\left(3+2i\right)[/latex]
[latex]\left(-4+4i\right)-\left(-6+9i\right)[/latex]
[latex]2-5i[/latex]
[/hidden-answer][latex]\left(2+3i\right)\left(4i\right)[/latex]
[latex]\left(5-2i\right)\left(3i\right)[/latex]
[latex]6+15i[/latex]
[/hidden-answer][latex]\left(6-2i\right)\left(5\right)[/latex]
[latex]\left(-2+4i\right)\left(8\right)[/latex]
[latex]-16+32i[/latex]
[/hidden-answer][latex]\left(2+3i\right)\left(4-i\right)[/latex]
[latex]\left(-1+2i\right)\left(-2+3i\right)[/latex]
[latex]-4-7i[/latex]
[/hidden-answer][latex]\left(4-2i\right)\left(4+2i\right)[/latex]
[latex]\left(3+4i\right)\left(3-4i\right)[/latex]
25
[/hidden-answer][latex]\frac{3+4i}{2}[/latex]
[latex]2-\frac{2}{3}i[/latex]
[/hidden-answer][latex]\frac{-5+3i}{2i}[/latex]
[latex]\frac{6+4i}{i}[/latex]
[latex]4-6i[/latex]
[/hidden-answer][latex]\frac{2-3i}{4+3i}[/latex]
[latex]\frac{3+4i}{2-i}[/latex]
[latex]\frac{2}{5}+\frac{11}{5}i[/latex]
[/hidden-answer][latex]\sqrt{-9}+3\sqrt{-16}[/latex]
[latex]15i[/latex]
[/hidden-answer][latex]-\sqrt{-4}-4\sqrt{-25}[/latex]
[latex]\frac{2+\sqrt{-12}}{2}[/latex]
[latex]1+i\sqrt{3}[/latex]
[/hidden-answer][latex]\frac{4+\sqrt{-20}}{2}[/latex]
[latex]{i}^{8}[/latex]
[latex]1[/latex]
[/hidden-answer][latex]{i}^{15}[/latex]
[latex]{i}^{22}[/latex]
For the following exercises, use a calculator to help answer the questions.
Evaluate[latex]\,{\left(1+i\right)}^{k}\,[/latex]for[latex]\,k=4,8,\text{and}\,12.\,[/latex]Predict the value if[latex]\,k=16.[/latex]
Evaluate[latex]\,{\left(1-i\right)}^{k}\,[/latex]for[latex]\,k=2,6,\text{and}\,10.\,[/latex]Predict the value if[latex]\,k=14.[/latex]
128i
[/hidden-answer]Evaluate[latex]{\left(\text{l}+i\right)}^{k}-{\left(\text{l}-i\right)}^{k}[/latex]for[latex]\,k=4,8,\text{and}\,12.\,[/latex]Predict the value for[latex]\,k=16.[/latex]
Show that a solution of[latex]\,{x}^{6}+1=0\,[/latex]is[latex]\,\frac{\sqrt{3}}{2}+\frac{1}{2}i.[/latex]
[latex]{\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)}^{6}=-1[/latex]
[/hidden-answer]Show that a solution of[latex]\,{x}^{8}-1=0\,[/latex]is[latex]\,\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i.[/latex]
For the following exercises, evaluate the expressions, writing the result as a simplified complex number.
[latex]\frac{1}{i}+\frac{4}{{i}^{3}}[/latex]
[latex]3i[/latex]
[/hidden-answer][latex]\frac{1}{{i}^{11}}-\frac{1}{{i}^{21}}[/latex]
[latex]{i}^{7}\left(1+{i}^{2}\right)[/latex]
0
[/hidden-answer][latex]{i}^{-3}+5{i}^{7}[/latex]
[latex]\frac{\left(2+i\right)\left(4-2i\right)}{\left(1+i\right)}[/latex]
[latex]5-5i[/latex]
[/hidden-answer][latex]\frac{\left(1+3i\right)\left(2-4i\right)}{\left(1+2i\right)}[/latex]
[latex]\frac{{\left(3+i\right)}^{2}}{{\left(1+2i\right)}^{2}}[/latex]
[latex]-2i[/latex]
[/hidden-answer][latex]\frac{3+2i}{2+i}+\left(4+3i\right)[/latex]
[latex]\frac{4+i}{i}+\frac{3-4i}{1-i}[/latex]
[latex]\frac{9}{2}-\frac{9}{2}i[/latex]
[/hidden-answer][latex]\frac{3+2i}{1+2i}-\frac{2-3i}{3+i}[/latex]
The computer monitor on the left in (Figure) is a 23.6-inch model and the one on the right is a 27-inch model. Proportionally, the monitors appear very similar. If there is a limited amount of space and we desire the largest monitor possible, how do we decide which one to choose? In this section, we will learn how to solve problems such as this using four different methods.
An equation containing a second-degree polynomial is called a quadratic equation. For example, equations such as[latex]\,2{x}^{2}+3x-1=0\,[/latex]and[latex]\,{x}^{2}-4=0\,[/latex]are quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological science, and, of course, mathematics.
Often the easiest method of solving a quadratic equation is factoring. Factoring means finding expressions that can be multiplied together to give the expression on one side of the equation.
If a quadratic equation can be factored, it is written as a product of linear terms. Solving by factoring depends on the zero-product property, which states that if[latex]\,a\cdot b=0,[/latex]then[latex]\,a=0\,[/latex]or[latex]\,b=0,[/latex]where a and b are real numbers or algebraic expressions. In other words, if the product of two numbers or two expressions equals zero, then one of the numbers or one of the expressions must equal zero because zero multiplied by anything equals zero.Multiplying the factors expands the equation to a string of terms separated by plus or minus signs. So, in that sense, the operation of multiplication undoes the operation of factoring. For example, expand the factored expression[latex]\,\left(x-2\right)\left(x+3\right)\,[/latex]by multiplying the two factors together.
The product is a quadratic expression. Set equal to zero,[latex]\,{x}^{2}+x-6=0\,[/latex]is a quadratic equation. If we were to factor the equation, we would get back the factors we multiplied.
The process of factoring a quadratic equation depends on the leading coefficient, whether it is 1 or another integer. We will look at both situations; but first, we want to confirm that the equation is written in standard form,[latex]\,a{x}^{2}+bx+c=0,[/latex]where a, b, and c are real numbers, and[latex]\,a\ne 0.\,[/latex]The equation[latex]\,{x}^{2}+x-6=0\,[/latex]is in standard form.
We can use the zero-product property to solve quadratic equations in which we first have to factor out the greatest common factor (GCF), and for equations that have special factoring formulas as well, such as the difference of squares, both of which we will see later in this section.
The zero-product property states
where a and b are real numbers or algebraic expressions.
A quadratic equation is an equation containing a second-degree polynomial; for example
where a, b, and c are real numbers, and if[latex]\,a\ne 0,[/latex]it is in standard form.
In the quadratic equation[latex]\,{x}^{2}+x-6=0,[/latex]the leading coefficient, or the coefficient of[latex]\,{x}^{2},[/latex]is 1. We have one method of factoring quadratic equations in this form.
Given a quadratic equation with the leading coefficient of 1, factor it.
Factor and solve the equation:[latex]\,{x}^{2}+x-6=0.[/latex]
To factor[latex]\,{x}^{2}+x-6=0,[/latex]we look for two numbers whose product equals[latex]\,-6\,[/latex]and whose sum equals 1. Begin by looking at the possible factors of[latex]\,-6.[/latex]
The last pair,[latex]\,3\cdot \left(-2\right)\,[/latex]sums to 1, so these are the numbers. Note that only one pair of numbers will work. Then, write the factors.
To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.
The two solutions are [latex]\,2\,[/latex] and [latex]\,-3.\,[/latex]We can see how the solutions relate to the graph in (Figure). The solutions are the x-intercepts of[latex]\,y={x}^{2}+x-6=0.[/latex]
Factor and solve the quadratic equation:[latex]\,{x}^{2}-5x-6=0.[/latex]
[latex]\left(x-6\right)\left(x+1\right)=0;x=6,x=-1[/latex]
[/hidden-answer]Solve the quadratic equation by factoring:[latex]\,{x}^{2}+8x+15=0.[/latex]
Find two numbers whose product equals[latex]\,15\,[/latex]and whose sum equals[latex]\,8.\,[/latex]List the factors of[latex]\,15.[/latex]
The numbers that add to 8 are 3 and 5. Then, write the factors, set each factor equal to zero, and solve.
The solutions are [latex]\,-3\,[/latex] and [latex]\,-5.[/latex][/hidden-answer]
Solve the quadratic equation by factoring:[latex]\,{x}^{2}-4x-21=0.[/latex]
[latex]\left(x-7\right)\left(x+3\right)=0,[/latex][latex]x=7,[/latex][latex]x=-3.[/latex]
[/hidden-answer]Solve the difference of squares equation using the zero-product property:[latex]\,{x}^{2}-9=0.[/latex]
Recognizing that the equation represents the difference of squares, we can write the two factors by taking the square root of each term, using a minus sign as the operator in one factor and a plus sign as the operator in the other. Solve using the zero-factor property.
The solutions are [latex]\,3\,[/latex] and [latex]\,-3.[/latex][/hidden-answer]
Solve by factoring:[latex]\,{x}^{2}-25=0.[/latex]
[latex]\left(x+5\right)\left(x-5\right)=0,[/latex][latex]x=-5,[/latex][latex]x=5.[/latex]
[/hidden-answer]When the leading coefficient is not 1, we factor a quadratic equation using the method called grouping, which requires four terms. With the equation in standard form, let’s review the grouping procedures:
Use grouping to factor and solve the quadratic equation:[latex]\,4{x}^{2}+15x+9=0.[/latex]
First, multiply[latex]\,ac:4\left(9\right)=36.\,[/latex]Then list the factors of[latex]\,36.[/latex]
The only pair of factors that sums to[latex]\,15\,[/latex]is[latex]\,3+12.\,[/latex]Rewrite the equation replacing the b term,[latex]\,15x,[/latex]with two terms using 3 and 12 as coefficients of x. Factor the first two terms, and then factor the last two terms.
Solve using the zero-product property.
Solve using factoring by grouping:[latex]\,12{x}^{2}+11x+2=0.[/latex]
[latex]\left(3x+2\right)\left(4x+1\right)=0,[/latex][latex]x=-\frac{2}{3},[/latex][latex]x=-\frac{1}{4}[/latex]
[/hidden-answer]Solve the equation by factoring:[latex]\,-3{x}^{3}-5{x}^{2}-2x=0.[/latex]
This equation does not look like a quadratic, as the highest power is 3, not 2. Recall that the first thing we want to do when solving any equation is to factor out the GCF, if one exists. And it does here. We can factor out[latex]\,-x\,[/latex]from all of the terms and then proceed with grouping.
Use grouping on the expression in parentheses.
The solutions are [latex]\,0,[/latex][latex]-\frac{2}{3},[/latex]and[latex]\,-1.[/latex][/hidden-answer]
Solve by factoring:[latex]\,{x}^{3}+11{x}^{2}+10x=0.[/latex]
[latex]x=0,x=-10,x=-1[/latex]
[/hidden-answer]When there is no linear term in the equation, another method of solving a quadratic equation is by using the square root property, in which we isolate the[latex]\,{x}^{2}\,[/latex]term and take the square root of the number on the other side of the equals sign. Keep in mind that sometimes we may have to manipulate the equation to isolate the[latex]\,{x}^{2}\,[/latex]term so that the square root property can be used.
With the[latex]\,{x}^{2}\,[/latex]term isolated, the square root property states that:
where k is a nonzero real number.
Given a quadratic equation with an[latex]\,{x}^{2}\,[/latex]term but no[latex]\,x\,[/latex]term, use the square root property to solve it.
Solve the quadratic using the square root property:[latex]\,{x}^{2}=8.[/latex]
Take the square root of both sides, and then simplify the radical. Remember to use a[latex]\,±\,[/latex]sign before the radical symbol.
The solutions are [latex]\,2\sqrt{2},[/latex][latex]-2\sqrt{2}.[/latex][/hidden-answer]
Solve the quadratic equation:[latex]\,4{x}^{2}+1=\text{7.}[/latex]
First, isolate the[latex]\,{x}^{2}\,[/latex]term. Then take the square root of both sides.
The solutions are [latex]\,\frac{\sqrt{6}}{2},[/latex][latex]\text{and}-\frac{\sqrt{6}}{2}.[/latex][/hidden-answer]
Solve the quadratic equation using the square root property:[latex]\,3{\left(x-4\right)}^{2}=15.[/latex]
[latex]x=4±\sqrt{5}[/latex]
[/hidden-answer]Not all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use a method for solving a quadratic equation known as completing the square. Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, a, must equal 1. If it does not, then divide the entire equation by a. Then, we can use the following procedures to solve a quadratic equation by completing the square.
We will use the example[latex]\,{x}^{2}+4x+1=0\,[/latex]to illustrate each step.
Given a quadratic equation that cannot be factored, and with[latex]\,a=1,[/latex]first add or subtract the constant term to the right sign of the equal sign.
Multiply the b term by[latex]\,\frac{1}{2}\,[/latex]and square it.
Add[latex]\,{\left(\frac{1}{2}b\right)}^{2}\,[/latex]to both sides of the equal sign and simplify the right side. We have
The left side of the equation can now be factored as a perfect square.
Use the square root property and solve.
The solutions are [latex]\,-2+\sqrt{3},[/latex][latex]\text{and}-2-\sqrt{3}.[/latex]
Solve the quadratic equation by completing the square:[latex]\,{x}^{2}-3x-5=0.[/latex]
First, move the constant term to the right side of the equal sign.
Then, take[latex]\,\frac{1}{2}\,[/latex]of the b term and square it.
Add the result to both sides of the equal sign.
Factor the left side as a perfect square and simplify the right side.
Use the square root property and solve.
The solutions are [latex]\,\frac{3}{2}+\frac{\sqrt{29}}{2},[/latex][latex]\text{and}\frac{3}{2}-\frac{\sqrt{29}}{2}.[/latex][/hidden-answer]
Solve by completing the square:[latex]\,{x}^{2}-6x=13.[/latex]
[latex]x=3±\sqrt{22}[/latex]
[/hidden-answer]The fourth method of solving a quadratic equation is by using the quadratic formula, a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number.
We can derive the quadratic formula by completing the square. We will assume that the leading coefficient is positive; if it is negative, we can multiply the equation by[latex]\,-1\,[/latex]and obtain a positive a. Given[latex]\,a{x}^{2}+bx+c=0,[/latex][latex]a\ne 0,[/latex]we will complete the square as follows:
First, move the constant term to the right side of the equal sign:
As we want the leading coefficient to equal 1, divide through by a:
Then, find[latex]\,\frac{1}{2}\,[/latex]of the middle term, and add[latex]\,{\left(\frac{1}{2}\frac{b}{a}\right)}^{2}=\frac{{b}^{2}}{4{a}^{2}}\,[/latex]to both sides of the equal sign:
Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:
Now, use the square root property, which gives
Finally, add[latex]\,-\frac{b}{2a}\,[/latex]to both sides of the equation and combine the terms on the right side. Thus,
Written in standard form,[latex]\,a{x}^{2}+bx+c=0,[/latex]any quadratic equation can be solved using the quadratic formula:
where a, b, and c are real numbers and[latex]\,a\ne 0.[/latex]
Given a quadratic equation, solve it using the quadratic formula
Solve the quadratic equation:[latex]\,{x}^{2}+5x+1=0.[/latex]
Identify the coefficients:[latex]\,a=1,b=5,c=1.\,[/latex]Then use the quadratic formula.
Use the quadratic formula to solve[latex]\,{x}^{2}+x+2=0.[/latex]
First, we identify the coefficients:[latex]\,a=1,b=1,[/latex]and[latex]\,c=2.[/latex]
Substitute these values into the quadratic formula.
The solutions to the equation are [latex]\,\frac{-1+i\sqrt{7}}{2}\,[/latex] and [latex]\,\frac{-1-i\sqrt{7}}{2}\,[/latex] or [latex]\,\frac{-1}{2}+\frac{i\sqrt{7}}{2}\,[/latex] and [latex]\,\frac{-1}{2}-\frac{i\sqrt{7}}{2}.[/latex][/hidden-answer]
Solve the quadratic equation using the quadratic formula:[latex]\,9{x}^{2}+3x-2=0.[/latex]
[latex]x=-\frac{2}{3},[/latex][latex]x=\frac{1}{3}[/latex]
[/hidden-answer]The quadratic formula not only generates the solutions to a quadratic equation, it tells us about the nature of the solutions when we consider the discriminant, or the expression under the radical,[latex]\,{b}^{2}-4ac.\,[/latex]The discriminant tells us whether the solutions are real numbers or complex numbers, and how many solutions of each type to expect. (Figure) relates the value of the discriminant to the solutions of a quadratic equation.
Value of Discriminant | Results |
---|---|
[latex]{b}^{2}-4ac=0[/latex] | One rational solution (double solution) |
[latex]{b}^{2}-4ac>0,[/latex]perfect square | Two rational solutions |
[latex]{b}^{2}-4ac>0,[/latex]not a perfect square | Two irrational solutions |
[latex]{b}^{2}-4ac<0[/latex] | Two complex solutions |
For[latex]\,a{x}^{2}+bx+c=0[/latex], where[latex]\,a[/latex], [latex]b[/latex], and[latex]\,c\,[/latex]are real numbers, the discriminant is the expression under the radical in the quadratic formula:[latex]\,{b}^{2}-4ac.\,[/latex]It tells us whether the solutions are real numbers or complex numbers and how many solutions of each type to expect.
Use the discriminant to find the nature of the solutions to the following quadratic equations:
Calculate the discriminant[latex]\,{b}^{2}-4ac\,[/latex]for each equation and state the expected type of solutions.
[latex]{x}^{2}+4x+4=0[/latex]
[latex]{b}^{2}-4ac={\left(4\right)}^{2}-4\left(1\right)\left(4\right)=0.\,[/latex]There will be one rational double solution.
[latex]8{x}^{2}+14x+3=0[/latex]
[latex]{b}^{2}-4ac={\left(14\right)}^{2}-4\left(8\right)\left(3\right)=100.\,[/latex]As[latex]\,100\,[/latex]is a perfect square, there will be two rational solutions.
[latex]3{x}^{2}-5x-2=0[/latex]
[latex]{b}^{2}-4ac={\left(-5\right)}^{2}-4\left(3\right)\left(-2\right)=49.\,[/latex]As[latex]\,49\,[/latex]is a perfect square, there will be two rational solutions.
[latex]3{x}^{2}-10x+15=0[/latex]
[latex]{b}^{2}-4ac={\left(-10\right)}^{2}-4\left(3\right)\left(15\right)=-80.\,[/latex]There will be two complex solutions.
One of the most famous formulas in mathematics is the Pythagorean Theorem. It is based on a right triangle, and states the relationship among the lengths of the sides as[latex]\,{a}^{2}+{b}^{2}={c}^{2},[/latex]where[latex]\,a\,[/latex]and[latex]\,b\,[/latex]refer to the legs of a right triangle adjacent to the[latex]\,90°\,[/latex]angle, and[latex]\,c\,[/latex]refers to the hypotenuse. It has immeasurable uses in architecture, engineering, the sciences, geometry, trigonometry, and algebra, and in everyday applications.
We use the Pythagorean Theorem to solve for the length of one side of a triangle when we have the lengths of the other two. Because each of the terms is squared in the theorem, when we are solving for a side of a triangle, we have a quadratic equation. We can use the methods for solving quadratic equations that we learned in this section to solve for the missing side.
The Pythagorean Theorem is given as
where[latex]\,a\,[/latex]and[latex]\,b\,[/latex]refer to the legs of a right triangle adjacent to the[latex]\,{90}^{\circ }\,[/latex]angle, and[latex]\,c\,[/latex]refers to the hypotenuse, as shown in (Figure).
Find the length of the missing side of the right triangle in (Figure).
As we have measurements for side b and the hypotenuse, the missing side is a.
Use the Pythagorean Theorem to solve the right triangle problem: Leg a measures 4 units, leg b measures 3 units. Find the length of the hypotenuse.
[latex]5\,[/latex]units
[/hidden-answer]Access these online resources for additional instruction and practice with quadratic equations.
quadratic formula | [latex]x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}[/latex] |
How do we recognize when an equation is quadratic?
It is a second-degree equation (the highest variable exponent is 2).
[/hidden-answer]When we solve a quadratic equation, how many solutions should we always start out seeking? Explain why when solving a quadratic equation in the form[latex]\,a{x}^{2}+bx+c=0\,[/latex]we may graph the equation[latex]\,y=a{x}^{2}+bx+c\,[/latex]and have no zeroes (x-intercepts).
When we solve a quadratic equation by factoring, why do we move all terms to one side, having zero on the other side?
We want to take advantage of the zero property of multiplication in the fact that if[latex]\,a\cdot b=0\,[/latex]then it must follow that each factor separately offers a solution to the product being zero:[latex]\,a=0\text{ }or\text{ b}=0.[/latex]
[/hidden-answer]In the quadratic formula, what is the name of the expression under the radical sign[latex]\,{b}^{2}-4ac,[/latex]and how does it determine the number of and nature of our solutions?
Describe two scenarios where using the square root property to solve a quadratic equation would be the most efficient method.
One, when no linear term is present (no x term), such as[latex]\,{x}^{2}=16.\,[/latex]Two, when the equation is already in the form[latex]\,{\left(ax+b\right)}^{2}=d.[/latex]
[/hidden-answer]For the following exercises, solve the quadratic equation by factoring.
[latex]{x}^{2}+4x-21=0[/latex]
[latex]{x}^{2}-9x+18=0[/latex]
[latex]x=6,[/latex][latex]x=3[/latex]
[/hidden-answer][latex]2{x}^{2}+9x-5=0[/latex]
[latex]6{x}^{2}+17x+5=0[/latex]
[latex]x=\frac{-5}{2},[/latex][latex]x=\frac{-1}{3}[/latex]
[/hidden-answer][latex]4{x}^{2}-12x+8=0[/latex]
[latex]3{x}^{2}-75=0[/latex]
[latex]x=5,[/latex][latex]x=-5[/latex]
[/hidden-answer][latex]8{x}^{2}+6x-9=0[/latex]
[latex]4{x}^{2}=9[/latex]
[latex]x=\frac{-3}{2},[/latex][latex]x=\frac{3}{2}[/latex]
[/hidden-answer][latex]2{x}^{2}+14x=36[/latex]
[latex]x=-2,3[/latex]
[/hidden-answer][latex]4{x}^{2}=5x[/latex]
[latex]7{x}^{2}+3x=0[/latex]
[latex]x=0,[/latex][latex]x=\frac{-3}{7}[/latex]
[/hidden-answer][latex]\frac{x}{3}-\frac{9}{x}=2[/latex]
For the following exercises, solve the quadratic equation by using the square root property.
[latex]{x}^{2}=36[/latex]
[latex]x=-6,[/latex][latex]x=6[/latex]
[/hidden-answer][latex]{x}^{2}=49[/latex]
[latex]{\left(x-1\right)}^{2}=25[/latex]
[latex]x=6,[/latex][latex]x=-4[/latex]
[/hidden-answer][latex]{\left(x-3\right)}^{2}=7[/latex]
[latex]{\left(2x+1\right)}^{2}=9[/latex]
[latex]x=1,[/latex][latex]x=-2[/latex]
[/hidden-answer][latex]{\left(x-5\right)}^{2}=4[/latex]
For the following exercises, solve the quadratic equation by completing the square. Show each step.
[latex]{x}^{2}-9x-22=0[/latex]
[latex]x=-2,[/latex][latex]x=11[/latex]
[/hidden-answer][latex]2{x}^{2}-8x-5=0[/latex]
[latex]{x}^{2}-6x=13[/latex]
[latex]x=3±\sqrt{22}[/latex]
[/hidden-answer][latex]{x}^{2}+\frac{2}{3}x-\frac{1}{3}=0[/latex]
[latex]2+z=6{z}^{2}[/latex]
[latex]z=\frac{2}{3},[/latex][latex]z=-\frac{1}{2}[/latex]
[/hidden-answer][latex]6{p}^{2}+7p-20=0[/latex]
[latex]2{x}^{2}-3x-1=0[/latex]
[latex]x=\frac{3±\sqrt{17}}{4}[/latex]
[/hidden-answer]For the following exercises, determine the discriminant, and then state how many solutions there are and the nature of the solutions. Do not solve.
[latex]2{x}^{2}-6x+7=0[/latex]
[latex]{x}^{2}+4x+7=0[/latex]
Not real
[/hidden-answer][latex]3{x}^{2}+5x-8=0[/latex]
[latex]9{x}^{2}-30x+25=0[/latex]
One rational
[/hidden-answer][latex]2{x}^{2}-3x-7=0[/latex]
[latex]6{x}^{2}-x-2=0[/latex]
Two real; rational
[/hidden-answer]For the following exercises, solve the quadratic equation by using the quadratic formula. If the solutions are not real, state No Real Solution.
[latex]2{x}^{2}+5x+3=0[/latex]
[latex]{x}^{2}+x=4[/latex]
[latex]x=\frac{-1±\sqrt{17}}{2}[/latex]
[/hidden-answer][latex]2{x}^{2}-8x-5=0[/latex]
[latex]3{x}^{2}-5x+1=0[/latex]
[latex]x=\frac{5±\sqrt{13}}{6}[/latex]
[/hidden-answer][latex]{x}^{2}+4x+2=0[/latex]
[latex]4+\frac{1}{x}-\frac{1}{{x}^{2}}=0[/latex]
[latex]x=\frac{-1±\sqrt{17}}{8}[/latex]
[/hidden-answer]For the following exercises, enter the expressions into your graphing utility and find the zeroes to the equation (the x-intercepts) by using 2^{nd} CALC 2:zero. Recall finding zeroes will ask left bound (move your cursor to the left of the zero,enter), then right bound (move your cursor to the right of the zero,enter), then guess (move your cursor between the bounds near the zero, enter). Round your answers to the nearest thousandth.
[latex]{\text{Y}}_{1}=4{x}^{2}+3x-2[/latex]
[latex]{\text{Y}}_{1}=-3{x}^{2}+8x-1[/latex]
[latex]x\approx 0.131\,[/latex]and[latex]\,x\approx 2.535[/latex]
[/hidden-answer][latex]{\text{Y}}_{1}=0.5{x}^{2}+x-7[/latex]
To solve the quadratic equation[latex]\,{x}^{2}+5x-7=4,[/latex]we can graph these two equations
[latex]\begin{array}{l}\hfill \\ \begin{array}{l}{\text{Y}}_{1}={x}^{2}+5x-7\hfill \\ {\text{Y}}_{2}=4\hfill \end{array}\hfill \end{array}[/latex]
and find the points of intersection. Recall 2^{nd} CALC 5:intersection. Do this and find the solutions to the nearest tenth.
[latex]x\approx -6.7\,[/latex]and[latex]\,x\approx 1.7[/latex]
[/hidden-answer]To solve the quadratic equation[latex]\,0.3{x}^{2}+2x-4=2,[/latex]we can graph these two equations
[latex]\begin{array}{l}\hfill \\ \begin{array}{l}{\text{Y}}_{1}=0.3{x}^{2}+2x-4\hfill \\ {\text{Y}}_{2}=2\hfill \end{array}\hfill \end{array}[/latex]
and find the points of intersection. Recall 2^{nd} CALC 5:intersection. Do this and find the solutions to the nearest tenth.
Beginning with the general form of a quadratic equation,[latex]\,a{x}^{2}+bx+c=0,[/latex]solve for x by using the completing the square method, thus deriving the quadratic formula.
[latex]\begin{array}{ccc}\hfill a{x}^{2}+bx+c& =& 0\hfill \\ \hfill {x}^{2}+\frac{b}{a}x& =& \frac{-c}{a}\hfill \\ \hfill {x}^{2}+\frac{b}{a}x+\frac{{b}^{2}}{4{a}^{2}}& =& \frac{-c}{a}+\frac{b}{4{a}^{2}}\hfill \\ \hfill {\left(x+\frac{b}{2a}\right)}^{2}& =& \frac{{b}^{2}-4ac}{4{a}^{2}}\hfill \\ \hfill x+\frac{b}{2a}& =& ±\sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}\hfill \\ \hfill x& =& \frac{-b±\sqrt{{b}^{2}-4ac}}{2a}\hfill \end{array}[/latex]
[/hidden-answer]Show that the sum of the two solutions to the quadratic equation is[latex]\,\frac{-b}{a}.[/latex]
A person has a garden that has a length 10 feet longer than the width. Set up a quadratic equation to find the dimensions of the garden if its area is 119 ft.^{2}. Solve the quadratic equation to find the length and width.
[latex]x\left(x+10\right)=119;[/latex]7 ft. and 17 ft.
[/hidden-answer]Abercrombie and Fitch stock had a price given as[latex]\,P=0.2{t}^{2}-5.6t+50.2,[/latex]where[latex]\,t\,[/latex]is the time in months from 1999 to 2001. ([latex]\,t=1\,[/latex]is January 1999). Find the two months in which the price of the stock was $30.
Suppose that an equation is given[latex]\,p=-2{x}^{2}+280x-1000,[/latex]where[latex]\,x\,[/latex]represents the number of items sold at an auction and[latex]\,p\,[/latex]is the profit made by the business that ran the auction. How many items sold would make this profit a maximum? Solve this by graphing the expression in your graphing utility and finding the maximum using 2^{nd} CALC maximum. To obtain a good window for the curve, set[latex]\,x\,[/latex][0,200] and[latex]\,y\,[/latex][0,10000].
maximum at[latex]\,x=70[/latex]
[/hidden-answer]A formula for the normal systolic blood pressure for a man age[latex]\,A,[/latex]measured in mmHg, is given as[latex]\,P=0.006{A}^{2}-0.02A+120.\,[/latex]Find the age to the nearest year of a man whose normal blood pressure measures 125 mmHg.
The cost function for a certain company is[latex]\,C=60x+300\,[/latex]and the revenue is given by[latex]\,R=100x-0.5{x}^{2}.\,[/latex]Recall that profit is revenue minus cost. Set up a quadratic equation and find two values of x (production level) that will create a profit of $300.
The quadratic equation would be[latex]\,\left(100x-0.5{x}^{2}\right)-\left(60x+300\right)=300.\,[/latex]The two values of[latex]\,x\,[/latex]are 20 and 60.
[/hidden-answer]A falling object travels a distance given by the formula[latex]\,d=5t+16{t}^{2}\,[/latex]ft, where[latex]\,t\,[/latex]is measured in seconds. How long will it take for the object to traveled 74 ft?
A vacant lot is being converted into a community garden. The garden and the walkway around its perimeter have an area of 378 ft^{2}. Find the width of the walkway if the garden is 12 ft. wide by 15 ft. long.
3 feet
[/hidden-answer]An epidemiological study of the spread of a certain influenza strain that hit a small school population found that the total number of students,[latex]\,P,[/latex]who contracted the flu[latex]\,t\,[/latex]days after it broke out is given by the model[latex]\,P=-{t}^{2}+13t+130,[/latex]where[latex]\,1\le t\le 6.\,[/latex]Find the day that 160 students had the flu. Recall that the restriction on[latex]\,t\,[/latex]is at most 6.
We have solved linear equations, rational equations, and quadratic equations using several methods. However, there are many other types of equations, and we will investigate a few more types in this section. We will look at equations involving rational exponents, polynomial equations, radical equations, absolute value equations, equations in quadratic form, and some rational equations that can be transformed into quadratics. Solving any equation, however, employs the same basic algebraic rules. We will learn some new techniques as they apply to certain equations, but the algebra never changes.
Rational exponents are exponents that are fractions, where the numerator is a power and the denominator is a root. For example,[latex]\,{16}^{\frac{1}{2}}\,[/latex]is another way of writing[latex]\,\sqrt{16};[/latex][latex]{8}^{\frac{1}{3}}\,[/latex]is another way of writing[latex]\text{}\,\sqrt[3]{8}.\,[/latex]The ability to work with rational exponents is a useful skill, as it is highly applicable in calculus.
We can solve equations in which a variable is raised to a rational exponent by raising both sides of the equation to the reciprocal of the exponent. The reason we raise the equation to the reciprocal of the exponent is because we want to eliminate the exponent on the variable term, and a number multiplied by its reciprocal equals 1. For example,[latex]\,\frac{2}{3}\left(\frac{3}{2}\right)=1,[/latex][latex]3\left(\frac{1}{3}\right)=1,[/latex]and so on.
A rational exponent indicates a power in the numerator and a root in the denominator. There are multiple ways of writing an expression, a variable, or a number with a rational exponent:
Evaluate[latex]\,{8}^{\frac{2}{3}}.[/latex]
Whether we take the root first or the power first depends on the number. It is easy to find the cube root of 8, so rewrite[latex]\,{8}^{\frac{2}{3}}\,[/latex]as[latex]\,{\left({8}^{\frac{1}{3}}\right)}^{2}.[/latex]
Evaluate[latex]\,{64}^{-\frac{1}{3}}.[/latex]
[latex]\frac{1}{4}[/latex]
[/hidden-answer]Solve the equation in which a variable is raised to a rational exponent:[latex]\,{x}^{\frac{5}{4}}=32.[/latex]
The way to remove the exponent on x is by raising both sides of the equation to a power that is the reciprocal of[latex]\,\frac{5}{4},[/latex]which is[latex]\,\frac{4}{5}.[/latex]
Solve the equation[latex]\,{x}^{\frac{3}{2}}=125.[/latex]
Solve[latex]\,3{x}^{\frac{3}{4}}={x}^{\frac{1}{2}}.[/latex]
Now, it looks like we should factor the left side, but what do we factor out? We can always factor the term with the lowest exponent. Rewrite[latex]\,{x}^{\frac{1}{2}}\,[/latex]as[latex]\,{x}^{\frac{2}{4}}.\,[/latex]Then, factor out[latex]\,{x}^{\frac{2}{4}}\,[/latex]from both terms on the left.
Where did[latex]\,{x}^{\frac{1}{4}}\,[/latex]come from? Remember, when we multiply two numbers with the same base, we add the exponents. Therefore, if we multiply[latex]\,{x}^{\frac{2}{4}}\,[/latex]back in using the distributive property, we get the expression we had before the factoring, which is what should happen. We need an exponent such that when added to[latex]\,\frac{2}{4}\,[/latex]equals[latex]\,\frac{3}{4}.\,[/latex]Thus, the exponent on x in the parentheses is[latex]\,\frac{1}{4}.\,[/latex]
Let us continue. Now we have two factors and can use the zero factor theorem.
The two solutions are [latex]\,0[/latex] and [latex]\frac{1}{81}.[/latex][/hidden-answer]
Solve:[latex]\,{\left(x+5\right)}^{\frac{3}{2}}=8.[/latex]
[latex]\left\{-1\right\}[/latex]
[/hidden-answer]We have used factoring to solve quadratic equations, but it is a technique that we can use with many types of polynomial equations, which are equations that contain a string of terms including numerical coefficients and variables. When we are faced with an equation containing polynomials of degree higher than 2, we can often solve them by factoring.
A polynomial of degree n is an expression of the type
where n is a positive integer and[latex]\,{a}_{n},\dots ,{a}_{0}\,[/latex]are real numbers and[latex]\,{a}_{n}\ne 0.[/latex]
Setting the polynomial equal to zero gives a polynomial equation. The total number of solutions (real and complex) to a polynomial equation is equal to the highest exponent n.
Solve the polynomial by factoring:[latex]\,5{x}^{4}=80{x}^{2}.[/latex]
Notice that we have the difference of squares in the factor[latex]\,{x}^{2}-16,[/latex]which we will continue to factor and obtain two solutions. The first term,[latex]\,5{x}^{2},[/latex]generates, technically, two solutions as the exponent is 2, but they are the same solution.
The solutions are [latex]\,0\text{ (double solution),}[/latex][latex]4,[/latex] and [latex]\,-4.[/latex][/hidden-answer]
We can see the solutions on the graph in (Figure). The x-coordinates of the points where the graph crosses the x-axis are the solutions—the x-intercepts. Notice on the graph that at the solution[latex]\,0,[/latex]the graph touches the x-axis and bounces back. It does not cross the x-axis. This is typical of double solutions.
Solve by factoring:[latex]\,12{x}^{4}=3{x}^{2}.[/latex]
[latex]x=0,[/latex][latex]x=\frac{1}{2},[/latex][latex]x=-\frac{1}{2}[/latex]
[/hidden-answer]Solve a polynomial by grouping:[latex]\,{x}^{3}+{x}^{2}-9x-9=0.[/latex]
This polynomial consists of 4 terms, which we can solve by grouping. Grouping procedures require factoring the first two terms and then factoring the last two terms. If the factors in the parentheses are identical, we can continue the process and solve, unless more factoring is suggested.
The grouping process ends here, as we can factor[latex]\,{x}^{2}-9\,[/latex] using the difference of squares formula.
The solutions are [latex]3,[/latex][latex]-3,[/latex] and [latex]\,-1.\,[/latex]Note that the highest exponent is 3 and we obtained 3 solutions. We can see the solutions, the x-intercepts, on the graph in (Figure).[/hidden-answer]
We looked at solving quadratic equations by factoring when the leading coefficient is 1. When the leading coefficient is not 1, we solved by grouping. Grouping requires four terms, which we obtained by splitting the linear term of quadratic equations. We can also use grouping for some polynomials of degree higher than 2, as we saw here, since there were already four terms.
Radical equations are equations that contain variables in the radicand (the expression under a radical symbol), such as
Radical equations may have one or more radical terms, and are solved by eliminating each radical, one at a time. We have to be careful when solving radical equations, as it is not unusual to find extraneous solutions, roots that are not, in fact, solutions to the equation. These solutions are not due to a mistake in the solving method, but result from the process of raising both sides of an equation to a power. However, checking each answer in the original equation will confirm the true solutions.
An equation containing terms with a variable in the radicand is called a radical equation.
Given a radical equation, solve it.
Solve[latex]\,\sqrt{15-2x}=x.[/latex]
The radical is already isolated on the left side of the equal side, so proceed to square both sides.
We see that the remaining equation is a quadratic. Set it equal to zero and solve.
The proposed solutions are [latex]-5\,[/latex] and [latex]3.\,[/latex]Let us check each solution back in the original equation. First, check[latex]\,x=-5.[/latex]
This is an extraneous solution. While no mistake was made solving the equation, we found a solution that does not satisfy the original equation.
Check[latex]\,x=3.[/latex]
Solve the radical equation:[latex]\,\sqrt{x+3}=3x-1[/latex]
[latex]x=1;[/latex]extraneous solution[latex]\,x=-\frac{2}{9}[/latex]
[/hidden-answer]Solve[latex]\,\sqrt{2x+3}+\sqrt{x-2}=4.[/latex]
As this equation contains two radicals, we isolate one radical, eliminate it, and then isolate the second radical.
Use the perfect square formula to expand the right side:[latex]\,{\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}.[/latex]
Now that both radicals have been eliminated, set the quadratic equal to zero and solve.
The proposed solutions are [latex]\,3\,[/latex] and [latex]\,83.\,[/latex]Check each solution in the original equation.
One solution is [latex]\,3.[/latex]
Check[latex]\,x=83.[/latex]
The only solution is [latex]\,3.\,[/latex]We see that[latex]\,x=83\,[/latex]is an extraneous solution.[/hidden-answer]
Solve the equation with two radicals:[latex]\,\sqrt{3x+7}+\sqrt{x+2}=1.[/latex]
[latex]x=-2;[/latex]extraneous solution[latex]\,x=-1[/latex]
[/hidden-answer]Next, we will learn how to solve an absolute value equation. To solve an equation such as[latex]\,|2x-6|=8,[/latex]we notice that the absolute value will be equal to 8 if the quantity inside the absolute value bars is[latex]\,8\,[/latex]or[latex]\,-8.\,[/latex]This leads to two different equations we can solve independently.
Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify numbers or points on a line that are at a specified distance from a given reference point.
The absolute value of x is written as[latex]\,|x|.\,[/latex]It has the following properties:
For real numbers[latex]\,A\,[/latex]and[latex]\,B,[/latex]an equation of the form[latex]\,|A|=B,[/latex]with[latex]\,B\ge 0,[/latex]will have solutions when[latex]\,A=B\,[/latex]or[latex]\,A=-B.\,[/latex]If[latex]\,B<0,[/latex]the equation[latex]\,|A|=B\,[/latex]has no solution.
An absolute value equation in the form[latex]\,|ax+b|=c\,[/latex]has the following properties:
Given an absolute value equation, solve it.
Solve the following absolute value equations:
(a) [latex]|6x+4|=8[/latex]
Write two equations and solve each:
The two solutions are [latex]\,\frac{2}{3}[/latex] and [latex]-2.[/latex]
(b) [latex]|3x+4|=-9[/latex]
There is no solution as an absolute value cannot be negative.
(c) [latex]|3x-5|-4=6[/latex]
Isolate the absolute value expression and then write two equations.
There are two solutions: [latex]\,5,[/latex] and [latex]-\frac{5}{3}.[/latex]
(d) [latex]|-5x+10|=0[/latex]
The equation is set equal to zero, so we have to write only one equation.There is one solution: [latex]\,2.[/latex][/hidden-answer]
Solve the absolute value equation:[latex]|1-4x|+8=13.[/latex]
[latex]x=-1,[/latex][latex]x=\frac{3}{2}[/latex]
[/hidden-answer]There are many other types of equations in addition to the ones we have discussed so far. We will see more of them throughout the text. Here, we will discuss equations that are in quadratic form, and rational equations that result in a quadratic.
Equations in quadratic form are equations with three terms. The first term has a power other than 2. The middle term has an exponent that is one-half the exponent of the leading term. The third term is a constant. We can solve equations in this form as if they were quadratic. A few examples of these equations include[latex]\,{x}^{4}-5{x}^{2}+4=0,{x}^{6}+7{x}^{3}-8=0,[/latex]and[latex]\,{x}^{\frac{2}{3}}+4{x}^{\frac{1}{3}}+2=0.\,[/latex]In each one, doubling the exponent of the middle term equals the exponent on the leading term. We can solve these equations by substituting a variable for the middle term.
If the exponent on the middle term is one-half of the exponent on the leading term, we have an equation in quadratic form, which we can solve as if it were a quadratic. We substitute a variable for the middle term to solve equations in quadratic form.
Given an equation quadratic in form, solve it.
Solve this fourth-degree equation:[latex]\,3{x}^{4}-2{x}^{2}-1=0.[/latex]
This equation fits the main criteria, that the power on the leading term is double the power on the middle term. Next, we will make a substitution for the variable term in the middle. Let[latex]\,u={x}^{2}.[/latex]Rewrite the equation in u.
Now solve the quadratic.
The solutions are [latex]\,±i\sqrt{\frac{1}{3}}\,[/latex] and [latex]\,±1.[/latex][/hidden-answer]
Solve using substitution:[latex]\,{x}^{4}-8{x}^{2}-9=0.[/latex]
[latex]x=-3,3,-i,i[/latex]
[/hidden-answer]Solve the equation in quadratic form:[latex]\,{\left(x+2\right)}^{2}+11\left(x+2\right)-12=0.[/latex]
This equation contains a binomial in place of the single variable. The tendency is to expand what is presented. However, recognizing that it fits the criteria for being in quadratic form makes all the difference in the solving process. First, make a substitution, letting[latex]\,u=x+2.\,[/latex]Then rewrite the equation in u.
Solve using the zero-factor property and then replace u with the original expression.
The second factor results in
We have two solutions: [latex]\,-14,[/latex] and [latex]-1.[/latex][/hidden-answer]
Solve:[latex]\,{\left(x-5\right)}^{2}-4\left(x-5\right)-21=0.[/latex]
[latex]x=2,x=12[/latex]
[/hidden-answer]Earlier, we solved rational equations. Sometimes, solving a rational equation results in a quadratic. When this happens, we continue the solution by simplifying the quadratic equation by one of the methods we have seen. It may turn out that there is no solution.
Solve the following rational equation:[latex]\,\frac{-4x}{x-1}+\frac{4}{x+1}=\frac{-8}{{x}^{2}-1}.[/latex]
We want all denominators in factored form to find the LCD. Two of the denominators cannot be factored further. However,[latex]\,{x}^{2}-1=\left(x+1\right)\left(x-1\right).\,[/latex]Then, the LCD is[latex]\,\left(x+1\right)\left(x-1\right).\,[/latex]Next, we multiply the whole equation by the LCD.
In this case, either solution produces a zero in the denominator in the original equation. Thus, there is no solution.[/hidden-answer]
Solve[latex]\,\frac{3x+2}{x-2}+\frac{1}{x}=\frac{-2}{{x}^{2}-2x}.[/latex]
[latex]x=-1,[/latex][latex]x=0[/latex]is not a solution.
[/hidden-answer]Access these online resources for additional instruction and practice with different types of equations.
In a radical equation, what does it mean if a number is an extraneous solution?
This is not a solution to the radical equation, it is a value obtained from squaring both sides and thus changing the signs of an equation which has caused it not to be a solution in the original equation.
[/hidden-answer]Explain why possible solutions must be checked in radical equations.
Your friend tries to calculate the value[latex]\,-{9}^{\frac{3}{2}}[/latex]and keeps getting an ERROR message. What mistake is he or she probably making?
He or she is probably trying to enter negative 9, but taking the square root of[latex]\,-9\,[/latex]is not a real number. The negative sign is in front of this, so your friend should be taking the square root of 9, cubing it, and then putting the negative sign in front, resulting in[latex]\,-27.[/latex]
[/hidden-answer]Explain why[latex]\,|2x+5|=-7\,[/latex]has no solutions.
Explain how to change a rational exponent into the correct radical expression.
A rational exponent is a fraction: the denominator of the fraction is the root or index number and the numerator is the power to which it is raised.
[/hidden-answer]For the following exercises, solve the rational exponent equation. Use factoring where necessary.
[latex]{x}^{\frac{2}{3}}=16[/latex]
[latex]{x}^{\frac{3}{4}}=27[/latex]
[latex]x=81[/latex]
[/hidden-answer][latex]2{x}^{\frac{1}{2}}-{x}^{\frac{1}{4}}=0[/latex]
[latex]{\left(x-1\right)}^{\frac{3}{4}}=8[/latex]
[latex]x=17[/latex]
[/hidden-answer][latex]{\left(x+1\right)}^{\frac{2}{3}}=4[/latex]
[latex]{x}^{\frac{2}{3}}-5{x}^{\frac{1}{3}}+6=0[/latex]
[latex]x=8, x=27[/latex]
[/hidden-answer][latex]{x}^{\frac{7}{3}}-3{x}^{\frac{4}{3}}-4{x}^{\frac{1}{3}}=0[/latex]
For the following exercises, solve the following polynomial equations by grouping and factoring.
[latex]{x}^{3}+2{x}^{2}-x-2=0[/latex]
[latex]x=-2,1,-1[/latex]
[/hidden-answer][latex]3{x}^{3}-6{x}^{2}-27x+54=0[/latex]
[latex]4{y}^{3}-9y=0[/latex]
[latex]y=0, \frac{3}{2}, \frac{-3}{2}[/latex]
[/hidden-answer][latex]{x}^{3}+3{x}^{2}-25x-75=0[/latex]
[latex]{m}^{3}+{m}^{2}-m-1=0[/latex]
[latex]m=1,-1[/latex]
[/hidden-answer][latex]2{x}^{5}-14{x}^{3}=0[/latex]
[latex]5{x}^{3}+45x=2{x}^{2}+18[/latex]
[latex]x=\frac{2}{5},±3i[/latex]
[/hidden-answer]For the following exercises, solve the radical equation. Be sure to check all solutions to eliminate extraneous solutions.
[latex]\sqrt{3x-1}-2=0[/latex]
[latex]\sqrt{x-7}=5[/latex]
[latex]x=32[/latex]
[/hidden-answer][latex]\sqrt{x-1}=x-7[/latex]
[latex]\sqrt{3t+5}=7[/latex]
[latex]t=\frac{44}{3}[/latex]
[/hidden-answer][latex]\sqrt{t+1}+9=7[/latex]
[latex]\sqrt{12-x}=x[/latex]
[latex]x=3[/latex]
[/hidden-answer][latex]\sqrt{2x+3}-\sqrt{x+2}=2[/latex]
[latex]\sqrt{3x+7}+\sqrt{x+2}=1[/latex]
[latex]x=-2[/latex]
[/hidden-answer][latex]\sqrt{2x+3}-\sqrt{x+1}=1[/latex]
For the following exercises, solve the equation involving absolute value.
[latex]|3x-4|=8[/latex]
[latex]x=4,\frac{-4}{3}[/latex]
[/hidden-answer][latex]|2x-3|=-2[/latex]
[latex]|1-4x|-1=5[/latex]
[latex]x=\frac{-5}{4},\frac{7}{4}[/latex]
[/hidden-answer][latex]|4x+1|-3=6[/latex]
[latex]|2x-1|-7=-2[/latex]
[latex]x=3,-2[/latex]
[/hidden-answer][latex]|2x+1|-2=-3[/latex]
[latex]|x+5|=0[/latex]
[latex]x=-5[/latex]
[/hidden-answer][latex]-|2x+1|=-3[/latex]
For the following exercises, solve the equation by identifying the quadratic form. Use a substitute variable and find all real solutions by factoring.
[latex]{x}^{4}-10{x}^{2}+9=0[/latex]
[latex]x=1,-1,3,-3[/latex]
[/hidden-answer][latex]4{\left(t-1\right)}^{2}-9\left(t-1\right)=-2[/latex]
[latex]{\left({x}^{2}-1\right)}^{2}+\left({x}^{2}-1\right)-12=0[/latex]
[latex]x=2,-2[/latex]
[/hidden-answer][latex]{\left(x+1\right)}^{2}-8\left(x+1\right)-9=0[/latex]
[latex]{\left(x-3\right)}^{2}-4=0[/latex]
[latex]x=1,5[/latex]
[/hidden-answer]For the following exercises, solve for the unknown variable.
[latex]{x}^{-2}-{x}^{-1}-12=0[/latex]
[latex]\sqrt{{|x|}^{2}}=x[/latex]
All real numbers
[/hidden-answer][latex]{t}^{10}-{t}^{5}+1=0[/latex]
[latex]|{x}^{2}+2x-36|=12[/latex]
[latex]x=4,6,-6,-8[/latex]
[/hidden-answer]For the following exercises, use the model for the period of a pendulum,[latex]\,T,[/latex]such that[latex]\,T=2\pi \sqrt{\frac{L}{g}},[/latex]where the length of the pendulum is L and the acceleration due to gravity is[latex]\,g.[/latex]
If the acceleration due to gravity is 9.8 m/s^{2} and the period equals 1 s, find the length to the nearest cm (100 cm = 1 m).
If the gravity is 32 ft/s^{2} and the period equals 1 s, find the length to the nearest in. (12 in. = 1 ft). Round your answer to the nearest in.
10 in.
[/hidden-answer]For the following exercises, use a model for body surface area, BSA, such that[latex]\,BSA=\sqrt{\frac{wh}{3600}},[/latex]where w = weight in kg and h = height in cm.
Find the height of a 72-kg female to the nearest cm whose[latex]\,BSA=1.8.[/latex]
Find the weight of a 177-cm male to the nearest kg whose[latex]\,BSA=2.1.[/latex]
90 kg
[/hidden-answer]It is not easy to make the honor role at most top universities. Suppose students were required to carry a course load of at least 12 credit hours and maintain a grade point average of 3.5 or above. How could these honor roll requirements be expressed mathematically? In this section, we will explore various ways to express different sets of numbers, inequalities, and absolute value inequalities.
Indicating the solution to an inequality such as[latex]\,x\ge 4\,[/latex]can be achieved in several ways.
We can use a number line as shown in (Figure). The blue ray begins at[latex]\,x=4\,[/latex]and, as indicated by the arrowhead, continues to infinity, which illustrates that the solution set includes all real numbers greater than or equal to 4.
We can use set-builder notation:[latex]\,\left\{x|x\ge 4\right\},[/latex]which translates to “all real numbers x such that x is greater than or equal to 4.” Notice that braces are used to indicate a set.The third method is interval notation, in which solution sets are indicated with parentheses or brackets. The solutions to[latex]\,x\ge 4\,[/latex]are represented as[latex]\,\left[4,\infty \right).\,[/latex]This is perhaps the most useful method, as it applies to concepts studied later in this course and to other higher-level math courses.
The main concept to remember is that parentheses represent solutions greater or less than the number, and brackets represent solutions that are greater than or equal to or less than or equal to the number. Use parentheses to represent infinity or negative infinity, since positive and negative infinity are not numbers in the usual sense of the word and, therefore, cannot be “equaled.” A few examples of an interval, or a set of numbers in which a solution falls, are[latex]\,\left[-2,6\right),[/latex]or all numbers between[latex]\,-2\,[/latex]and[latex]\,6,[/latex]including[latex]\,-2,[/latex]but not including[latex]\,6;[/latex][latex]\left(-1,0\right),[/latex]all real numbers between, but not including[latex]\,-1\,[/latex]and[latex]\,0;[/latex]and[latex]\,\left(-\infty ,1\right],[/latex]all real numbers less than and including[latex]\,1.\,[/latex](Figure) outlines the possibilities.
Set Indicated | Set-Builder Notation | Interval Notation |
---|---|---|
All real numbers between a and b, but not including a or b | [latex]\left\{x|a<x<b\right\}[/latex] | [latex]\left(a,b\right)[/latex] |
All real numbers greater than a, but not including a | [latex]\left\{x|x>a\right\}[/latex] | [latex]\left(a,\infty \right)[/latex] |
All real numbers less than b, but not including b | [latex]\left\{x|x<b\right\}[/latex] | [latex]\left(-\infty ,b\right)[/latex] |
All real numbers greater than a, including a | [latex]\left\{x|x\ge a\right\}[/latex] | [latex]\left[a,\infty \right)[/latex] |
All real numbers less than b, including b | [latex]\left\{x|x\le b\right\}[/latex] | [latex]\left(-\infty ,b\right][/latex] |
All real numbers between a and b, including a | [latex]\left\{x|a\le x<b\right\}[/latex] | [latex]\left[a,b\right)[/latex] |
All real numbers between a and b, including b | [latex]\left\{x|a<x\le b\right\}[/latex] | [latex]\left(a,b\right][/latex] |
All real numbers between a and b, including a and b | [latex]\left\{x|a\le x\le b\right\}[/latex] | [latex]\left[a,b\right][/latex] |
All real numbers less than a or greater than b | [latex]\left\{x|x<a\,\text{or}\,x>b\right\}[/latex] | [latex]\left(-\infty ,a\right)\cup \left(b,\infty \right)[/latex] |
All real numbers | [latex]\left\{x|x\text{ is all real numbers}\right\}[/latex] | [latex]\left(-\infty ,\infty \right)[/latex] |
Use interval notation to indicate all real numbers greater than or equal to[latex]\,-2.[/latex]
Use a bracket on the left of[latex]\,-2\,[/latex]and parentheses after infinity:[latex]\,\left[-2,\infty \right).[/latex]The bracket indicates that[latex]\,-2\,[/latex]is included in the set with all real numbers greater than[latex]\,-2\,[/latex]to infinity.
[/hidden-answer]Use interval notation to indicate all real numbers between and including[latex]\,-3\,[/latex]and[latex]\,5.[/latex]
[latex]\left[-3,5\right][/latex]
[/hidden-answer]Write the interval expressing all real numbers less than or equal to[latex]\,-1\,[/latex]or greater than or equal to[latex]\,1.[/latex]
We have to write two intervals for this example. The first interval must indicate all real numbers less than or equal to 1. So, this interval begins at[latex]\,-\infty \,[/latex]and ends at[latex]\,-1,[/latex]which is written as[latex]\,\left(-\infty ,-1\right].[/latex]
The second interval must show all real numbers greater than or equal to[latex]\,1,[/latex]which is written as[latex]\,\left[1,\infty \right).\,[/latex]However, we want to combine these two sets. We accomplish this by inserting the union symbol,[latex]\cup ,[/latex]between the two intervals.
Express all real numbers less than[latex]\,-2\,[/latex]or greater than or equal to 3 in interval notation.
[latex]\left(-\infty ,-2\right)\cup \left[3,\infty \right)[/latex]
[/hidden-answer]When we work with inequalities, we can usually treat them similarly to but not exactly as we treat equalities. We can use the addition property and the multiplication property to help us solve them. The one exception is when we multiply or divide by a negative number; doing so reverses the inequality symbol.
These properties also apply to[latex]\,a\le b,[/latex][latex]a>b,[/latex]and[latex]\,a\ge b.[/latex]
Illustrate the addition property for inequalities by solving each of the following:
The addition property for inequalities states that if an inequality exists, adding or subtracting the same number on both sides does not change the inequality.
Solve:[latex]\,3x-2<1.[/latex]
[latex]x<1[/latex]
[/hidden-answer]Illustrate the multiplication property for inequalities by solving each of the following:
Solve:[latex]\,4x+7\ge 2x-3.[/latex]
[latex]x\ge -5[/latex]
[/hidden-answer]As the examples have shown, we can perform the same operations on both sides of an inequality, just as we do with equations; we combine like terms and perform operations. To solve, we isolate the variable.
Solve the inequality:[latex]\,13-7x\ge 10x-4.[/latex]
Solving this inequality is similar to solving an equation up until the last step.
Solve the inequality and write the answer using interval notation:[latex]\,-x+4<\frac{1}{2}x+1.[/latex]
[latex]\left(2,\infty \right)[/latex]
[/hidden-answer]We begin solving in the same way we do when solving an equation.
The solution set is the interval[latex]\,\left(-\infty ,\frac{15}{34}\right].[/latex][/hidden-answer]
Solve the inequality and write the answer in interval notation:[latex]\,-\frac{5}{6}x\le \frac{3}{4}+\frac{8}{3}x.[/latex]
[latex]\left[-\frac{3}{14},\infty \right)[/latex]
[/hidden-answer]A compound inequality includes two inequalities in one statement. A statement such as[latex]\,4<x\le 6\,[/latex]means[latex]\,4<x\,[/latex]and[latex]\,x\le 6.\,[/latex]There are two ways to solve compound inequalities: separating them into two separate inequalities or leaving the compound inequality intact and performing operations on all three parts at the same time. We will illustrate both methods.
Solve the compound inequality:[latex]\,3\le 2x+2<6.[/latex]
The first method is to write two separate inequalities:[latex]\,3\le 2x+2\,[/latex]and[latex]\,2x+2<6.\,[/latex]We solve them independently.
Then, we can rewrite the solution as a compound inequality, the same way the problem began.
In interval notation, the solution is written as[latex]\,\left[\frac{1}{2},2\right).[/latex]
The second method is to leave the compound inequality intact, and perform solving procedures on the three parts at the same time.
We get the same solution:[latex]\,\left[\frac{1}{2},2\right).[/latex][/hidden-answer]
Solve the compound inequality:[latex]\,4<2x-8\le 10.[/latex]
[latex]6<x\le 9\text{}\text{}\,\,\text{or}\,\,\left(6,9\right][/latex]
[/hidden-answer]Solve the compound inequality with variables in all three parts:[latex]\,3+x>7x-2>5x-10.[/latex]
Solve the compound inequality:[latex]\,3y<4-5y<5+3y.[/latex]
[latex]\left(-\frac{1}{8},\frac{1}{2}\right)[/latex]
[/hidden-answer]As we know, the absolute value of a quantity is a positive number or zero. From the origin, a point located at[latex]\,\left(-x,0\right)\,[/latex]has an absolute value of[latex]\,x,[/latex]as it is x units away. Consider absolute value as the distance from one point to another point. Regardless of direction, positive or negative, the distance between the two points is represented as a positive number or zero.
An absolute value inequality is an equation of the form
Where A, and sometimes B, represents an algebraic expression dependent on a variable x. Solving the inequality means finding the set of all [latex]\,x[/latex]-values that satisfy the problem. Usually this set will be an interval or the union of two intervals and will include a range of values.
There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two equations. The advantage of the algebraic approach is that solutions are exact, as precise solutions are sometimes difficult to read from a graph.
Suppose we want to know all possible returns on an investment if we could earn some amount of money within $200 of $600. We can solve algebraically for the set of x-values such that the distance between[latex]\,x\,[/latex]and 600 is less than 200. We represent the distance between[latex]\,x\,[/latex]and 600 as[latex]\,|x-600|,[/latex]and therefore,[latex]\,|x-600|\le 200\,[/latex]orThis means our returns would be between $400 and $800.
To solve absolute value inequalities, just as with absolute value equations, we write two inequalities and then solve them independently.
For an algebraic expression X, and[latex]\,k>0,[/latex]an absolute value inequality is an inequality of the form
These statements also apply to[latex]\,|X|\le k\,[/latex]and[latex]\,|X|\ge k.[/latex]
Describe all values[latex]\,x\,[/latex]within a distance of 4 from the number 5.
The distance from[latex]\,x\,[/latex]to 5 can be represented using an absolute value symbol,[latex]\,|x-5|.\,[/latex]Write the values of[latex]\,x\,[/latex]that satisfy the condition as an absolute value inequality.
We need to write two inequalities as there are always two solutions to an absolute value equation.
If the solution set is[latex]\,x\le 9\,[/latex]and[latex]\,x\ge 1,[/latex]then the solution set is an interval including all real numbers between and including 1 and 9.
So[latex]\,|x-5|\le 4\,[/latex]is equivalent to[latex]\,\left[1,9\right]\,[/latex]in interval notation.[/hidden-answer]
Describe all x-values within a distance of 3 from the number 2.
[latex]|x-2|\le 3[/latex]
[/hidden-answer]Solve [latex]|x-1|\le 3[/latex].
Given the equation [latex]y=-\frac{1}{2}|4x-5|+3,[/latex]determine the x-values for which the y-values are negative.
We are trying to determine where[latex]\,y<0,[/latex]which is when[latex]\,-\frac{1}{2}|4x-5|+3<0.\,[/latex]We begin by isolating the absolute value.
Next, we solve for the equality [latex]|4x-5|=6.[/latex]
Solve[latex]\,-2|k-4|\le -6.[/latex]
[latex]k\le 1\,[/latex]or[latex]\,k\ge 7;[/latex]in interval notation, this would be[latex]\,\left(-\infty ,1\right]\cup \left[7,\infty \right).[/latex]
[/hidden-answer]Access these online resources for additional instruction and practice with linear inequalities and absolute value inequalities.
When solving an inequality, explain what happened from Step 1 to Step 2:
[latex]\begin{array}{ll}\text{Step 1}\hfill & \phantom{\rule{2em}{0ex}}-2x>6\hfill \\ \text{Step 2}\hfill & \phantom{\rule{3em}{0ex}}x<-3\hfill \end{array}[/latex]
When we divide both sides by a negative it changes the sign of both sides so the sense of the inequality sign changes.
[/hidden-answer]When solving an inequality, we arrive at:
[latex]\begin{array}{l}x+2<x+3\hfill \\ \phantom{\rule{1.2em}{0ex}}2<3\hfill \end{array}[/latex]
Explain what our solution set is.
When writing our solution in interval notation, how do we represent all the real numbers?
[latex]\left(-\infty ,\infty \right)[/latex]
[/hidden-answer]When solving an inequality, we arrive at:
[latex]\begin{array}{l}x+2>x+3\hfill \\ \phantom{\rule{1.2em}{0ex}}2>3\hfill \end{array}[/latex]
Explain what our solution set is.
Describe how to graph[latex]\,y=|x-3|[/latex]
We start by finding the x-intercept, or where the function = 0. Once we have that point, which is[latex]\,\left(3,0\right),[/latex]we graph to the right the straight line graph[latex]\,y=x-3,[/latex]and then when we draw it to the left we plot positive y values, taking the absolute value of them.
[/hidden-answer]For the following exercises, solve the inequality. Write your final answer in interval notation.
[latex]4x-7\le 9[/latex]
[latex]3x+2\ge 7x-1[/latex]
[latex]\left(-\infty ,\frac{3}{4}\right][/latex]
[/hidden-answer][latex]-2x+3>x-5[/latex]
[latex]4\left(x+3\right)\ge 2x-1[/latex]
[latex]\left[\frac{-13}{2},\infty \right)[/latex]
[/hidden-answer][latex]-\frac{1}{2}x\le \frac{-5}{4}+\frac{2}{5}x[/latex]
[latex]-5\left(x-1\right)+3>3x-4-4x[/latex]
[latex]\left(-\infty ,3\right)[/latex]
[/hidden-answer][latex]-3\left(2x+1\right)>-2\left(x+4\right)[/latex]
[latex]\frac{x+3}{8}-\frac{x+5}{5}\ge \frac{3}{10}[/latex]
[latex]\left(-\infty ,-\frac{37}{3}\right][/latex]
[/hidden-answer][latex]\frac{x-1}{3}+\frac{x+2}{5}\le \frac{3}{5}[/latex]
For the following exercises, solve the inequality involving absolute value. Write your final answer in interval notation.
[latex]|x+9|\ge -6[/latex]
All real numbers[latex]\,\left(-\infty ,\infty \right)[/latex]
[/hidden-answer][latex]|2x+3|<7[/latex]
[latex]|3x-1|>11[/latex]
[latex]\left(-\infty ,\frac{-10}{3}\right)\cup \left(4,\infty \right)[/latex]
[/hidden-answer][latex]|2x+1|+1\le 6[/latex]
[latex]|x-2|+4\ge 10[/latex]
[latex]\left(-\infty ,-4\right]\cup \left[8,+\infty \right)[/latex]
[/hidden-answer][latex]|-2x+7|\le 13[/latex]
[latex]|x-7|<-4[/latex]
No solution
[/hidden-answer][latex]|x-20|>-1[/latex]
[latex]|\frac{x-3}{4}|<2[/latex]
[latex]\left(-5,11\right)[/latex]
[/hidden-answer]For the following exercises, describe all the x-values within or including a distance of the given values.
Distance of 5 units from the number 7
Distance of 3 units from the number 9
[latex]\left[6,12\right][/latex]
[/hidden-answer]Distance of10 units from the number 4
Distance of 11 units from the number 1
[latex]\left[-10,12\right][/latex]
[/hidden-answer]For the following exercises, solve the compound inequality. Express your answer using inequality signs, and then write your answer using interval notation.
[latex]-4<3x+2\le 18[/latex]
[latex]3x+1>2x-5>x-7[/latex]
[latex]\begin{array}{ll}x>-6\text{ and }x>-2\hfill & \phantom{\rule{2em}{0ex}}\text{Take the intersection of two sets}.\hfill \\ x>-2,\text{ }\left(-2,+\infty \right)\hfill & \hfill \end{array}[/latex]
[/hidden-answer][latex]3y<5-2y<7+y[/latex]
[latex]2x-5<-11\text{ or }5x+1\ge 6[/latex]
[latex]\begin{array}{ll}x<-3\text{ }\mathrm{or}\text{ }x\ge 1\hfill & \phantom{\rule{2em}{0ex}}\text{Take the union of the two sets}.\hfill \\ \left(-\infty ,-3\right){{\cup }^{\text{}}}^{\text{}}\left[1,\infty \right)\hfill & \hfill \end{array}[/latex]
[/hidden-answer][latex]x+7<x+2[/latex]
For the following exercises, graph the function. Observe the points of intersection and shade the x-axis representing the solution set to the inequality. Show your graph and write your final answer in interval notation.
[latex]|x-1|>2[/latex]
[latex]\left(-\infty ,-1\right)\cup \left(3,\infty \right)[/latex]
[/hidden-answer][latex]|x+3|\ge 5[/latex]
[latex]|x+7|\le 4[/latex]
[latex]\left[-11,-3\right][/latex]
[/hidden-answer][latex]|x-2|<7[/latex]
[latex]|x-2|<0[/latex]
It is never less than zero. No solution.
[/hidden-answer]For the following exercises, graph both straight lines (left-hand side being y1 and right-hand side being y2) on the same axes. Find the point of intersection and solve the inequality by observing where it is true comparing the y-values of the lines.
[latex]x+3<3x-4[/latex]
[latex]x-2>2x+1[/latex]
Where the blue line is above the orange line; point of intersection is[latex]\,x=-3.[/latex]
[latex]\left(-\infty ,-3\right)[/latex]
[/hidden-answer][latex]x+1>x+4[/latex]
[latex]\frac{1}{2}x+1>\frac{1}{2}x-5[/latex]
Where the blue line is above the orange line; always. All real numbers.
[latex]\left(-\infty ,-\infty \right)[/latex]
[/hidden-answer][latex]4x+1<\frac{1}{2}x+3[/latex]
For the following exercises, write the set in interval notation.
[latex]\left\{x|-1<x<3\right\}[/latex]
[latex]\left(-1,3\right)[/latex]
[/hidden-answer][latex]\left\{x|x\ge 7\right\}[/latex]
[latex]\left\{x|x<4\right\}[/latex]
[latex]\left(-\infty ,4\right)[/latex]
[/hidden-answer][latex]\left\{\,x|\,x\text{ is all real numbers}\right\}[/latex]
For the following exercises, write the interval in set-builder notation.
[latex]\left(-\infty ,6\right)[/latex]
[latex]\left\{x|x<6\right\}[/latex]
[/hidden-answer][latex]\left(4,+\infty \right)[/latex]
[latex]\left[-3,5\right)[/latex]
[latex]\left\{x|-3\le x<5\right\}[/latex]
[/hidden-answer][latex]\left[-4,1\right]\cup \left[9,\infty \right)[/latex]
For the following exercises, write the set of numbers represented on the number line in interval notation.
[latex]\left(-2,1\right][/latex]
[/hidden-answer][latex]\left(-\infty ,4\right][/latex]
[/hidden-answer]For the following exercises, input the left-hand side of the inequality as a Y1 graph in your graphing utility. Enter y2 = the right-hand side. Entering the absolute value of an expression is found in the MATH menu, Num, 1:abs(. Find the points of intersection, recall (2^{nd} CALC 5:intersection, 1^{st} curve, enter, 2^{nd} curve, enter, guess, enter). Copy a sketch of the graph and shade the x-axis for your solution set to the inequality. Write final answers in interval notation.
[latex]|x+2|-5<2[/latex]
[latex]\frac{-1}{2}|x+2|<4[/latex]
Where the blue is below the orange; always. All real numbers.[latex]\,\left(-\infty ,+\infty \right).[/latex]
[/hidden-answer][latex]|4x+1|-3>2[/latex]
[latex]|x-4|<3[/latex]
Where the blue is below the orange;[latex]\,\left(1,7\right).[/latex]
[/hidden-answer][latex]|x+2|\ge 5[/latex]
Solve[latex]\,|3x+1|=|2x+3|[/latex]
[latex]x=2,\frac{-4}{5}[/latex]
[/hidden-answer]Solve[latex]{x}^{2}-x>12[/latex]
[latex]\frac{x-5}{x+7}\le 0,[/latex][latex]x\ne -7[/latex]
[latex]\left(-7,5\right][/latex]
[/hidden-answer][latex]p=-{x}^{2}+130x-3000\,[/latex]is a profit formula for a small business. Find the set of x-values that will keep this profit positive.
In chemistry the volume for a certain gas is given by[latex]\,V=20T,[/latex]where V is measured in cc and T is temperature in ºC. If the temperature varies between 80ºC and 120ºC, find the set of volume values.
[latex]\begin{array}{l}80\le T\le 120\\ 1,600\le 20T\le 2,400\end{array}[/latex]
[latex]\left[1,600, 2,400\right][/latex]
[/hidden-answer]A basic cellular package costs $20/mo. for 60 min of calling, with an additional charge of $.30/min beyond that time.. The cost formula would be[latex]\,C=\text{\$}20+.30\left(x-60\right).\,[/latex]If you have to keep your bill lower than $50, what is the maximum calling minutes you can use?
For the following exercises, find the x-intercept and the y-intercept without graphing.
[latex]4x-3y=12[/latex]
x-intercept:[latex]\,\left(3,0\right);[/latex]y-intercept:[latex]\,\left(0,-4\right)[/latex]
[/hidden-answer][latex]2y-4=3x[/latex]
For the following exercises, solve for y in terms of x, putting the equation in slope–intercept form.
[latex]5x=3y-12[/latex]
[latex]y=\frac{5}{3}x+4[/latex]
[/hidden-answer][latex]2x-5y=7[/latex]
For the following exercises, find the distance between the two points.
[latex]\left(-2,5\right)\left(4,-1\right)[/latex]
[latex]\sqrt{72}=6\sqrt{2}[/latex]
[/hidden-answer][latex]\left(-12,-3\right)\left(-1,5\right)[/latex]
Find the distance between the two points[latex]\,\left(-71,432\right)\,[/latex]and[latex]\,\text{(511,218)}\,[/latex]using your calculator, and round your answer to the nearest thousandth.
[latex]620.097[/latex]
[/hidden-answer]For the following exercises, find the coordinates of the midpoint of the line segment that joins the two given points.
[latex]\left(-1,5\right)\text{ and }\left(4,6\right)[/latex]
[latex]\left(-13,5\right)\text{ and }\left(17,18\right)[/latex]
midpoint is[latex]\,\left(2,\frac{23}{2}\right)[/latex]
[/hidden-answer]For the following exercises, construct a table and graph the equation by plotting at least three points.
[latex]y=\frac{1}{2}x+4[/latex]
[latex]4x-3y=6[/latex]
x | y |
0 | −2 |
3 | 2 |
6 | 6 |
For the following exercises, solve for[latex]\,x.[/latex]
[latex]5x+2=7x-8[/latex]
[latex]3\left(x+2\right)-10=x+4[/latex]
[latex]x=4[/latex]
[/hidden-answer][latex]7x-3=5[/latex]
[latex]12-5\left(x+1\right)=2x-5[/latex]
[latex]x=\frac{12}{7}[/latex]
[/hidden-answer][latex]\frac{2x}{3}-\frac{3}{4}=\frac{x}{6}+\frac{21}{4}[/latex]
For the following exercises, solve for[latex]\,x.\,[/latex]State all x-values that are excluded from the solution set.
[latex]\frac{x}{{x}^{2}-9}+\frac{4}{x+3}=\frac{3}{{x}^{2}-9}\,[/latex][latex]x\ne 3,-3[/latex]
No solution
[/hidden-answer][latex]\frac{1}{2}+\frac{2}{x}=\frac{3}{4}[/latex]
For the following exercises, find the equation of the line using the point-slope formula.
Passes through these two points:[latex]\,\left(-2,1\right)\text{,}\left(4,2\right).[/latex]
[latex]y=\frac{1}{6}x+\frac{4}{3}[/latex]
[/hidden-answer]Passes through the point[latex]\,\left(-3,4\right)\,[/latex]and has a slope of[latex]\,\frac{-1}{3}.[/latex]
Passes through the point[latex]\,\left(-3,4\right)\,[/latex]and is parallel to the graph[latex]\,y=\frac{2}{3}x+5.[/latex]
[latex]y=\frac{2}{3}x+6[/latex]
[/hidden-answer]Passes through these two points:[latex]\,\left(5,1\right)\text{,}\left(5,7\right).[/latex]
For the following exercises, write and solve an equation to answer each question.
The number of males in the classroom is five more than three times the number of females. If the total number of students is 73, how many of each gender are in the class?
females 17, males 56
[/hidden-answer]A man has 72 ft. of fencing to put around a rectangular garden. If the length is 3 times the width, find the dimensions of his garden.
A truck rental is $25 plus $.30/mi. Find out how many miles Ken traveled if his bill was $50.20.
84 mi
[/hidden-answer]For the following exercises, use the quadratic equation to solve.
[latex]{x}^{2}-5x+9=0[/latex]
[latex]2{x}^{2}+3x+7=0[/latex]
[latex]x=\frac{-3}{4}±\frac{i\sqrt{47}}{4}[/latex]
[/hidden-answer]For the following exercises, name the horizontal component and the vertical component.
[latex]4-3i[/latex]
[latex]-2-i[/latex]
horizontal component[latex]\,-2;[/latex]vertical component[latex]\,-1[/latex]
[/hidden-answer]For the following exercises, perform the operations indicated.
[latex]\left(9-i\right)-\left(4-7i\right)[/latex]
[latex]\left(2+3i\right)-\left(-5-8i\right)[/latex]
[latex]7+11i[/latex]
[/hidden-answer][latex]2\sqrt{-75}+3\sqrt{25}[/latex]
[latex]\sqrt{-16}+4\sqrt{-9}[/latex]
[latex]16i[/latex]
[/hidden-answer][latex]-6i\left(i-5\right)[/latex]
[latex]{\left(3-5i\right)}^{2}[/latex]
[latex]-16-30i[/latex]
[/hidden-answer][latex]\sqrt{-4}·\sqrt{-12}[/latex]
[latex]\sqrt{-2}\left(\sqrt{-8}-\sqrt{5}\right)[/latex]
[latex]-4-i\sqrt{10}[/latex]
[/hidden-answer][latex]\frac{2}{5-3i}[/latex]
[latex]\frac{3+7i}{i}[/latex]
[latex]x=7-3i[/latex]
[/hidden-answer]For the following exercises, solve the quadratic equation by factoring.
[latex]2{x}^{2}-7x-4=0[/latex]
[latex]3{x}^{2}+18x+15=0[/latex]
[latex]x=-1,-5[/latex]
[/hidden-answer][latex] 25{x}^{2}-9=0[/latex]
[latex] 7{x}^{2}-9x=0[/latex]
[latex]x=0,\frac{9}{7}[/latex]
[/hidden-answer]For the following exercises, solve the quadratic equation by using the square-root property.
[latex]{x}^{2}=49[/latex]
[latex]{\left(x-4\right)}^{2}=36[/latex]
[latex]x=10,-2[/latex]
[/hidden-answer]For the following exercises, solve the quadratic equation by completing the square.
[latex]{x}^{2}+8x-5=0[/latex]
[latex]4{x}^{2}+2x-1=0[/latex]
[latex]x=\frac{-1±\sqrt{5}}{4}[/latex]
[/hidden-answer]For the following exercises, solve the quadratic equation by using the quadratic formula. If the solutions are not real, state No real solution.
[latex]2{x}^{2}-5x+1=0[/latex]
[latex]15{x}^{2}-x-2=0[/latex]
[latex]x=\frac{2}{5},\frac{-1}{3}[/latex]
[/hidden-answer]For the following exercises, solve the quadratic equation by the method of your choice.
[latex]{\left(x-2\right)}^{2}=16[/latex]
[latex]{x}^{2}=10x+3[/latex]
[latex]x=5±2\sqrt{7}[/latex]
[/hidden-answer]For the following exercises, solve the equations.
[latex]{x}^{\frac{3}{2}}=27[/latex]
[latex]{x}^{\frac{1}{2}}-4{x}^{\frac{1}{4}}=0[/latex]
[latex]x=0,256[/latex]
[/hidden-answer][latex]4{x}^{3}+8{x}^{2}-9x-18=0[/latex]
[latex]3{x}^{5}-6{x}^{3}=0[/latex]
[latex]x=0,±\sqrt{2}[/latex]
[/hidden-answer][latex]\sqrt{x+9}=x-3[/latex]
[latex]\sqrt{3x+7}+\sqrt{x+2}=1[/latex]
[latex]x=-2[/latex]
[/hidden-answer][latex]|3x-7|=5[/latex]
[latex]|2x+3|-5=9[/latex]
[latex]x=\frac{11}{2},\frac{-17}{2}[/latex]
[/hidden-answer]For the following exercises, solve the inequality. Write your final answer in interval notation.
[latex]5x-8\le 12[/latex]
[latex]-2x+5>x-7[/latex]
[latex]\left(-\infty ,4\right)[/latex]
[/hidden-answer][latex]\frac{x-1}{3}+\frac{x+2}{5}\le \frac{3}{5}[/latex]
[latex]|3x+2|+1\le 9[/latex]
[latex]\left[\frac{-10}{3},2\right][/latex]
[/hidden-answer][latex]|5x-1|>14[/latex]
[latex]|x-3|<-4[/latex]
No solution
[/hidden-answer]For the following exercises, solve the compound inequality. Write your answer in interval notation.
[latex]-4<3x+2\le 18[/latex]
[latex]3y<1-2y<5+y[/latex]
[latex]\left(-\frac{4}{3},\frac{1}{5}\right)[/latex]
[/hidden-answer]For the following exercises, graph as described.
Graph the absolute value function and graph the constant function. Observe the points of intersection and shade the x-axis representing the solution set to the inequality. Show your graph and write your final answer in interval notation.
[latex]|x+3|\ge 5[/latex]
Graph both straight lines (left-hand side being y1 and right-hand side being y2) on the same axes. Find the point of intersection and solve the inequality by observing where it is true comparing the y-values of the lines. See the interval where the inequality is true.
[latex]x+3<3x-4[/latex]
Where the blue is below the orange line; point of intersection is[latex]\,x=3.5.[/latex]
[latex]\left(3.5,\infty \right)[/latex]
[/hidden-answer]Graph the following:[latex]\,2y=3x+4.[/latex]
[latex]y=\frac{3}{2}x+2[/latex]
x | y |
---|---|
0 | 2 |
2 | 5 |
4 | 8 |
Find the x- and y-intercepts for the following:
[latex]2x-5y=6[/latex]
[/hidden-answer]Find the x- and y-intercepts of this equation, and sketch the graph of the line using just the intercepts plotted.
[latex]3x-4y=12[/latex]
[latex]\left(0,-3\right)[/latex][latex]\left(4,0\right)[/latex]
[/hidden-answer]Find the exact distance between[latex]\,\left(5,-3\right)\,[/latex]and[latex]\,\left(-2,8\right).\,[/latex]Find the coordinates of the midpoint of the line segment joining the two points.
Write the interval notation for the set of numbers represented by[latex]\,\left\{x|x\le 9\right\}.[/latex]
[latex]\left(-\infty ,9\right][/latex]
[/hidden-answer]Solve for x:[latex]\,5x+8=3x-10.[/latex]
Solve for x:[latex]\,3\left(2x-5\right)-3\left(x-7\right)=2x-9.[/latex]
[latex]x=-15[/latex]
[/hidden-answer]Solve for x:[latex]\,\frac{x}{2}+1=\frac{4}{x}[/latex]
Solve for x:[latex]\,\frac{5}{x+4}=4+\frac{3}{x-2}.[/latex]
[latex]x\ne -4,2;[/latex][latex]x=\frac{-5}{2},1[/latex]
[/hidden-answer]The perimeter of a triangle is 30 in. The longest side is 2 less than 3 times the shortest side and the other side is 2 more than twice the shortest side. Find the length of each side.
Solve for x. Write the answer in simplest radical form.
[latex]\frac{{x}^{2}}{3}-x=\frac{-1}{2}[/latex]
[latex]x=\frac{3±\sqrt{3}}{2}[/latex]
[/hidden-answer]Solve:[latex]\,3x-8\le 4.[/latex]
Solve:[latex]\,|2x+3|<5.[/latex]
[latex]\left(-4,1\right)[/latex]
[/hidden-answer]Solve:[latex]\,|3x-2|\ge 4.[/latex]
For the following exercises, find the equation of the line with the given information.
Passes through the points[latex]\,\left(-4,2\right)\,[/latex]and[latex]\,\left(5,-3\right).[/latex]
[latex]y=\frac{-5}{9}x-\frac{2}{9}[/latex]
[/hidden-answer]Has an undefined slope and passes through the point[latex]\,\left(4,3\right).[/latex]
Passes through the point[latex]\,\left(2,1\right)\,[/latex]and is perpendicular to[latex]\,y=\frac{-2}{5}x+3.[/latex]
[latex]y=\frac{5}{2}x-4[/latex]
[/hidden-answer]Add these complex numbers:[latex]\,\left(3-2i\right)+\left(4-i\right).[/latex]
Simplify:[latex]\,\sqrt{-4}+3\sqrt{-16}.[/latex]
[latex]14i[/latex]
[/hidden-answer]Multiply:[latex]\,5i\left(5-3i\right).[/latex]
Divide:[latex]\,\frac{4-i}{2+3i}.[/latex]
[latex]\frac{5}{13}-\frac{14}{13}i[/latex]
[/hidden-answer]Solve this quadratic equation and write the two complex roots in[latex]\,a+bi\,[/latex]form:[latex]\,{x}^{2}-4x+7=0.[/latex]
Solve:[latex]\,{\left(3x-1\right)}^{2}-1=24.[/latex]
[latex]x=2,\frac{-4}{3}[/latex]
[/hidden-answer]Solve:[latex]\,{x}^{2}-6x=13.[/latex]
Solve:[latex]\,4{x}^{2}-4x-1=0[/latex]
[latex]x=\frac{1}{2}±\frac{\sqrt{2}}{2}[/latex]
[/hidden-answer]Solve:
[latex]\sqrt{x-7}=x-7[/latex]
Solve:[latex]\,2+\sqrt{12-2x}=x[/latex]
[latex]4[/latex]
[/hidden-answer]Solve:[latex]\,{\left(x-1\right)}^{\frac{2}{3}}=9[/latex]
For the following exercises, find the real solutions of each equation by factoring.
[latex]2{x}^{3}-{x}^{2}-8x+4=0[/latex]
[latex]x=\frac{1}{2},2,-2[/latex]
[/hidden-answer][latex]{\left(x+5\right)}^{2}-3\left(x+5\right)-4=0[/latex]
Toward the end of the twentieth century, the values of stocks of Internet and technology companies rose dramatically. As a result, the Standard and Poor’s stock market average rose as well. (Figure) tracks the value of that initial investment of just under $100 over the 40 years. It shows that an investment that was worth less than $500 until about 1995 skyrocketed up to about $1100 by the beginning of 2000. That five-year period became known as the “dot-com bubble” because so many Internet startups were formed. As bubbles tend to do, though, the dot-com bubble eventually burst. Many companies grew too fast and then suddenly went out of business. The result caused the sharp decline represented on the graph beginning at the end of 2000.
Notice, as we consider this example, that there is a definite relationship between the year and stock market average. For any year we choose, we can determine the corresponding value of the stock market average. In this chapter, we will explore these kinds of relationships and their properties.
]]>A jetliner changes altitude as its distance from the starting point of a flight increases. The weight of a growing child increases with time. In each case, one quantity depends on another. There is a relationship between the two quantities that we can describe, analyze, and use to make predictions. In this section, we will analyze such relationships.
A relation is a set of ordered pairs. The set consisting of the first components of each ordered pair is called the domain and the set consisting of the second components of each ordered pair is called the range. Consider the following set of ordered pairs. The first numbers in each pair are the first five natural numbers. The second number in each pair is twice that of the first.
The domain is [latex]\left\{1,\,2,\,3,\,4,\,5\right\}.[/latex] The range is [latex]\left\{2,\,4,\,6,\,8,\,10\right\}.[/latex]
Note that each value in the domain is also known as an input value, or independent variable, and is often labeled with the lowercase letter[latex]\,x.\,[/latex]Each value in the range is also known as an output value, or dependent variable, and is often labeled lowercase letter[latex]\,y.[/latex]
A function[latex]\,f\,[/latex]is a relation that assigns a single element in the range to each element in the domain. In other words, no x-values are repeated. For our example that relates the first five natural numbers to numbers double their values, this relation is a function because each element in the domain, [latex]\left\{1,\,2,\,3,\,4,\,5\right\},[/latex] is paired with exactly one element in the range, [latex]\left\{2,\,4,\,6,\,8,\,10\right\}.[/latex]
Now let’s consider the set of ordered pairs that relates the terms “even” and “odd” to the first five natural numbers. It would appear as
Notice that each element in the domain, [latex]\left\{\text{even,}\,\text{odd}\right\}[/latex] is not paired with exactly one element in the range, [latex]\left\{1,\,2,\,3,\,4,\,5\right\}.[/latex] For example, the term “odd” corresponds to three values from the domain, [latex]\left\{1,\,3,\,5\right\}[/latex] and the term “even” corresponds to two values from the range, [latex]\left\{2,\,4\right\}.[/latex] This violates the definition of a function, so this relation is not a function.
(Figure) compares relations that are functions and not functions.
A function is a relation in which each possible input value leads to exactly one output value. We say “the output is a function of the input.”
The input values make up the domain, and the output values make up the range.
Given a relationship between two quantities, determine whether the relationship is a function.
The coffee shop menu, shown in (Figure) consists of items and their prices.
Each item on the menu has only one price, so the price is a function of the item.
Therefore, the item is a not a function of price.
[/hidden-answer]
In a particular math class, the overall percent grade corresponds to a grade-point average. Is grade-point average a function of the percent grade? Is the percent grade a function of the grade-point average? (Figure) shows a possible rule for assigning grade points.
Percent grade | 0–56 | 57–61 | 62–66 | 67–71 | 72–77 | 78–86 | 87–91 | 92–100 |
Grade-point average | 0.0 | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 | 3.5 | 4.0 |
For any percent grade earned, there is an associated grade-point average, so the grade-point average is a function of the percent grade. In other words, if we input the percent grade, the output is a specific grade point average.
In the grading system given, there is a range of percent grades that correspond to the same grade-point average. For example, students who receive a grade point average of 3.0 could have a variety of percent grades ranging from 78 all the way to 86. Thus, percent grade is not a function of grade-point average.
[/hidden-answer](Figure)^{1} lists the five greatest baseball players of all time in order of rank.
Player | Rank |
---|---|
Babe Ruth | 1 |
Willie Mays | 2 |
Ty Cobb | 3 |
Walter Johnson | 4 |
Hank Aaron | 5 |
a. yes; b. yes. (Note: If two players had been tied for, say, 4th place, then the name would not have been a function of rank.)
[/hidden-answer]Once we determine that a relationship is a function, we need to display and define the functional relationships so that we can understand and use them, and sometimes also so that we can program them into graphing calculators and computers. There are various ways of representing functions. A standard function notation is one representation that facilitates working with functions.
To represent “height is a function of age,” we start by identifying the descriptive variables [latex]h[/latex] for height and [latex]a[/latex] for age. The letters[latex]\,f,\,g,[/latex]and[latex]\,h\,[/latex]are often used to represent functions just as we use [latex]x,\,y,[/latex]and[latex]z[/latex] to represent numbers and [latex]A,\,B,[/latex] and [latex]C[/latex] to represent sets.
Remember, we can use any letter to name the function; the notation[latex]\,h\left(a\right)\,[/latex]shows us that[latex]\,h\,[/latex]depends on[latex]\,a.\,[/latex]The value[latex]\,a\,[/latex]must be put into the function[latex]\,h\,[/latex]to get a result. The parentheses indicate that age is input into the function; they do not indicate multiplication.
We can also give an algebraic expression as the input to a function. For example[latex]\,f\left(a+b\right)\,[/latex]means “first add a and b, and the result is the input for the function f.” The operations must be performed in this order to obtain the correct result.
The notation[latex]\,y=f\left(x\right)\,[/latex]defines a function named[latex]\,f.\,[/latex]This is read as[latex]\,“y\,[/latex]is a function of[latex]\,x.”\,[/latex]The letter[latex]\,x\,[/latex]represents the input value, or independent variable. The letter[latex]\,y\text{, }[/latex]or[latex]\,f\left(x\right),\,[/latex]represents the output value, or dependent variable.
Use function notation to represent a function whose input is the name of a month and output is the number of days in that month. Assume that the domain does not include leap years.
The number of days in a month is a function of the name of the month, so if we name the function [latex]f,[/latex] we write [latex]\text{days}=f\left(\text{month}\right)[/latex] or [latex]d=f\left(m\right).[/latex] The name of the month is the input to a “rule” that associates a specific number (the output) with each input.
For example,[latex]\,f\left(\text{March}\right)=31,\,[/latex]because March has 31 days. The notation[latex]\,d=f\left(m\right)\,[/latex]reminds us that the number of days,[latex]\,d\,[/latex](the output), is dependent on the name of the month,[latex]\,m\,[/latex](the input).[/hidden-answer]Note that the inputs to a function do not have to be numbers; function inputs can be names of people, labels of geometric objects, or any other element that determines some kind of output. However, most of the functions we will work with in this book will have numbers as inputs and outputs.
A function[latex]\,N=f\left(y\right)\,[/latex]gives the number of police officers,[latex]\,N,\,[/latex]in a town in year[latex]\,y.\,[/latex]What does[latex]\,f\left(2005\right)=300\,[/latex]represent?
When we read[latex]\,f\left(2005\right)=300,\,[/latex]we see that the input year is 2005. The value for the output, the number of police officers[latex]\,\left(N\right),\,[/latex]is 300. Remember,[latex]\,N=f\left(y\right).\,[/latex]The statement[latex]\,f\left(2005\right)=300\,[/latex]tells us that in the year 2005 there were 300 police officers in the town.
[/hidden-answer]Use function notation to express the weight of a pig in pounds as a function of its age in days[latex]\,d\text{.}[/latex]
[latex]w=f\left(d\right)[/latex]
[/hidden-answer]Instead of a notation such as[latex]\,y=f\left(x\right),\,[/latex]could we use the same symbol for the output as for the function, such as[latex]\,y=y\left(x\right),\,[/latex]meaning “[latex]y[/latex] is a function of [latex]x[/latex]?”
Yes, this is often done, especially in applied subjects that use higher math, such as physics and engineering. However, in exploring math itself we like to maintain a distinction between a function such as[latex]\,f,\,[/latex]which is a rule or procedure, and the output[latex]\,y\,[/latex]we get by applying[latex]\,f\,[/latex]to a particular input[latex]\,x.\,[/latex]This is why we usually use notation such as[latex]\,y=f\left(x\right),P=W\left(d\right),\,[/latex]and so on.
A common method of representing functions is in the form of a table. The table rows or columns display the corresponding input and output values. In some cases, these values represent all we know about the relationship; other times, the table provides a few select examples from a more complete relationship.
(Figure) lists the input number of each month (January = 1, February = 2, and so on) and the output value of the number of days in that month. This information represents all we know about the months and days for a given year (that is not a leap year). Note that, in this table, we define a days-in-a-month function[latex]\,f\,[/latex]where[latex]\,D=f\left(m\right)\,[/latex]identifies months by an integer rather than by name.
Month number,[latex]\,m\,[/latex](input) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
Days in month,[latex]\,D\,[/latex](output) | 31 | 28 | 31 | 30 | 31 | 30 | 31 | 31 | 30 | 31 | 30 | 31 |
(Figure) defines a function[latex]\,Q=g\left(n\right).\,[/latex]Remember, this notation tells us that[latex]\,g\,[/latex]is the name of the function that takes the input[latex]\,n\,[/latex]and gives the output[latex]\,Q\text{\hspace{0.17em}.}[/latex]
[latex]n[/latex] | 1 | 2 | 3 | 4 | 5 |
[latex]Q[/latex] | 8 | 6 | 7 | 6 | 8 |
(Figure) displays the age of children in years and their corresponding heights. This table displays just some of the data available for the heights and ages of children. We can see right away that this table does not represent a function because the same input value, 5 years, has two different output values, 40 in. and 42 in.
Age in years,[latex]\text{ }a\text{ }[/latex](input) | 5 | 5 | 6 | 7 | 8 | 9 | 10 |
Height in inches,[latex]\text{ }h\text{ }[/latex](output) | 40 | 42 | 44 | 47 | 50 | 52 | 54 |
Given a table of input and output values, determine whether the table represents a function.
Which table, (Figure), (Figure), or (Figure), represents a function (if any)?
Input | Output |
---|---|
2 | 1 |
5 | 3 |
8 | 6 |
Input | Output |
---|---|
–3 | 5 |
0 | 1 |
4 | 5 |
Input | Output |
---|---|
1 | 0 |
5 | 2 |
5 | 4 |
(Figure) and (Figure) define functions. In both, each input value corresponds to exactly one output value. (Figure) does not define a function because the input value of 5 corresponds to two different output values.
When a table represents a function, corresponding input and output values can also be specified using function notation.
The function represented by (Figure) can be represented by writing
Similarly, the statements
represent the function in (Figure).
(Figure) cannot be expressed in a similar way because it does not represent a function.[/hidden-answer]Does (Figure) represent a function?
Input | Output |
---|---|
1 | 10 |
2 | 100 |
3 | 1000 |
yes
[/hidden-answer]When we know an input value and want to determine the corresponding output value for a function, we evaluate the function. Evaluating will always produce one result because each input value of a function corresponds to exactly one output value.
When we know an output value and want to determine the input values that would produce that output value, we set the output equal to the function’s formula and solve for the input. Solving can produce more than one solution because different input values can produce the same output value.
When we have a function in formula form, it is usually a simple matter to evaluate the function. For example, the function[latex]\,f\left(x\right)=5-3{x}^{2}\,[/latex]can be evaluated by squaring the input value, multiplying by 3, and then subtracting the product from 5.
Given the formula for a function, evaluate.
Evaluate[latex]\,f\left(x\right)={x}^{2}+3x-4\,[/latex]at
Replace the[latex]\,x\,[/latex] in the function with each specified value.
and we know that
Now we combine the results and simplify.
Given the function[latex]\,h\left(p\right)={p}^{2}+2p,\,[/latex]evaluate[latex]\,h\left(4\right).\,[/latex]
To evaluate[latex]\,h\left(4\right),\,[/latex]we substitute the value 4 for the input variable[latex]\,p\,[/latex]in the given function.
Therefore, for an input of 4, we have an output of 24.[/hidden-answer]
Given the function[latex]\,g\left(m\right)=\sqrt{m-4},\,[/latex]evaluate[latex]\,g\left(5\right).[/latex]
[latex]\,g\left(5\right)=1\,[/latex]
[/hidden-answer]Given the function[latex]\,h\left(p\right)={p}^{2}+2p,\,[/latex]solve for[latex]\,h\left(p\right)=3.[/latex]
If[latex]\,\left(p+3\right)\left(p-1\right)=0,\,[/latex]either[latex]\,\left(p+3\right)=0\,[/latex]or[latex]\,\left(p-1\right)=0\,[/latex](or both of them equal 0). We will set each factor equal to 0 and solve for[latex]\,p\,[/latex]in each case.
[/hidden-answer]
Given the function[latex]\,g\left(m\right)=\sqrt{m-4},\,[/latex]solve[latex]\,g\left(m\right)=2.[/latex]
[latex]\,m=8\,[/latex]
[/hidden-answer]Some functions are defined by mathematical rules or procedures expressed in equation form. If it is possible to express the function output with a formula involving the input quantity, then we can define a function in algebraic form. For example, the equation[latex]\,2n+6p=12\,[/latex]expresses a functional relationship between[latex]\,n\,[/latex] and[latex]\,p.\,[/latex]We can rewrite it to decide if[latex]\,p\,[/latex]is a function of[latex]\,n.\,[/latex]
Given a function in equation form, write its algebraic formula.
Express the relationship[latex]\,2n+6p=12\,[/latex]as a function[latex]\,p=f\left(n\right),\,[/latex]if possible.
To express the relationship in this form, we need to be able to write the relationship where[latex]\,p\,[/latex]is a function of[latex]\,n,\,[/latex]which means writing it as[latex]\,p=\left[\text{expression}\,\text{involving}\,n\right].[/latex]
Therefore,[latex]\,p\,[/latex]as a function of[latex]\,n\,[/latex]is written as
Does the equation[latex]\,{x}^{2}+{y}^{2}=1\,[/latex]represent a function with[latex]\,x\,[/latex]as input and[latex]\,y\,[/latex]as output? If so, express the relationship as a function[latex]\,y=f\left(x\right).[/latex]
First we subtract[latex]\,{x}^{2}\,[/latex]from both sides.
We now try to solve for[latex]\,y\,[/latex]in this equation.
We get two outputs corresponding to the same input, so this relationship cannot be represented as a single function[latex]\,y=f\left(x\right).[/latex] If we graph both functions on a graphing calculator, we will get the upper and lower semicircles.[/hidden-answer]
If[latex]\,x-8{y}^{3}=0,\,[/latex]express[latex]\,y\,[/latex]as a function of[latex]\,x.[/latex]
[latex]y=f\left(x\right)=\frac{\sqrt[3]{x}}{2}[/latex]
[/hidden-answer]Are there relationships expressed by an equation that do represent a function but that still cannot be represented by an algebraic formula?
Yes, this can happen. For example, given the equation[latex]\,x=y+{2}^{y},\,[/latex]if we want to express[latex]\,y\,[/latex]as a function of[latex]\,x,\,[/latex]there is no simple algebraic formula involving only[latex]\,x\,[/latex]that equals[latex]\,y.\,[/latex]However, each[latex]\,x\,[/latex]does determine a unique value for[latex]\,y,\,[/latex]and there are mathematical procedures by which[latex]\,y\,[/latex]can be found to any desired accuracy. In this case, we say that the equation gives an implicit (implied) rule for[latex]\,y\,[/latex]as a function of[latex]\,x,\,[/latex]even though the formula cannot be written explicitly.
As we saw above, we can represent functions in tables. Conversely, we can use information in tables to write functions, and we can evaluate functions using the tables. For example, how well do our pets recall the fond memories we share with them? There is an urban legend that a goldfish has a memory of 3 seconds, but this is just a myth. Goldfish can remember up to 3 months, while the beta fish has a memory of up to 5 months. And while a puppy’s memory span is no longer than 30 seconds, the adult dog can remember for 5 minutes. This is meager compared to a cat, whose memory span lasts for 16 hours.
The function that relates the type of pet to the duration of its memory span is more easily visualized with the use of a table. See (Figure).^{2}
Pet | Memory span in hours |
---|---|
Puppy | 0.008 |
Adult dog | 0.083 |
Cat | 16 |
Goldfish | 2160 |
Beta fish | 3600 |
At times, evaluating a function in table form may be more useful than using equations. Here let us call the function [latex]P.[/latex] The domain of the function is the type of pet and the range is a real number representing the number of hours the pet’s memory span lasts. We can evaluate the function[latex]\,P\,[/latex]at the input value of “goldfish.” We would write [latex]P\left(\text{goldfish}\right)=2160.[/latex] Notice that, to evaluate the function in table form, we identify the input value and the corresponding output value from the pertinent row of the table. The tabular form for function[latex]\,P\,[/latex]seems ideally suited to this function, more so than writing it in paragraph or function form.
Given a function represented by a table, identify specific output and input values.
Using (Figure),
[latex]n[/latex] | 1 | 2 | 3 | 4 | 5 |
[latex]g\left(n\right)[/latex] | 8 | 6 | 7 | 6 | 8 |
[latex]n[/latex] | 1 | 2 | 3 | 4 | 5 |
[latex]g\left(n\right)[/latex] | 8 | 6 | 7 | 6 | 8 |
When we input 2 into the function[latex]\,g,\,[/latex]our output is 6. When we input 4 into the function[latex]\,g,\,[/latex]our output is also 6.
[/hidden-answer]Using (Figure), evaluate[latex]\,g\left(1\right).[/latex]
[latex]g\left(1\right)=8[/latex]
[/hidden-answer]Evaluating a function using a graph also requires finding the corresponding output value for a given input value, only in this case, we find the output value by looking at the graph. Solving a function equation using a graph requires finding all instances of the given output value on the graph and observing the corresponding input value(s).
Given the graph in (Figure),
Using (Figure), solve[latex]\,f\left(x\right)=1.[/latex]
[latex]x=0\,[/latex]or[latex]\,x=2\,[/latex]
[/hidden-answer]Some functions have a given output value that corresponds to two or more input values. For example, in the stock chart shown in (Figure) at the beginning of this chapter, the stock price was $1000 on five different dates, meaning that there were five different input values that all resulted in the same output value of $1000.
However, some functions have only one input value for each output value, as well as having only one output for each input. We call these functions one-to-one functions. As an example, consider a school that uses only letter grades and decimal equivalents, as listed in (Figure).
Letter grade | Grade point average |
---|---|
A | 4.0 |
B | 3.0 |
C | 2.0 |
D | 1.0 |
This grading system represents a one-to-one function because each letter input yields one particular grade-point average output and each grade-point average corresponds to one input letter.
To visualize this concept, let’s look again at the two simple functions sketched in (Figure)(a) and (Figure)(b). The function in part (a) shows a relationship that is not a one-to-one function because inputs[latex]\,q\,[/latex]and[latex]\,r\,[/latex]both give output[latex]\,n.\,[/latex]The function in part (b) shows a relationship that is a one-to-one function because each input is associated with a single output.
A one-to-one function is a function in which each output value corresponds to exactly one input value. There are no repeated x- or y-values.
Is the area of a circle a function of its radius? If yes, is the function one-to-one?
A circle of radius [latex]\,r\,[/latex]has a unique area measure given by[latex]\,A=\pi {r}^{2},[/latex]so for any input,[latex]\,r,\,[/latex]there is only one output, [latex]A.[/latex] The area is a function of radius[latex]\,r.[/latex]
If the function is one-to-one, the output value, the area, must correspond to a unique input value, the radius. Any area measure[latex]\,A\,[/latex]is given by the formula[latex]\,A=\pi {r}^{2}.\,[/latex]Because areas and radii are positive numbers, there is exactly one solution:[latex]\sqrt{\frac{A}{\pi }}.[/latex]So the area of a circle is a one-to-one function of the circle’s radius.
[/hidden-answer]a. yes, because each bank account has a single balance at any given time; b. no, because several bank account numbers may have the same balance; c. no, because the same output may correspond to more than one input.
[/hidden-answer]As we have seen in some examples above, we can represent a function using a graph. Graphs display a great many input-output pairs in a small space. The visual information they provide often makes relationships easier to understand. By convention, graphs are typically constructed with the input values along the horizontal axis and the output values along the vertical axis.
The most common graphs name the input value[latex]\,x\,[/latex]and the output value[latex]\,y,\,[/latex]and we say[latex]\,y\,[/latex]is a function of[latex]\,x,\,[/latex]or[latex]\,y=f\left(x\right)\,[/latex]when the function is named[latex]\,f.\,[/latex]The graph of the function is the set of all points[latex]\,\left(x,y\right)\,[/latex]in the plane that satisfies the equation[latex]y=f\left(x\right).\,[/latex]If the function is defined for only a few input values, then the graph of the function consists of only a few points, where the x-coordinate of each point is an input value and the y-coordinate of each point is the corresponding output value. For example, the black dots on the graph in (Figure) tell us that[latex]\,f\left(0\right)=2\,[/latex]and[latex]\,f\left(6\right)=1.\,[/latex]However, the set of all points[latex]\,\left(x,y\right)\,[/latex]satisfying[latex]\,y=f\left(x\right)\,[/latex]is a curve. The curve shown includes[latex]\,\left(0,2\right)\,[/latex]and[latex]\,\left(6,1\right)\,[/latex]because the curve passes through those points.
The vertical line test can be used to determine whether a graph represents a function. If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because a function has only one output value for each input value. See (Figure).
Given a graph, use the vertical line test to determine if the graph represents a function.
Which of the graphs in (Figure) represent(s) a function[latex]\,y=f\left(x\right)?[/latex]
If any vertical line intersects a graph more than once, the relation represented by the graph is not a function. Notice that any vertical line would pass through only one point of the two graphs shown in parts (a) and (b) of (Figure). From this we can conclude that these two graphs represent functions. The third graph does not represent a function because, at most x-values, a vertical line would intersect the graph at more than one point, as shown in (Figure).
Does the graph in (Figure) represent a function?
yes
[/hidden-answer]Once we have determined that a graph defines a function, an easy way to determine if it is a one-to-one function is to use the horizontal line test. Draw horizontal lines through the graph. If any horizontal line intersects the graph more than once, then the graph does not represent a one-to-one function.
Given a graph of a function, use the horizontal line test to determine if the graph represents a one-to-one function.
Consider the functions shown in (Figure)(a) and (Figure)(b). Are either of the functions one-to-one?
The function in (Figure)(a) is not one-to-one. The horizontal line shown in (Figure) intersects the graph of the function at two points (and we can even find horizontal lines that intersect it at three points.)
The function in (Figure)(b) is one-to-one. Any horizontal line will intersect a diagonal line at most once.[/hidden-answer]
Is the graph shown in (Figure) one-to-one?
No, because it does not pass the horizontal line test.
[/hidden-answer]In this text, we will be exploring functions—the shapes of their graphs, their unique characteristics, their algebraic formulas, and how to solve problems with them. When learning to read, we start with the alphabet. When learning to do arithmetic, we start with numbers. When working with functions, it is similarly helpful to have a base set of building-block elements. We call these our “toolkit functions,” which form a set of basic named functions for which we know the graph, formula, and special properties. Some of these functions are programmed to individual buttons on many calculators. For these definitions we will use[latex]\,x\,[/latex]as the input variable and[latex]\,y=f\left(x\right)\,[/latex]as the output variable.
We will see these toolkit functions, combinations of toolkit functions, their graphs, and their transformations frequently throughout this book. It will be very helpful if we can recognize these toolkit functions and their features quickly by name, formula, graph, and basic table properties. The graphs and sample table values are included with each function shown in (Figure).
Toolkit Functions | ||
---|---|---|
Name | Function | Graph |
Constant | [latex]f\left(x\right)=c,\,[/latex]where [latex]c[/latex] is a constant | |
Identity | [latex]f\left(x\right)=x[/latex] | |
Absolute value | [latex]f\left(x\right)=|x|[/latex] | |
Quadratic | [latex]f\left(x\right)={x}^{2}[/latex] | |
Cubic | [latex]f\left(x\right)={x}^{3}[/latex] | |
Reciprocal | [latex]f\left(x\right)=\frac{1}{x}[/latex] | |
Reciprocal squared | [latex]f\left(x\right)=\frac{1}{{x}^{2}}[/latex] | |
Square root | [latex]f\left(x\right)=\sqrt{x}[/latex] | |
Cube root | [latex]f\left(x\right)=\sqrt[3]{x}[/latex] |
Access the following online resources for additional instruction and practice with functions.
Constant function | [latex]f\left(x\right)=c,[/latex]where[latex]\,c\,[/latex]is a constant |
Identity function | [latex]f\left(x\right)=x[/latex] |
Absolute value function | [latex]f\left(x\right)=|x|[/latex] |
Quadratic function | [latex]f\left(x\right)={x}^{2}[/latex] |
Cubic function | [latex]f\left(x\right)={x}^{3}[/latex] |
Reciprocal function | [latex]f\left(x\right)=\frac{1}{x}[/latex] |
Reciprocal squared function | [latex]f\left(x\right)=\frac{1}{{x}^{2}}[/latex] |
Square root function | [latex]f\left(x\right)=\sqrt{x}[/latex] |
Cube root function | [latex]f\left(x\right)=\sqrt[3]{x}[/latex] |
What is the difference between a relation and a function?
A relation is a set of ordered pairs. A function is a special kind of relation in which no two ordered pairs have the same first coordinate.
[/hidden-answer]What is the difference between the input and the output of a function?
Why does the vertical line test tell us whether the graph of a relation represents a function?
When a vertical line intersects the graph of a relation more than once, that indicates that for that input there is more than one output. At any particular input value, there can be only one output if the relation is to be a function.
[/hidden-answer]How can you determine if a relation is a one-to-one function?
Why does the horizontal line test tell us whether the graph of a function is one-to-one?
When a horizontal line intersects the graph of a function more than once, that indicates that for that output there is more than one input. A function is one-to-one if each output corresponds to only one input.
[/hidden-answer]For the following exercises, determine whether the relation represents a function.
[latex]\left\{\left(a,b\right),\text{ }\left(c,d\right),\text{ }\left(a,c\right)\right\}[/latex]
[latex]\left\{\left(a,b\right),\left(b,c\right),\left(c,c\right)\right\}[/latex]
function
[/hidden-answer]For the following exercises, determine whether the relation represents[latex]\,y\,[/latex]as a function of[latex]\,x.\,[/latex]
[latex]5x+2y=10[/latex]
[latex]y={x}^{2}[/latex]
function
[/hidden-answer][latex]x={y}^{2}[/latex]
[latex]3{x}^{2}+y=14[/latex]
function
[/hidden-answer][latex]2x+{y}^{2}=6[/latex]
[latex]y=-2{x}^{2}+40x[/latex]
function
[/hidden-answer][latex]y=\frac{1}{x}[/latex]
[latex]x=\frac{3y+5}{7y-1}[/latex]
function
[/hidden-answer][latex]x=\sqrt{1-{y}^{2}}[/latex]
[latex]y=\frac{3x+5}{7x-1}[/latex]
function
[/hidden-answer][latex]{x}^{2}+{y}^{2}=9[/latex]
[latex]2xy=1[/latex]
function
[/hidden-answer][latex]x={y}^{3}[/latex]
[latex]y={x}^{3}[/latex]
function
[/hidden-answer][latex]y=\sqrt{1-{x}^{2}}[/latex]
[latex]x=±\sqrt{1-y}[/latex]
function
[/hidden-answer][latex]y=±\sqrt{1-x}[/latex]
[latex]{y}^{2}={x}^{2}[/latex]
not a function
[/hidden-answer][latex]{y}^{3}={x}^{2}[/latex]
For the following exercises, evaluate the function[latex]\,f\,[/latex]at the indicated values[latex]\text{ }f\left(-3\right),f\left(2\right),f\left(-a\right),-f\left(a\right),f\left(a+h\right).[/latex]
[latex]f\left(x\right)=2x-5[/latex]
[latex]\begin{array}{cccc}f\left(-3\right)=-11;& f\left(2\right)=-1;& f\left(-a\right)=-2a-5;& -f\left(a\right)=-2a+5;\,\,\,\,f\left(a+h\right)=2a+2h-5\end{array}[/latex]
[/hidden-answer][latex]f\left(x\right)=-5{x}^{2}+2x-1[/latex]
[latex]f\left(x\right)=\sqrt{2-x}+5[/latex]
[latex]\begin{array}{cccc}f\left(-3\right)=\sqrt{5}+5;& f\left(2\right)=5;& f\left(-a\right)=\sqrt{2+a}+5;& -f\left(a\right)=-\sqrt{2-a}-5;\,\,\,\,\,f\left(a+h\right)=\end{array}[/latex][latex]\sqrt{2-a-h}+5[/latex]
[/hidden-answer][latex]f\left(x\right)=\frac{6x-1}{5x+2}[/latex]
[latex]f\left(x\right)=|x-1|-|x+1|[/latex]
[latex]\begin{array}{cccc}f\left(-3\right)=2;& f\left(2\right)=1-3=-2;& f\left(-a\right)=|-a-1|-|-a+1|;& -f\left(a\right)=-|a-1|\,+|a+1|;\text{ }\text{ }f\left(a+h\right)=\,|a+h-1|-|a+h+1|\end{array}[/latex]
[/hidden-answer]Given the function[latex]\,g\left(x\right)=5-{x}^{2},\,[/latex]simplify[latex]\,\frac{g\left(x+h\right)-g\left(x\right)}{h},\,h\ne 0.[/latex]
Given the function[latex]\,g\left(x\right)={x}^{2}+2x,\,[/latex]simplify[latex]\,\frac{g\left(x\right)-g\left(a\right)}{x-a},\,x\ne a.[/latex]
[latex]\frac{g\left(x\right)-g\left(a\right)}{x-a}=x+a+2,\,x\ne a[/latex]
[/hidden-answer]Given the function[latex]\,k\left(t\right)=2t-1\text{:}[/latex]
Given the function[latex]\,f\left(x\right)=8-3x\text{:}[/latex]
a.[latex]\,f\left(-2\right)=14;\,[/latex]b.[latex]\,x=3[/latex]
[/hidden-answer]Given the function[latex]\,p\left(c\right)={c}^{2}+c\text{:}[/latex]
Given the function[latex]\,f\left(x\right)={x}^{2}-3x\text{:}[/latex]
a.[latex]\,f\left(5\right)=10;\,[/latex]b.[latex]\,x=-1\text{ }[/latex]or[latex]\text{ }x=4[/latex]
[/hidden-answer]Given the function[latex]\,f\left(x\right)=\sqrt{x+2}\text{:}[/latex]
Consider the relationship[latex]\,3r+2t=18.[/latex]
a.[latex]\,f\left(t\right)=6-\frac{2}{3}t;\,[/latex]b.[latex]\,f\left(-3\right)=8;\,[/latex]c.[latex]\,t=6\,[/latex]
[/hidden-answer]For the following exercises, use the vertical line test to determine which graphs show relations that are functions.
not a function
[/hidden-answer]function
[/hidden-answer]function
[/hidden-answer]function
[/hidden-answer]function
[/hidden-answer]function
[/hidden-answer]Given the following graph,
Given the following graph,
a.[latex]\,f\left(0\right)=1;\,[/latex]b.[latex]\,f\left(x\right)=-3,\,x=-2\text{ }[/latex]or[latex]\text{ }x=2\,[/latex]
[/hidden-answer]Given the following graph,
For the following exercises, determine if the given graph is a one-to-one function.
not a function so it is also not a one-to-one function
[/hidden-answer]one-to- one function
[/hidden-answer]function, but not one-to-one
[/hidden-answer]For the following exercises, determine whether the relation represents a function.
[latex]\left\{\left(-1,-1\right),\left(-2,-2\right),\left(-3,-3\right)\right\}[/latex]
[latex]\left\{\left(3,4\right),\left(4,5\right),\left(5,6\right)\right\}[/latex]
function
[/hidden-answer][latex]\left\{\left(2,5\right),\left(7,11\right),\left(15,8\right),\left(7,9\right)\right\}[/latex]
For the following exercises, determine if the relation represented in table form represents[latex]\,y\,[/latex]as a function of[latex]\,x.[/latex]
[latex]x[/latex] | 5 | 10 | 15 |
[latex]y[/latex] | 3 | 8 | 14 |
function
[/hidden-answer][latex]x[/latex] | 5 | 10 | 15 |
[latex]y[/latex] | 3 | 8 | 8 |
[latex]x[/latex] | 5 | 10 | 10 |
[latex]y[/latex] | 3 | 8 | 14 |
not a function
[/hidden-answer]For the following exercises, use the function[latex]\,f\,[/latex]represented in (Figure).
[latex]x[/latex] | [latex]f\left(x\right)[/latex] |
0 | 74 |
1 | 28 |
2 | 1 |
3 | 53 |
4 | 56 |
5 | 3 |
6 | 36 |
7 | 45 |
8 | 14 |
9 | 47 |
Evaluate[latex]\,f\left(3\right).[/latex]
Solve[latex]\,f\left(x\right)=1.[/latex]