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9.1

9.2

9.3

9.4

9.5

9.6

Radicals

Rational Exponents

Adding, Subtracting, and Multiplying Radicals

Quotients, Powers, and Rationalizing Denominators

Solving Equations with Radicals and Exponents

Complex Numbers

9 Radicals and Rational Exponents Just how cold is it in Fargo, North Dakota, in winter? According to local meteorol

ogists, the mercury hit a low of –33°F on January 18, 1994. But air temperature

alone is not always a reliable indicator of how cold you feel. On the same date,

the average wind velocity was 13.8 miles per hour. This dramatically affected how

cold people felt when they stepped outside. High winds along with cold temper

atures make exposed skin feel colder because the wind significantly speeds up

the loss of body heat. Meteorologists use the terms “wind chill factor,”“wind chill

index,” and “wind chill temperature” to take into account both air temperature

and wind velocity.

Through experimentation in Ant

arctica, Paul A. Siple developed a

formula in the 1940s that measures the

wind chill from the velocity of the wind

and the air temperature. His complex

formula involving the square root of

the velocity of the wind is still used

today to calculate wind chill temper

atures. Siple’s formula is unlike most

scientific formulas in that it is not

based on theory. Siple experimented

with various formulas involving wind

velocity and temperature until he

found a formula that seemed to predict

how cold the air felt.

W in

d ch

ill te

m pe

ra tu

re ( F

) fo

r 25

F a

ir te

m pe

ra tu

Wind velocity (mph)

5 10 15 20 25 30

Siple s formula is stated and used in Exercises 111

and 112 of Section 9.1.

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→

558 Chapter 9 Radicals and Rational Exponents 9-2

9.1 Radicals

In Section 4.1, you learned the basic facts about powers. In this section, you will study roots and see how powers and roots are related.

In This Section

U1V Roots

U2V Roots and Variables

U3V Product Rule for Radicals

U4V Quotient Rule for Radicals U1V Roots U5V Domain of a Radical We use the idea of roots to reverse powers. Because 32 = 9 and (-3)2 = 9, both 3 andExpression or Function

-3 are square roots of 9. Because 24 = 16 and (-2)4 = 16, both 2 and -2 are fourth roots of 16. Because 23 = 8 and (-2)3 = -8, there is only one real cube root of 8 and only one real cube root of -8. The cube root of 8 is 2 and the cube root of -8 is -2.

nth Roots

If a = bn for a positive integer n, then b is an nth root of a. If a = b2, then b is a square root of a. If a = b3, then b is the cube root of a.

If n is a positive even integer and a is positive, then there are two real nth roots of a. We call these roots even roots. The positive even root of a positive number is called the principal root. The principal square root of 9 is 3 and the principal fourth root of 16 is 2, and these roots are even roots.

If n is a positive odd integer and a is any real number, there is only one real nth root of a. We call that root an odd root. Because 25 = 32, the fifth root of 32 is 2 and 2 is an odd root.

We use the radical symbol Va to signify roots.

U Helpful Hint V V n aa The parts of a radical:

If n is a positive even integer and a is positive, then V n a a denotes the principal nth Radical root of a.Index

V n symbol

aa If n is a positive odd integer, then V n a a denotes the nth root of a. Radicand If n is any positive integer, then V

n a 0 = 0.

We read V n a a as “the nth root of a.” In the notation V n a a, n is the index of the radical and a is the radicand. For square roots the index is omitted, and we simply write Va a.

E X A M P L E 1 Evaluating radical expressions Find the following roots:

a) Va25 3

-b) Va27

c) V6 64a

d) -Va 4

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9-3 9.1 Radicals 559

Solution a) Because 52 = 25, V25a = 5. b) Because (-3)3 = -27, V3 -27a = -3. c) Because 26 = 64, V6 64a = 2. d) Because V4a = 2, -V4a = -(V4a) = -2.

U Calculator Close-Up V

We can use the radical symbol to find a square root on a graphing calcula tor, but for other roots we use the xth root symbol as shown. The xth root symbol is in the MATH menu.

Now do Exercises 1–16

CAUTION In radical notation, V4a represents the principal square root of 4, so V4a = 2. Note that -2 is also a square root of 4, but Va4 * -2.

Note that even roots of negative numbers are omitted from the definition of nth roots because even powers of real numbers are never negative. So no real number can be an even root of a negative number. Expressions such as

V-a9, V-81 and V-64 4 a, 6 a

are not real numbers. Square roots of negative numbers will be discussed in Section 9.6 when we discuss the imaginary numbers.

U2V Roots and Variables A whole number is a perfect square if it is the square of another whole number. So 9 is a perfect square because 32 = 9. Likewise, an exponential expression is a perfect square if it is the square of another exponential expression. So x10 is a perfect square because (x5)2 = x10. The exponent in a perfect square must be divisible by 2. An expo nential expression is a perfect cube if it is the cube of another exponential expression. So x21 is a perfect cube because (x7)3 = x21. The exponent in a perfect cube must be divisible by 3. The exponent in a perfect fourth power is divisible by 4, and so on.

2 4 6 8 10 12 Perfect squares x , x , x , x , x , x , . . . Exponent divisible by 2 3 6 9 12 15 18 Perfect cubes x , x , x , x , x , x , . . . Exponent divisible by 3 4 8 12 16 20 24 Perfect fourth powers x , x , x , x , x , x , . . . Exponent divisible by 4

To find the square root of a perfect square, divide the exponent by 2. If x is non negative, we have

2 4 2 6 3 Vx Vx , a = x and so on.a = x, a = x Vx , We specified that x was nonnegative because Vx a = x3 are not correcta2 = x and Vx6 if x is negative. If x is negative, x and x3 are negative but the radical symbol with an even root must be a positive number. Using absolute value symbols we can say that

2 6 Vx a = lx3l for any real numbers.a = lxl and Vx To find the cube root of a perfect cube, divide the exponent by 3. If x is any real

number, we have

3 3 3 3 6 2 9 3 Vx Vx , Vx , and so on.a = x, a = x a = x Note that both sides of each of these equations have the same sign whether x is posi tive or negative. For cube roots and other odd roots, we will not need absolute value symbols to make statements that are true for any real numbers. We need absolute value

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560 Chapter 9 Radicals and Rational Exponents 9-4

symbols only when the result of an even root has an odd exponent. For example, Vm30 = lm6 a 5l for any real number m.

E X A M P L E 2 Roots of exponential expressions Find each root. Assume that the variables can represent any real numbers. Use absolute value symbols when necessary.

a) Va2a b) Vx22a c) V4 w40a d) V3 t18a e) V5 s30a

Solution a) For a square root, divide the exponent by 2. But if a is negative, Va2a = a is not

correct, because the square root symbol represents the nonnegative square root. So if a is any real number, Va2a = la l.

b) Divide the exponent by 2. But if x is negative, Vx22a = x11 is not correct because x11 is negative and Vx22a is positive. So if x is any real number, Vx22a = l x11 l.

c) For a fourth root, divide the exponent by 4. So V4 w40a = w10. We don’t need absolute value symbols because both sides of this equation have the same sign whether w is positive or negative.

d) For a cube root, divide the exponent by 3. So V3 t18a = t6. We don’t need absolute value symbols because both sides of this equation have the same sign whether t is positive or negative.

e) For a fifth root, divide the exponent by 5. So V5 s30a = s6. We don’t need absolute value symbols because both sides of this equation have the same sign whether s is positive or negative.

U Calculator Close-Up V

You can illustrate the product rule for radicals with a calculator.

Now do Exercises 17–32

U3V Product Rule for Radicals Consider the expression V2a · V3a. If we square this product, we get

(V2a · Va3)2 = (V2a)2(Va3)2 Power of a product rule = 2 · 3 (V2a)2 = 2 and (V3a)2 = 3 = 6.

The number V6a is the unique positive number whose square is 6. Because we squared V2a · V3a and obtained 6, we must have Va6 = Va2 · Va3. This example illustrates the product rule for radicals.

Product Rule for Radicals

The nth root of a product is equal to the product of the nth roots. In symbols,

n n nVab aa · Va,a = V b

provided all of these roots are real numbers.

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9-5 9.1 Radicals 561

E X A M P L E 3 Using the product rule for radicals to simplify Simplify each radical. Assume that all variables represent nonnegative real numbers.

a) V4ya b) V3y8a c) V3 125w2a

Solution a) V4ya = V4a · Vya Product rule for radicals

= 2Vya Simplify.

b) V3y8a = V3a · Vy8a Product rule for radicals = V3a · y4 Simplify. = y4V3a A radical is usually written last in a product.

c) V3 125w2a = V3 125a · V3 w2a = 5V3 w2a Now do Exercises 33–44

In Example 4, we simplify by factoring the radicand before applying the product rule.

E X A M P L E 4 Using the product rule to simplify Simplify each radical.

a) V12a b) V3 54a c) V4 80a d) V5 64a

Solution a) Since 12 = 4 · 3 and 4 is a perfect square, we can factor and then apply the

product rule:

V12a = V4 · 3a = V4a · V3a = 2V3a

b) Since 54 = 27 · 2 and 27 is a perfect cube, we can factor and then apply the product rule:

V3 54a = V3 27 · 2a = V3 27a · V3 2a = 3V3 2a

c) Since 80 = 16 · 5 and 16 is a perfect fourth power, we can factor and then apply the product rule:

V4 80a = V4 16 · 5a = V4 16a · V4 5a = 2V4 5a

d) V5 64a = V5 32 · 2a = V5 32a · V5 2a = 2V5 2a Now do Exercises 45–58

In general, we simplify radical expressions of index n by using the product rule to remove any perfect nth powers from the radicand. In Example 5, we use the product rule to simplify more radicals involving variables. Remember xn is a perfect square if n is divisible by 2, a perfect cube if n is divisible by 3, and so on.

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562 Chapter 9 Radicals and Rational Exponents 9-6

E X A M P L E 5

U Calculator Close-Up V

You can illustrate the quotient rule for radicals with a calculator.

Using the product rule to simplify Simplify each radical. Assume that all variables represent nonnegative real numbers.

3 4 53 8 4b1 7a) V20ax b) V40a c) V48a a d) Vwa a1 a

Solution a) Factor 20x3 so that all possible perfect squares are inside one radical:

3 2 · 5V20ax = V4x aax Factor out perfect squares.

= V4x2 a Product rulea · V5x

= 2xV5x Simplify.a

b) Factor 40a8 so that all possible perfect cubes are inside one radical:

3 a 3 aa8 6 · 5 2V40a = V8a a Factor out perfect cubes. 3 3 a= V8aa6 · V5a2 Product rule

= 2a2V5a23 a Simplify.

c) Factor 48a4b11 so that all possible perfect fourth powers are inside one radical:

4 4a4b11 a4b8 · 3b3V48a a = V16a a Factor out perfect fourth powers. 4 44b8= V16a 3b3 Product rulea · Va

4 a= 2ab2V3b3 Simplify.

2wa Now do Exercises 59–72

d) V5 w7a = V5 wa2 = Vw5 a = wV5 · wa 5 a · V5 w2 5

U4V Quotient Rule for Radicals Because V2a · Va3 = Va6, we have V6a - Va3 = Va2, or

6 Va6V2a = J = . 3 V3a This example illustrates the quotient rule for radicals.

Quotient Rule for Radicals

The nth root of a quotient is equal to the quotient of the nth roots. In symbols,

n a Vaa

n = , nJ b Vba provided that all of these roots are real numbers and b * 0.

E X A M P L E 6 Using the quotient rule for radicals Simplify each radical. Assume that all variables represent positive real numbers.

25 V15 b x21a 3 3a) J b) c) J d) J 9 V3a 125 y6

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9-7 9.1 Radicals 563

Solution

a) J2 9 5 = Quotient rule for radicals =

5 3

Simplify.

b) = J1 3 5 Quotient rule for radicals = V5a Simplify.

c) J3 1 b 25 = =

d) J3 x y 2

= = x y

2 V3 x21a V3 y6a

V3 ba 5

V3 ba V3 125a

V15a V3a

V25a V9a

Now do Exercises 73–84

In Example 7, we use the product and quotient rules to simplify radical expressions.

E X A M P L E 7 Using the product and quotient rules for radicals Simplify each radical. Assume that all variables represent positive real numbers.

a) J5 4 0 9 b) J3 x 8 5 c) J4 a b 5 8 Solution

a) J5 4 0 9 = Product and quotient rules for radicals = Simplify.

b) J3 x 8 5 = = c) J4 a b 5 8 = = aV

aa b2

V4 a4a · V4 aa V4 b8a

xV3 x2a 2

V3 x3a · V3 x2a V3 8a

5V2a 7

V25a · V2a V49a

Now do Exercises 85–96

U5V Domain of a Radical Expression or Function The domain of any expression involving one variable is the set of all real numbers that can be used in place of the variable. For many expressions the domain of the expres sion is the set of all real numbers. For example, any real number can be used in place of x in the expression 2x + 3 and its domain is the set of all real numbers, (-o, o).

For a radical expression the domain depends on the radicand and whether the root 3is even or odd. Since every real number has an odd root, the domain of Vxa is (-o, o).

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564 Chapter 9 Radicals and Rational Exponents 9-8

Since there are no real even roots of negative numbers, the domain of Vxa is the set of nonnegative real numbers or [0, o).

E X A M P L E 8 Finding the domain of a radical expression Find the domain of each expression. Express the answer in interval notation.

a) Vx - 5a b) V3 x + 7a c) V4 2x + 6a

Solution a) Since the radicand in a square root must be nonnegative, x - 5 must be

nonnegative:

x - 5 2 0

x 2 5

So only values of x that are 5 or larger can be used for x. The domain is [5, o).

b) Since any real number has a cube root, any real number can be used in place of x in V3 x + 7a. So the domain is (-o, o).

c) Since the radicand in a fourth root must be nonnegative, 2x + 6 must be nonnegative:

2x + 6 2 0

2x 2 -6

x 2 -3

So the domain of V4 2x + 6a is [-3, o). Now do Exercises 97–110

If a radical expression is used to determine the value of a second variable y, then we have a radical function. For example,

R(x) = Vx – 5 V(x) = Vx + 7 T(x) = V2x + 6a, 3 a, and 4 a

are radical functions. The domain of a radical function is the domain of the radical expression. Since these are the radical expressions of Example 8, the domain for R(x) is [5, o), the domain for V(x) is (-o, o), and the domain for T(x) is [-3, o).

Warm-Ups ▼

Fill in the blank. 1. If bn = a, then b is an of a.

2. If n is even and a > 0, then V n

aa is the nth root of a.

3. According to the rule for radicals V n

aa · V n

ba = V

n aba provided all of the roots are real.

4. According to the rule for radicals V n

aa/V n

ba = Vn a/ba provided all of the roots are real.

True or false? 5. V2a · V2a = 2

6. V 3

2a · V 3

2a = 2

7. V 3

-27a = -3

8. V 4

16a = 2

9. V9a = ±3

10. V2a · V7a = V14a

11. V V

6a 2a

= V3a

12. V

2 10a

= V5a

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Exercises

U Study Tips V • If you have a choice, sit at the front of the class. It is easier to stay alert when you are at the front. • If you miss what is going on in class, you miss what your instructor feels is important and most likely to appear on tests and quizzes.

9 .1

U1V Roots

Find each root. See Example 1.

1. V36 3. V100 5. fV 9

3 7. V 8 3

f89. V 5 11. V32

13. V 3 1000 4 15. Vf16

U2V Roots and Variables

2. V49 4. V81 6. fV25

3 8. V27 3 10. Vf 1 4 12. V81 4 14. V16

16. Vf1

Find each root. See Example 2. All variables represent real numbers. Use absolute value when necessary.

2 6 17. Vm 18. Vm 16 36 19. Vx 20. Vy

5 4 15 8 21. V 22. V y m 3 15 8 23. Vy 24. Vm 3 4 3 x425. Vm 26. V 4 5 12 30 27. Vw 28. V a

m 6 42 29. Vb18 30. V 4 24 31. Vy 32. Vt44

U3V Product Rule for Radicals

Use the product rule for radicals to simplify each expression. See Example 3. All variables represent nonnegative real numbers.

33. V 9y 34. V16n 2 2 35. V4 a 36. V36 n

4 2 6t237. Vx y 38. Vw 12 16 39. V5 m 40. V7z

41. V 42. V 3 8y 3 27z2 3 3 6 43. V3 a 44. V5 b9

Use the product rule to simplify. See Example 4.

45. V 20 46. V18 47. V50 48. V45

49. V 72 50. V98 3 40 3 24 51. V 52. V 3 3 250 53. V81 54. V 4 4 55. V48 56. V 32

57. V96 58. V 5 5 2430

Use the product rule to simplify. See Example 5. All variables represent nonnegative real numbers.

3 59. Va 60. Vb 5

6 8 61. V 18a 62. V12 x 5 3 3 63. V y 8 w y20x 64. V

3 3 65. V24m 66. V 4 54ab5

67. V 68. V 4 32a5 4 162b4 5 5 6 8 69. V64x 70. V 96a

3 8 7 3 8 7 71. V48x y 72. V y z 3 48 x z

U4V Quotient Rule for Radicals

Simplify each radical. See Example 6. All variables represent positive real numbers.

t 73. 4

625 75. 16

V30 77.

t 3 79. 8

6 8x f 3 81. 3 y

6 4a83. 9

w 74. 36

9 76. 144

V50 78.

V 2

a 3 80. 27

f27y3 3 6

82. 1000 2 9a84. 4 49b

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566 Chapter 9 Radicals and Rational Exponents 9-10

Use the product and quotient rules to simplify. See Example 7. All variables represent positive real numbers.

2 8 85. J1 86. J825 1

7 8 87. J2 88. J9 16 9

4 7a b 3 389. J 90. J125 1000

3 481 a b 3 391. J 92. J38b 125

the wind velocity v. Through experimentation in Antarctica, Paul Siple developed a formula for W:

(10.5 + 6.7Vva - 0.45v)(457 - 5t) W = 91.4 - ,

110

where W and t are in degrees Fahrenheit and v is in miles per hour (mph).

a) Find W to the nearest whole degree when t = 25°F and v = 20 mph.

b) Use the accompanying graph to estimate W when t = 25°F and v = 30 mph.

W in

d ch

ill te

m pe

ra tu

re (

� F )

fo r

25 � F

a ir

te m

pe ra

tu re

Wind velocity (mph)

5 10 15 20 25 30

7 5 4 25

493. J 494. Jx x y 8 12y z 5 7a a b

4 495. J 96. J1216b 81c16 -5

-10

-15 U5V Domain of a Radical Expression or Function

Find the domain of each radical expression. See Example 8.

97. Vx - 2a

98. V2 - xa 399. V3x - 7a 3100. V5 - 4xa 4101. V9 - 3xa 4102. V4x - 8a

103. V2x + 1a

104. V4x - 1a

Find the domain of each radical function.

105. R(x) = Vx - 6a

Figure for Exercise 111

112. Comparing wind chills. Use the formula from Exercise 111 to determine who will feel colder: a person in Minneapolis at 10°F with a 15-mph wind or a person in Chicago at 20°F with a 25-mph wind.

113. Diving time. The time t (in seconds) that it takes for a cliff diver to reach the water is a function of the height h (in feet) from which he dives:

t = J1 h 6

T im

e (s

ec on

ds )

0 20 40 60 80 100 Height (feet)

106. V(x) = V7 - xa 3

107. y = Vx + 1a 5

108. y = V3x - 2a 4

109. S(x) = V9 - xa 4T(x) = Vx - 9a110.

0Applications

Solve each problem.

111. Wind chill. The wind chill temperature W (how cold the air feels) is determined by the air temperature t and Figure for Exercise 113

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9-11 9.1 Radicals 567

a) Use the properties of radicals to simplify this 118. Landing speed and weight. Because the gross weight formula.

b) Find the exact time (according to the formula) that it takes for a diver to hit the water when diving from a height of 40 feet.

c) Use the graph to estimate the height if a diver takes 2.5 seconds to reach the water.

114. Sky diving. The formula in Exercise 113 accounts for the effect of gravity only on a falling object. According to that formula, how long would it take a sky diver to reach the earth when jumping from 17,000 feet? (A sky diver can actually get about twice as much falling time by spreading out and using the air to slow the fall.)

115. Maximum sailing speed. To find the maximum possi ble speed in knots (nautical miles per hour) for a sail boat, sailors use the function M = 1.3Vwa, where w is the length of the waterline in feet. If the waterline for the sloop Golden Eye is 20 feet, then what is the maxi mum speed of the Golden Eye?

116. America’s Cup. Since 1988 basic yacht dimensions for the America’s Cup competition have satisfied the inequality

3 DL + 1.25VSa - 9.8Va - 16.296,

where L is the boat’s length in meters (m), S is the sail area in square meters (m2), and D is the displacement in cubic meters (www.sailing.com). A team of naval architects is planning to build a boat with a displacement of 21.44 cubic meters (m3), a sail area of 320.13 m2, and a length of 21.22 m. Does this boat satisfy the inequality? If the length and displacement of this boat cannot be changed, then how many square meters of sail area must be removed so that the boat satisfies the inequality?

117. Landing speed. The proper landing speed for an airplane V (in feet per second) is determined from the gross weight of the aircraft L (in pounds), the coefficient of lift C, and the wing surface area S (in square feet), by the formula

41L V = J8 . CS

a) Find V (to the nearest tenth) for the Piper Cheyenne, for which L = 8700 lb, C = 2.81, and S = 200 ft2.

b) Find V in miles per hour (to the nearest tenth).

of the Piper Cheyenne depends on how much fuel and cargo are on board, the proper landing speed (from Exercise 117) is not always the same. The formula V = V1.496L gives the landing speed in terms ofa the gross weight only.

a) Find the landing speed if the gross weight is 7000 lb. b) What gross weight corresponds to a landing speed of

115 ft/sec?

Getting More Involved

119. Cooperative learning

Work in a group to determine whether each equation is an identity. Explain your answers.

2 3 3 a) Vxa = x b) Vxa = x 44 2 4c) Vxa = x d) Vxa = x

n For which values of n is Vxan = x an identity?

120. Cooperative learning

Work in a group to determine whether each inequality is correct.

a) V0.9a > 0.9

b) V1.01a > 1.01

c) V0.993 a > 0.99

d) V1.0013 a > 1.001

nFor which values of x and n is Vxa > x?

121. Discussion

If your test scores are 80 and 100, then the arithmetic mean of your scores is 90. The geometric mean of the scores is a number h such that

80 h = .

h 100

Are you better off with the arithmetic mean or the geometric mean?

http:www.sailing.com

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568 Chapter 9 Radicals and Rational Exponents 9-12

Math at Work Deficit and Debt

Have you ever heard politicians talk about budget surpluses and lowering the deficit, while the national debt keeps increasing? The national debt has increased every year since 1967 and stood at $11.3 trillion in 2009. Confusing? Not if you know the definitions of these words. If the federal government spends more than it collects in taxes in a particular year, then it has a deficit. The amount that is overspent must be borrowed, and that adds to the national debt, which is the total amount that the federal government owes. Interest alone on the national debt was $676 billion in 2009 and is the second largest expense in the federal budget.

To get an idea of the size of the national debt, divide the $11.3 trillion debt in 2009 by the U.S. population of 306 million to get about $37,000 per person. The national debt went

from $2.4 trillion in 1987 to $11.3 trillion in 2009. We can calculate the average annual percentage increase in the debt for these 22 years using the formula i = n A/P - 1, which yields i = 22 11.3/2 .4 - 1 = 7.3%. With the U.S. population increasing an average of 1% per year and the debt increasing 7.3% per year, in 25 years the debt will be 11.3(1 + 0.073)25 or about $65.8 trillion while the population will increase to 306(1 + 0.01)25 or about 392 million. See the accompanying figure. So in 25 years the debt will be about $168,000 per person. Since only one person in three is a wage earner, the debt will be about one-half of a million dollars per wage earner!

50 10 15 20 25

10 20 30 40 50

Years since 2006

N at

io na

l d eb

t ($

tr ill

io n)

9.2 Rational Exponents

You have learned how to use exponents to express powers of numbers and radicals to express roots. In this section, you will see that roots can be expressed with exponents also. The advantage of using exponents to express roots is that the rules of exponents can be applied to the expressions.

In This Section

U1V Rational Exponents

U2V Using the Rules of Exponents

U3V Simplifying Expressions Involving Variables

U Calculator Close-Up V

You can find the fifth root of 2 using radical notation or exponent nota tion. Note that the fractional expo nent 1/5 must be in parentheses.

U1V Rational Exponents Cubing and cube root are inverse operations. For example, if we start with 2 and apply

3 23both operations we get back 2: = 2. If we were to use an exponent for cube root, then we must have (23)? = 2. The only exponent that is consistent with the power of

1a power rule is because (23)1/3 = 21 = 2. So we make the following definition. 3

1/nDefinition of a

If n is any positive integer, then

1/n = a n a , nprovided that a is a real number.

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9-13 9.2 Rational Exponents 569

Later in this section we will see that using exponent 1/n for the nth root is com patible with the rules for integral exponents that we already know.

E X A M P L E 1 Radicals or exponents Write each radical expression using exponent notation and each exponential expression using radical notation.

a) V3 35a b) V4 xya c) 51/2 d) a1/5

Solution a) V3 35a = 351/3 b) V4 xya = (xy)1/4

c) 51/2 = V5a d) a1/5 = V5 aa

U Helpful Hint V m/nNote that in a we do not require

m/n to be reduced. As long as the nth m/nroot of a is real, then the value of a

is the same whether or not m/n is in lowest terms.

Now do Exercises 1–8

In Example 2, we evaluate some exponential expressions.

E X A M P L E 2 Finding roots Evaluate each expression.

a) 41/2 b) (-8)1/3 c) 811/4

d) (-9)1/2 e) -91/2

Solution a) 41/2 = V4a = 2 b) (-8)1/3 = V3 -8a = -2 c) 811/4 = V4 81a = 3 d) Because (-9)1/2 or V-9a is an even root of a negative number, it is not

a real number.

e) Because the exponent in -an is applied only to the base a (Section 1.5), we have -91/2 = -V9a = -3.

Now do Exercises 9–16

We now extend the definition of exponent 1/n to include any rational number as an exponent. The numerator of the rational number indicates the power, and the denominator indicates the root. For example, the expression

Power↓ 82/3 ← Root

represents the square of the cube root of 8. So we have

82/3 = (81/3)2 = (2)2 = 4.

m/nDefinition of a

If m and n are positive integers and a1/n is a real number, then

am/n = (a1/n)m . n

Using radical notation, am/n = (Vaa)m.

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570 Chapter 9 Radicals and Rational Exponents 9-14

By definition am/n is the mth power of the nth root of a. However, am/n is also equal to the nth root of the mth power of a. For example,

82/3 = (82)1/3 = 641/3 = 4.

Evaluating am/n in Either Order

If m and n are positive integers and a1/n is a real number, then

am/n = (a1/n)m = (am)1/n . n nUsing radical notation, am/n = (Vaa)m = Vaam.

A negative rational exponent indicates a reciprocal:

-m/nDefinition of a

If m and n are positive integers, a * 0, and a1/n is a real number, then

1 a -m/n = /n . am

Using radical notation, a -m/n = n 1

. (Vaa)m

E X A M P L E 3 Radicals to exponents Write each radical expression using exponent notation.

a) V3 x2a b)

Solution

a) V3 x2a = x2/3 b) = m

1 3/4 = m

-3/41

V4 m3a

Now do Exercises 17–20

E X A M P L E 4 Exponents to radicals Write each exponential expression using radicals.

a) 52/3 b) a -2/5

Solution

a) 52/3 = V3 52a b) a -2/5 = 1 V5 a2a

Now do Exercises 21–24

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9-15 9.2 Rational Exponents 571

To evaluate an expression with a negative rational exponent, remember that the denominator indicates root, the numerator indicates power, and the negative sign indicates reciprocal:

a -m/n ↑ ↑ ↑

Root Power Reciprocal

The root, power, and reciprocal can be evaluated in any order. However, it is usually simplest to use the following strategy.

Strategy for Evaluating a�m/n

1. Find the nth root of a.

2. Raise your result to the mth power.

3. Find the reciprocal.

For example, to evaluate 8-2/3, we find the cube root of 8 (which is 2), square 2 to get 4,

then find the reciprocal of 4 to get 1. In print 8-2/3 could be written for evaluation as 1((81/3)2)-1 or

(81/3)2 .

E X A M P L E 5 Rational exponents Evaluate each expression.

a) 272/3 b) 4-3/2 c) 81-3/4 d) (-8)-5/3

Solution a) Because the exponent is 2/3, we find the cube root of 27 and then square it:

272/3 = (271/3)2 = 32 = 9

b) Because the exponent is -3/2, we find the square root of 4, cube it, and find the reciprocal:

4-3/2 = (41

1 /2)3

= 2

1 3 =

c) Because the exponent is -3/4, we find the fourth root of 81, cube it, and find the reciprocal:

81-3/4 = (81

1 1/4)3

= 3 1 3 = 2

1 7

Definition of negative exponent

d) (-8)-5/3 = ((-8 1 )1/3)5 = (-

1 2)5

= -

1 32

= - 3 1 2

U Calculator Close-Up V

A negative fractional exponent indi cates a reciprocal, a root, and a power. To find 4-3/2 you can find the recipro cal first, the square root first, or the third power first as shown here.

Now do Exercises 25–36

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572 Chapter 9 Radicals and Rational Exponents 9-16

CAUTION An expression with a negative base and a negative exponent can have a positive or a negative value. For example,

1 15/3 2/3(-8)- = - and (-8)- = . 32 4

U2V Using the Rules of Exponents All of the rules for integral exponents that you learned in Sections 4.1 and 4.2 hold for rational exponents as well. We restate those rules in the following box. Note that some expressions with rational exponents [such as (-3)3/4] are not real numbers and the rules do not apply to such expressions.

Rules for Rational Exponents

The following rules hold for any nonzero real numbers a and b and rational num bers r and s for which the expressions represent real numbers.

saras = ar+ r

1. Product rule a s2. = ar- Quotient rulesa

= ars3. (ar)s Power of a power rule

4. (ab)r = arbr Power of a product rule r a

5. �ab� r

= Power of a quotient rulerb

We can use the product rule to add rational exponents. For example,

161/4 · 161/4 = 162/4.

The fourth root of 16 is 2, and 2 squared is 4. So 162/4 = 4. Because we also have 161/2 = 4, we see that a rational exponent can be reduced to its lowest terms. If an exponent can be reduced, it is usually simpler to reduce the exponent before we eval uate the expression. We can simplify 161/4 · 161/4 as follows:

161/4 · 161/4 = 162/4 = 161/2 = 4

E X A M P L E 6 Using the product and quotient rules with rational exponents Simplify each expression.

a) 271/6 · 271/2 b) 5 5

Solution a) 271/6 · 271/2 = 271/6+1/2 Product rule for exponents

= 272/3

= 9

b) 5 5

4 = 5 3/4-1/4 = 52/4 = 51/2 = V5a We used the quotient rule to

subtract the exponents.

Now do Exercises 37–44

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9-17 9.2 Rational Exponents 573

E X A M P L E 7 Using the power rules with rational exponents Simplify each expression.

a) 31/2 · 121/2 b) (310)1/2 c) �2 3 6

9� -1/3

Solution a) Because the bases 3 and 12 are different, we cannot use the product rule to add the

exponents. Instead, we use the power of a product rule to place the 1/2 power outside the parentheses:

31/2 · 121/2 = (3 · 12)1/2 = 361/2 = 6

b) Use the power of a power rule to multiply the exponents:

(310)1/2 = 35

c) �2 3 6

9� -1/3

= ( ( 2 3

) ) -

3 Power of a quotient rule

= 2 3

3 Power of a power rule

= 3 2

2 Definition of negative exponent

= 2 4 7

U Helpful Hint V

We usually think of squaring and tak ing a square root as inverse operations, which they are as long as we stick to positive numbers. We can square 3 to get 9, and then find the square root of 9 to get 3—what we started with. We don’t get back to where we began if we start with -3.

Now do Exercises 45–54

U3V Simplifying Expressions Involving Variables When simplifying expressions involving rational exponents and variables, we must be careful to write equivalent expressions. For example, in the equation

(x2)1/2 = x

it looks as if we are correctly applying the power of a power rule. However, this state ment is false if x is negative because the 1/2 power on the left-hand side indicates the positive square root of x2. For example, if x = -3, we get

[(-3)2]1/2 = 91/2 = 3,

which is not equal to -3. To write a simpler equivalent expression for (x2)1/2, we use absolute value as follows.

2Square Root of x

For any real number x,

(x = x Vx2)1/2 and a2 = x .

2)1/2 2Note that both (x = and Vx x are identities. They are true whether x isx a = positive, negative, or zero.

It is also necessary to use absolute value when writing identities for other even roots of expressions involving variables.

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574 Chapter 9 Radicals and Rational Exponents 9-18

E X A M P L E 8 Using absolute value symbols with roots Simplify each expression. Assume the variables represent any real numbers and use absolute value symbols as necessary.

a) (x8y4)1/4 b) �x 8 9

� 1/3

Solution a) Apply the power of a product rule to get the equation (x8y4)1/4 = x2y. The left-hand

side is nonnegative for any choices of x and y, but the right-hand side is negative when y is negative. So for any real values of x and y we have

(x8y4)1/4 = x2 y .

Note that the absolute value symbols could also be placed around the entire expression:

(x8y4)1/4 = x2 y . b) Using the power of a quotient rule, we get

�x 8 9

� 1/3

= x 2

This equation is valid for every real number x, so no absolute value signs are used.

Now do Exercises 55–64

Because there are no real even roots of negative numbers, the expressions 1/2 x -3/4 1/6a , , and y

are not real numbers if the variables have negative values. To simplify matters, we sometimes assume the variables represent only positive numbers when we are work ing with expressions involving variables with rational exponents. That way we do not have to be concerned with undefined expressions and absolute value.

E X A M P L E 9 Expressions involving variables with rational exponents Use the rules of exponents to simplify the following. Write your answers with positive exponents. Assume all variables represent positive real numbers.

a1/2 x2 a) x2/3x4/3 b) c) (x1/2 y -3)1/2 d) �y /3�

-1/2

a1/4 1

Solution 2/3 4/3 6/3a) x x = x Use the product rule to add the exponents.

= x2 Reduce the exponent.

a1/2 1/2-1/4b) = a Use the quotient rule to subtract the exponents. a1/4

1/4= a Simplify.

1/2 1/2)1/2(y -3)1/2c) (x y -3)1/2 = (x Power of a product rule 1/4y -3/2= x Power of a power rule

x1/4 = Definition of negative exponent3/2y

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9-19 9.2 Rational Exponents 575

Warm-Ups ▼

Fill in the blank. 1. The notation a1/n represents the of a.

2. The notation am/n represents the of the nth root.

3. The expression a-m/n is the of am/n .

4. The expression a-m/n is a real number except when n is and a is , or when a = 0.

True or false? 5. 91/3 = V

3 96

6. 85/3 = V 5

836 7. (-16)1/2 = -161/2

8. 9-3/2 = 2 1 7

9. 6-1/2 = V 6 66

10. 21/2 . 21/2 = 41/2

11. 61/6 . 61/6 = 61/3

12. (28)3/4 = 26

9 .2

d) Because this expression is a negative power of a quotient, we can first find the reciprocal of the quotient and then apply the power of a power rule:

( y x 1 2

/3 ) -1/2

= ( y x 1/

) 1/2

= y1

3 .

2 =

Now do Exercises 65–76

Exercises

U Study Tips V • Avoid cramming. When you have limited time to study for a test, start with class notes and homework assignments. Work one or two

problems of each type. • Don’t get discouraged if you cannot work the hardest problems. Instructors often ask some relatively easy questions to see if you

understand the basics.

1/2 U1V Rational Exponents 7. a

8. (-b)1/5 Write each radical expression using exponent notation. See Example 1. Evaluate each expression. See Example 2.

1. V7 2. V64 6 3 cbs 9. 251/2 10. 161/2 3. V65x 4. V3y6

11. (-125)1/3 12. (-32)1/5 Write each exponential expression using radical notation.

13. 161/4 14. 81/3See Example 1.

5. 91/5 15. (-4)1/2

6. 31/2 16. (-16)1/4

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9-20 576 Chapter 9 Radicals and Rational Exponents

Write each radical expression using exponent notation. See Example 3.

3 17. Vwa7 18. Vaa5

1 19. 20. J3 1 --3 210Va a2

Write each exponential expression using radical notation. See Example 4.

w -3/421. 22. 6-5/3

23. (ab)3/2 24. (3m)-1/5

Evaluate each expression. See Example 5. See the Strategy for Evaluating a-m/n box on page 571.

25. 1252/3 26. 10002/3

27. 253/2 28. 163/2

29. 27-4/3 30. 16-3/4

31. 16-3/2 32. 25-3/2

33. (-27)-1/3 34. (-8)-4/3

35. (-16)-1/4

36. (-100)-3/2

U2V Using the Rules of Exponents

Use the rules of exponents to simplify each expression. See Examples 6 and 7.

37. 31/331/4 38. 21/221/3

39. 31/33-1/3 40. 51/45-1/4

1/3 -2/38 27 41. 42.2/3 -1/38 27

43. 43/4 - 41/4 44. 91/4 - 93/4

45. 181/221/2 46. 81/221/2

47. (26)1/3 48. (310)1/5

49. (38)1/2 50. (3-6)1/3

51. (2-4)1/2 52. (54)1/2

4 1/2 4 1/2 53. 54.�2

3 � �3 5 �6 6

U3V Simplifying Expressions Involving Variables

Simplify each expression. Assume the variables represent any real numbers and use absolute value as necessary. See Example 8.

55. (x4)1/4 56. (y6)1/6

57. (a8)1/2 58. (b10)1/2

59. (y3)1/3 60. (w9)1/3 6 2)1/2 8b4)1/461. (9x y 62. (16a

12 1/4 8 1/281x 144a 63. 64.20 18� y � � 9y �

Simplify. Assume all variables represent positive numbers. Write answers with positive exponents only. See Example 9.

1/2 1/4 1/3 1/365. x x 66. y y

1/2 1/2)67. (x y)(x -3/4y 68. (a1/2b-1/3)(ab)

w1/3 a1/2 69. 70.3 2w a

16)1/2 8)1/371. (144x 72. (125a -1/2 -4 1/2 6a 2a

73. 74.�b � � b �-1/4 1/3 1/3 3 -1/2 -3

75. 76.�2w -3/ � �a 2/3�4w 3a

Miscellaneous

Simplify each expression. Write your answers with positive exponents. Assume that all variables represent positive real numbers.

77. (92)1/2 78. (416)1/2

79. -16-3/4 80. -25-3/2

81. 125-4/3 82. 27-2/3

83. 21/22-1/4 84. 9-191/2

85. 30.2630.74 86. 21.520.5

87. 31/4271/4 88. 32/392/3

8 8 89. �- 7�

2/3 90. �- 7�

-1/3

2 2

1 5 91. �- 6�

-3/4 92. �-9�

-7/2

9 6 93. �16�

-1/2 94. �11�

-1/4

http:30.2630.74

oscar.vasquezmolina

Highlight

oscar.vasquezmolina

Sticky Note

solve this problem

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9-21

5 27� -4/3

95. -�26� -3/2

96. �-3 8 97. (9x9)1/2 98. (-27x9)1/3

99. (3a -2/3)-3 100. (5x -1/2)-2

1/4 1/2)2(m2 3)1/2101. (a1/2b)1/2(ab1/2) 102. (m n n

103. (km1/2)3(k3m5)1/2 104. (tv1/3)2(t2v -3)-1/2

Use a scientific calculator with a power key (xy) to find the decimal value of each expression. Round approximate answers to four decimal places.

21/3 51/2105. 106.

107. -21/2 108. (-3)1/3

109. 10241/10 110. 77760.2

-3/564 32 111. 112.�15,625�

-1/6 �243�

Simplify each expression. Assume a and b are positive real numbers and m and n are rational numbers.

am/2 · b-n/3113. · am/4 114. bn/2

-m/5 -n/4a b 115. 116. -m/3 -n/3a b

(a -1/mb-1/n)-mn (a -m/2b-n/3)-6117. 118.

m -6n -3/m 6/n

�a -3 b �

-1/3 b n�

-1/3

�a 119. 120.9m -6/m 9/a a b

Applications

Solve each problem. Round answers to two decimal places when necessary.

121. Falling object. The time in seconds t that it takes for a ball to fall to the earth from a height of h feet is given by the function

t(h) = 0.25h1/2.

Find t(1), t(16), and t(36).

9.2 Rational Exponents 577

122. Sailboat speed. The maximum speed for a sailboat in knots M is a function of the length of the waterline in feet w, given by

1/2M(w) = 1.3w .

Find M(19), M(24), and M(30) to the nearest hundredth.

123. Diagonal of a box. The length of the diagonal of a box D is a function of its length L, width W, and height H:

D = (L2 + W2 + H2)1/2

a) Find D for the box shown in the accompanying figure.

b) Find D if L = W = H = 1 inch.

3 in. 4 in.

12 in.

Figure for Exercise 123

124. Radius of a sphere. The radius of a sphere is given by the function

� � � 1/30.75V

r = ,

where V is its volume. Find the radius of a spherical tank that has a volume of 32� cubic meters.

Figure for Exercise 124

125. Maximum sail area. According to the new International America’s Cup Class Rules, the maximum sail area in square meters for a yacht in the America’s Cup race is given by the function

S = (13.0368 + 7.84D1/3 - 0.8L)2,

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578 Chapter 9 Radicals and Rational Exponents 9-22

where D is the displacement in cubic meters (m3), and L (www.money.com). Find the 5-year average annual is the length in meters (m) (www.sailing.com). Find the return. maximum sail area for a boat that has a displacement of 18.42 m3 and a length of 21.45 m.

128. Top bond fund. An investment of $10,000 in the Templeton Global Bond Fund in 2004 was worth $14,789 in 2009 (www.money.com). Use the formula from Exercise 127 to find the 5-year average annual return.

129. Overdue loan payment. In 1777 a wealthy Pennsylvania

S (m2)

D (m3) L (m)

merchant, Jacob DeHaven, lent $450,000 to the Continental Congress to rescue the troops at Valley Forge. The loan was not repaid. In 1990 DeHaven’s Figure for Exercise 125 descendants filed suit for $141.6 billion (New York Times, May 27, 1990). What average annual rate of return were they using to calculate the value of the debt after

126. Orbits of the planets. According to Kepler’s third law of planetary motion, the average radius R of the orbit of a planet around the sun is determined by R = T 2/3, where T is the number of years for one orbit and R is measured in astronomical units or AUs (Windows to the Universe,

213 years? (See Exercise 127.)

130. California growin’. The population of California grew www.windows.umich.edu). from 19.9 million in 1970 to 32.5 million in 2000 a) It takes Mars 1.881 years to make one orbit of the sun.

What is the average radius (in AUs) of the orbit of Mars? b) The average radius of the orbit of Saturn is 9.5 AU.

Use the accompanying graph to estimate the number of years it takes Saturn to make one orbit of the sun.

(U.S. Census Bureau, www.census.gov). Find the average annual rate of growth for that time period. (Use the formula from Exercise 127 with P being the initial population and S being the population n years later.)

C al

if or

ni a

po pu

la tio

n (m

ill io

ns o

f pe

op le

)

R ad

iu s

of o

rb it

(A U

)

R = T 2/3

0 10 20 30

10 2 1970 1980 1990 2000

Year

258 20

6 15

Time for one orbit (years) Figure for Exercise 130

Figure for Exercise 126

127. Top stock fund. The average annual return r is a function of the initial investment P, the number of years n, and the amount S that it is worth after n years:

�P S�

1/n r = - 1

An investment of $10,000 in the T. Rowe Price Latin America Fund in 2004 was worth $20,733 in 2009

Getting More Involved

131. Discussion

Determine whether each equation is an identity. Explain.

2 2)1/2a) (w x = w · x 2 2)1/2b) (w x = wx 2c) (w x2)1/2 = w x

http:www.census.gov

http:www.windows.umich.edu

http:www.money.com

http:www.sailing.com

http:www.money.com

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9-23 9.3 Adding, Subtracting, and Multiplying Radicals 579

9.3 Adding, Subtracting, and Multiplying Radicals

In this section, we will use the ideas of Section 9.1 in performing arithmetic operations with radical expressions.

In This Section

U1V Adding and Subtracting Radicals

U2V Multiplying Radicals

U3V Conjugates

U4V Multiplying Radicals with Different Indices

U1V Adding and Subtracting Radicals To find the sum of V2a and V3a, we can use a calculator to get Va2 � 1.414 and Va3 � 1.732. (The symbol � means “is approximately equal to.”) We can then add the decimal numbers and get

V2a + V3a � 1.414 + 1.732 = 3.146.

We cannot write an exact decimal form for V2a + V3a; the number 3.146 is an approx imation of V2a + Va3. To represent the exact value of V2a + Va3, we just use the form V2a + V3a. This form cannot be simplified any further. However, a sum of like radicals can be simplified. Like radicals are radicals that have the same index and the same radicand.

To simplify the sum 3V2a + 5V2a, we can use the fact that 3x + 5x = 8x is true for any value of x. Substituting V2a for x gives us 3V2a + 5Va2 = 8Va2. So like radicals can be combined just as like terms are combined.

E X A M P L E 1 Adding and subtracting like radicals Simplify the following expressions. Assume the variables represent positive numbers.

a) 3V5a + 4V5a b) V4 wa - 6V4 wa c) V3a + V5a - 4V3a + 6V5a d) 3V3 6xa + 2V3 xa + V3 6xa + V3 xa

Solution a) 3V5a + 4V5a = 7V5a b) V4 wa - 6V4 wa = -5V4 wa c) V3a + V5a - 4V3a + 6V5a = -3V3a + 7V5a Only like radicals are combined. d) 3V3 6xa + 2V3 xa + V3 6xa + V3 xa = 4V3 6xa + 3V3 xa

Now do Exercises 1–12

Remember that only radicals with the same index and same radicand can be combined by addition or subtraction. If the radicals are not in simplified form, then they must be simplified before you can determine whether they can be combined.

E X A M P L E 2 Simplifying radicals before combining Perform the indicated operations. Assume the variables represent positive numbers.

3 2 3a) V8a + V18 b) V2x - V4x aa a a + 5V18x 3 34 3 4 3c) V16ax y - V54ax y

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580 Chapter 9 Radicals and Rational Exponents 9-24

Solution a) V8a + V18a = V4a · V2a + V9a · V2a

= 2V2a + 3V2a Simplify each radical.

= 5V2a Add like radicals.

Note that V8a + V18a * V26a.

b) V2x3a - V4x2a + 5V18x3a = Vx2a · V2xa - 2x + 5 · V9x2a · V2xa = xV2xa - 2x + 15xV2xa Simplify each radical.

= 16xV2xa - 2x Add like radicals only.

c) V3 16x4y3a - V3 54x4y3a = V3 8x3y3a · V3 2xa - V3 27x3y3a · V3 2xa = 2xyV3 2xa - 3xyV3 2xa Simplify each radical.

= -xyV3 2xa

U Calculator Close-Up V

Check that V8a + V18a = 5V2a.

Now do Exercises 13–28

U2V Multiplying Radicals n n nThe product rule for radicals, Vaa · Vb = Va, allows multiplication of radicals witha ab

the same index, such as

3 3 3 5 2 5 5 3V5a a a, V2 · Va = V10 and Vx x = Va.· V3 = V15 a 5 a, a · Va x

CAUTION The product rule does not allow multiplication of radicals that have dif 3ferent indices. We cannot use the product rule to multiply V2a and Va5.

E X A M P L E 3 Multiplying radicals with the same index Multiply and simplify the following expressions. Assume the variables represent positive numbers.

a) 5V6a · 4V3a b) V3a2a · V6aa

c) V3 4a · V3 4a d) J4 x 2 3

· J4 x 8 2

Solution a) 5V6a · 4V3a = 5 · 4 · V6a · V3a

= 20V18a Product rule for radicals

= 20 · 3V2a V18a = V9a · V2a = 3V2a

= 60V2a

b) V3a2a · V6aa = V18a3a Product rule for radicals = V9a2a · V2aa = 3aV2aa Simplify.

U Helpful Hint V

Students often write

V15a · V15a = V225a = 15.

Although this is correct, you should get used to the idea that

V15a · V15a = 15.

Because of the definition of a square root, Vaa · Vaa = a for any positive number a.

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9-25 9.3 Adding, Subtracting, and Multiplying Radicals 581

c) V3 4s . V3 4s = V3 16s = V3 8s . V3 2s Simplify. = 2V3 2s

d) #4 x 2 3

. #4 x 8 2

= #4 1 x 6 5

Product rule for radicals = Product and quotient rules for radicals

= xV

4 xs Simplify.

V4 x4s . V4 xs

V4 16s

Now do Exercises 29-42

We find a product such as 3V2s(4V2s - Vs3) by using the distributive prop erty as we do when multiplying a monomial and a binomial. A product such as (2V3s + Vs5)(3Vs3 - 2Vs5) can be found by using FOIL as we do for the product of two binomials.

E X A M P L E 4 Multiplying radicals Multiply and simplify.

a) 3V2s (4V2s - V3s ) b) V3 as (V3 as - V3 a2s) c) (2V3s + V5s )(3V3s - 2V5s ) d) (3 + Vx - 9s )2

Solution a) 3V2s (4V2s - V3s) = 3V2s . 4V2s - 3V2s . V3s

= 12 . 2 - 3V6s

= 24 - 3V6s

b) V3 as (V3 as - V3 a2s) = V3 a2s - V3 a3s Distributive property = V3 a2s - a

c) (2V3s + V5s )(3V3s - 2V5s ) F O I L

= 2V3s . 3V3s - 2V3s . 2V5s + V5s . 3V3s - V5s . 2V5s = 18 - 4V15s + 3V15s - 10 = 8 - V15s Combine like radicals.

d) To square a sum, we use (a + b)2 = a2 + 2ab + b2:

(3 +Vx - 9s)2 = 32 + 2 . 3Vx - 9s + (V x - 9s)2

= 9 + 6Vx - 9s + x - 9 = x + 6Vx - 9s

Distributive property Because V2s . V2s = 2 and V2s . V3s = V6s

��

Now do Exercises 43-56

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582 Chapter 9 Radicals and Rational Exponents 9-26

CAUTION We can’t simplify Vxa- 9 in Example 4(d), because in general Va - ba * Vaa - Vb a6 = V9 = 3 and V25 a = 1.a. For example, V 25 - 1a a a - V16 Find an example where Vaa + b * Vaa + Vab.

U3V Conjugates 2Recall the special product rule (a + b)(a - b) = a - b2. The product of the sum

4 + V3a and the difference 4 - V3a can be found by using this rule:

(4 + V3a)(4 - V3a) = 42 - (V3a)2 = 16 - 3 = 13

The product of the irrational number 4 + V3a and the irrational number 4 - V3a is the rational number 13. For this reason the expressions 4 + V3a and 4 - V3a are called conjugates of one another. We will use conjugates in Section 9.4 to rationalize some denominators.

E X A M P L E 5 Multiplying conjugates Find the products. Assume the variables represent positive real numbers.

a) (2 + 3V5a )(2 - 3V5a )

b) (V3a - V2a )(V3a + V2a )

c) (V2xa - Vya )(V2xa + Vya )

Solution a) (2 + 3V5a )(2 - 3V5a ) = 22 - (3V5a )2 (a + b)(a - b) = a2 - b2

= 4 - 45 (3V5a)2 = 9 · 5 = 45

= -41

b) (V3a - V2a )(V3a + V2a ) = 3 - 2 = 1

c) (V2xa - Vya )(V2xa + Vya ) = 2x - y Now do Exercises 57–66

U4V Multiplying Radicals with Different Indices The product rule for radicals applies only to radicals with the same index. To multiply radicals with different indices we convert the radicals into exponential expressions with rational exponents. If the exponential expressions have the same base, apply the product rule for exponents (am · an = am+n) to get a single exponential expression and then convert back to a radical [Example 6(a)]. If the bases of the exponential expression are different, get a common denominator for the rational exponents,

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9-27 9.3 Adding, Subtracting, and Multiplying Radicals 583

n nconvert back to radicals and then apply the product rule for radicals (Vaa · Vab = nVaba) to get a single radical expression [Example 6(b)].

Multiplying radicals with different indices Write each product as a single radical expression.

a) V3 2a · V4 2a b) V3 2a · V3a

Solution a) V3 2a · V4 2a = 21/3 · 21/4 Write in exponential notation.

= 27/12 Product rule for exponents: 1 3

+ 1 4

= 1 7 2

= V12 27a Write in radical notation. = V12 128a

b) V3 2a · V3a = 21/3 · 31/2 Write in exponential notation.

= 22/6 · 33/6 Write the exponents with the LCD of 6.

= V6 22a · V6 33a Write in radical notation. = V6 22 · 33a Product rule for radicals = V6 108a 22 · 33 = 4 · 27 = 108

E X A M P L E 6

U Calculator Close-Up V

Check that

V3 2a · V4 2a = V12 128a.

Now do Exercises 67–74

CAUTION Because the bases in 21/3 · 21/4 are identical, we can add the exponents [Example 6(a)]. Because the bases in 22/6 · 33/6 are not the same, we cannot add the exponents [Example 6(b)]. Instead, we write each factor as a sixth root and use the product rule for radicals.

Warm-Ups ▼

Fill in the blank. 1. radicals have the same index and the same

radicand.

2. The property is used to combine like radicals.

3. The product rule for radicals is used to multiply radicals with the same .

4. The of 2 - V3a is 2 + V3a.

True or false? 5. V3a + V3a = V6a

6. V8a + V2a = 3V2a

7. 2V3a · 3V3a = 6V3a

8. 2V5a · 3V2a = 6V10a

9. V 3

2a · V 3

2a = 2

10. V2a(V3a - V2a) = V6a - 2 11. (V2a + V3a)2 = 2 + 3 12. (V3a - V2a)(V3a + V2a) = 1

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9 .3584 Chapter 9 Radicals and Rational Exponents 9-28Exercises

U Study Tips V • If you must miss class, let your instructor know. Be sure to get notes from a reliable classmate. • Take good notes in class for yourself and your classmates.You never know when a classmate will ask to see your notes.

U1V Adding and Subtracting Radicals

All variables in the following exercises represent positive numbers. Simplify the sums and differences. Give exact answers. See Example 1.

1. Vs3 - 2Vs3 2. Vs5 - 3Vs5

3. 5Vs + 4V7x 4. 3Vs + 7Vs7x s 6a 6a

3 3 3 35. 2Vs2 + 3Vs2 6. Vs4 + 4Vs4

7. Vs3 - Vs5 + 3Vs3 - Vs5

8. Vs2 - 5Vs3 - 7Vs2 + 9Vs3

3 3 3 39. Vs2 + Vsx - Vs2 + 4Vsx 3 3 3 310. V5y 5y x xs - 4Vs + Vs + Vs 3 3s + V11. Vsx - V2x sx 3 312. Vab s + 5Va abs + Va s + Vs

Simplify each expression. Give exact answers. See Example 2.

13. Vs8 + V28s

14. V12 ss + V24

15. Vs8 + V18 16. V12 ss s + V27

17. 2Vs45 - 3Vs20 s - 2V3218. 3V50 s

19. Vs2 - V8 20. V20 ss s - V125

3 2 2 321. V45x s + V50x ss - V18x s - V20x 5 522. V12x s - Vs + Vss - V18x 300x 98x

3 323. 2V24 81s + Vs

3 324. 5V24 375s + 2Vs

4 425. V48 243s - 2Vs

5 526. V64 2s + 7Vs 3 33 327. V y 16t4y54t4 ss - V 3 3s2 5 z528. V2000wsz - V16sw2

U2V Multiplying Radicals

Simplify the products. Give exact answers. See Examples 3 and 4.

29. V3s . V5s 30. V5s . V7s

31. 2V5s . 3V10s 32. (3V2s)(-4V10s)

33. 2V7as . 3V2as 34. 2V5cs . 5V5s

35. V4 9s . V4 27s 36. V3 5s . V3 100s

37. (2V3s )2 38. (-4V2s )2

39. V5x3s . V8x4s

40. V3b3s . V6b5s

41. f4 ix 3 5

i . f4 i2 x 7 2 i

42. f3 ia 2 4

i . f3 ia 4 3

i 43. 2V3s(V6s + 3V3s)

44. 2V5s(V3s + 3V5s)

45. V5s(V10s - 2)

46. V6s(V15s - 1)

47. V3 s3t(V3 s9t - V3 t2s)

48. V3 2s(V3 12xs - V3 2xs)

49. (V3s + 2)(V3s - 5)

50. (V5s + 2)(V5s - 6)

51. (V11s - 3)(V11s + 3)

52. (V2s + 5)(V2s + 5)

53. (2V5s - 7)(2V5s + 4)

54. (2V6s - 3)(2V6s + 4)

55. (2Vs3 - Vs6)(Vs3 + 2Vs6)

56. (3Vs3 - Vs2)(Vs2 + Vs3)

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9-29

U3V Conjugates

Find the product of each pair of conjugates. See Example 5.

57. (V3a - 2)(Va3 + 2) 58. (7 - V3a)(7 + V3a) 59. (V5a + Va2)(V5a - V2a) 60. (V6a + Va5)(V6a - V5a) 61. (2V5a + 1)(2V5a - 1) 62. (3V2a - 4)(3Va2 + 4) 63. (3V2a + Va5)(3V2a - V5a) 64. (2V3a - V7a)(2Va3 + V7a) 65. (5 - 3Vxa)(5 + 3Vxa) 66. (4Vya + 3Vaz)(4Vya - 3Vza)

U4V Multiplying Radicals with Different Indices

Write each product as a single radical expression. See Example 6.

367. Va3 · Va3 3 469. Va5 · V5a 371. Va2 · Va5 3 473. Va2 · V3a

Miscellaneous

Simplify each expression.

75. V300 aa + V3 77. 2V5a · 5Va6

79. (3 + 2V7a)(V7a - 2) 80. (2 + V7a)(V7a - 2) 82. 3Vma · 5Vma

84. V2at5 · V10t4a 85. (2V5a + Va2)(3V5a - V2a) 86. (3V2a - V3a)(2Va2 + 3V3a)

Va2 Va2 87. +

3 5

Va2 Va3 88. +

4 5

89. (5 + 2V2a)(5 - 2V2a)

90. (3 - 2V7a)(3 + 2V7a)

91. (3 + Vxa)2

92. (1 - Vxa)2

93. (5Vxa - 3)2

468. V3a · Va3 3 570. Va2 · V2a

372. V6a · Va2 3 474. Va3 · V2a

76. V50 aa + V2 78. 3V6a a· 5V10

81. 4Vwa · 4Vwa

83. V3ax3 · V6ax2

9.3 Adding, Subtracting, and Multiplying Radicals 585

94. (3Vaa + 2)2

a)295. (1 + Vx + 2

a + 1)296. (Vx - 1

97. V4w aa - V9w

98. 10Vm - V16ma a

99. 2Va3 a3 - 2aV4aa + 3Va a 2 2 2100. 5Vway - 7Vway + 6Vway

101. Vxa5 + 2xVxa3

102. V8ax3 + V50ax3 - xV2ax 3103. V3 -a16x4 + 5xV54ax

5 7 5 7104. V3 3ax y - V3 24ax y 3 3 4105. Va · V2x 106. Va2m · Va2x a 2n

Applications

Solve each problem.

107. Area of a rectangle. Find the exact area of a rectangle that has a length of V6a feet and a width of V3a feet.

108. Volume of a cube. Find the exact volume of a cube with sides of length V3a meters.

109. Area of a trapezoid. Find the exact area of a trapezoid with a height of Va6 feet and bases of V3 a feet.a feet and V12

�3 ft

�3 m

�6 ft

�12 ft �3 m

Figure for Exercise 108 Figure for Exercise 109

110. Area of a triangle. Find the exact area of a triangle with a base of V30 a meters.a meters and a height of V6

�6 m

�30 m

Figure for Exercise 110

oscar.vasquezmolina

Highlight

oscar.vasquezmolina

Sticky Note

solve this problem.

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586 Chapter 9 Radicals and Rational Exponents 9-30

Getting More Involved 113. Exploration

Because 3 is the square of V3a, a binomial such as y2 - 3 is a difference of two squares.

111. Discussion

Is Vaa + Vba a= Va + b for all values of a and b? a) Factor y2 - 3 and 2a2 - 7 using radicals. b) Use factoring with radicals to solve the equation

112. Discussion 2x - 8 = 0. Which of the following equations are identities? c) Assuming a is a positive real number, solve the Explain your answers. equation x2 - a = 0.

a) V9x a a aa = 3Vx b) V9 + x = 3 + Vx x Vxa

c) Vx - 4 = Vx - 2 d) J4 =a a 2

Mid-Chapter Quiz Sections 9.1 through 9.3 Chapter 9

Simplify each radical expression. 15. 3V10a · 2Va14 a)(8 - V1016. (8 + V10 a) 3

1. V64 2. V-27a a 3

3. V120 4. V56 5 5a a 3 317. V8x 27xa + Va 7 3 b15. V12x 6. Vaa3a 3 24a

Miscellaneous. 3w x

7. J1 8. J8 18. Find the domain of the expression V6 - 3xa.6 9 9. 811/2 10. 1003/2

19. Find the solution set to Vx2a = x. 1/35

11. -163/2 12. �5-2/3� -3

4)1/420. Find the solution set to (x = lx l. 3

Perform the indicated operations. 21. Write the product V2a · Va2 as a single radical expression.

13. 2V3a a a + V6a 14. 9V20 a- 5V6 - 4V3 a - 3V45 22. Suppose that h(t) = 5t2/3. Find h(8).

9.4 Quotients, Powers, and Rationalizing Denominators In this section, we will continue studying operations with radicals. We will first learn how to rationalize denominators, and then we will find quotients and powers with radicals.

In This Section

U1V Rationalizing the Denominator

U2V Simplifying Radicals

U3V Dividing Radicals

U4V Rationalizing Denominators U1V Rationalizing the DenominatorUsing Conjugates

U5V Powers of Radical Square roots such as V2a, V3a, and Va5 are irrational numbers. If roots of this type Expressions appear in the denominator of a fraction, it is customary to rewrite the fraction with a

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9-31

E X A M P L E 1

U Helpful Hint V

If you are going to compute the value of a radical expression with a calculator, it does not matter if the denominator is rational. However, rationalizing the denominator pro vides another opportunity to prac tice building up the denominator of a fraction and multiplying radicals.

9.4 Quotients, Powers, and Rationalizing Denominators 587

rational number in the denominator, or rationalize it. We rationalize a denominator by multiplying both the numerator and denominator by another radical that makes the denominator rational.

You can find products of radicals in two ways. By definition, V2a is the positive number that you multiply by itself to get 2. So,

V2a · V2a = 2.

3 3 3By the product rule, V2a · V2a = V4a = 2. Note that Va2 · V2a = Va4 by the product 3rule, but V4a * 2. By definition of a cube root,

3 3 3Va2 · V2a · Va2 = 2.

Rationalizing the denominator Rewrite each expression with a rational denominator.

V3a 3 a) b) 3V5a V2a

Solution a) Because V5a · V5a = 5, multiplying both the numerator and denominator by V5a

will rationalize the denominator:

V3 V3 V5 V15a a a a = · By the product rule, V3 · V5a = V15= a a. V5a Va5 Va5 5

b) We must build up the denominator to be the cube root of a perfect cube. So we 3 3 3 3multiply by Va4 to get Va4 · Va2 = Va8:

3 3 33 3 V4a 3V4a 3V4a = · = = 3 3 3 3Va2 V2a Va4 Va8 2

Now do Exercises 1–8

CAUTION To rationalize a denominator with a single square root, you simply multiply by that square root. If the denominator has a cube root, you build the denominator to a cube root of a perfect cube, as in Example 1(b). For a fourth root you build to a fourth root of a perfect fourth power, and so on.

U2V Simplifying Radicals When simplifying a radical expression, we have three specific conditions to satisfy. First, we use the product rule to factor out perfect nth powers from the radicand in nth roots. That is, we factor out perfect squares in square roots, perfect cubes in cube roots, and so on. For example,

3 3 3 3a a a = V 3 = 2Va.V72a = Va36 · V2 = 6V2 and V24 a8 · Va 3

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588 Chapter 9 Radicals and Rational Exponents 9-32

Second, we use the quotient rule to remove all fractions from inside a radical. For example,

� 2 V26 .3 V63 Third, we remove radicals from denominators by rationalizing the denominator:

� 2 V62 . V63 V66 3 V36 . V36 3 A radical expression that satisfies the following three conditions is in simplified radical form.

Simplified Radical Form for Radicals of Index n

A radical expression of index n is in simplified radical form if it has

1. no perfect nth powers as factors of the radicand, 2. no fractions inside the radical, and 3. no radicals in the denominator.

E X A M P L E 2 Writing radical expressions in simplified radical form Simplify.

V10 6a) V66

5 3b) � 9

Solution a) To rationalize the denominator, multiply the numerator and denominator by V66:

6 6 V66V10 V10 . Rationalize the denominator. V66 V66 V66

V606

V64V156 6.Remove the perfect square from V60

2V156

V15 26 1 to 6 � 3 * V65.Reduce . Note that V15 3 6 3

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�

� � �

�

9-33 9.4 Quotients, Powers, and Rationalizing Denominators 589

b) To rationalize the denominator, build up the denominator to a cube root of a 3 3 3 3perfect cube. Because V9 3 V6 3, we multiply by V36 . V6 27 6:

5 3 3 V65

3 Quotient rule for radicals9 V96 3 3V56 V63 3 . 3 Rationalize the denominator. V96 V63 3 6V15 3 6V27

3 6V15 3

Now do Exercises 9–18

E X A M P L E 3 Rationalizing the denominator with variables Simplify each expression. Assume all variables represent positive real numbers.

3a xx 3a) b) c) 5b y y

Solution a V6a

a) Quotient rule for radicals b V6b

V6a . V6b Rationalize the denominator.

V6b . V6b

Vab6 b

6x3 Vx3 b) 5 Quotient rule for radicals y Vy56

Vx2 66 . Vx Product rule for radicals

Vy4 66 . Vy

xVx6 Simplify.

y 2Vy6

xV6x . V6y Rationalize the denominator.

y2Vy6 . Vy6

xVxy xVxy6 6 2 3y . y y

3 2c) Multiply by Vy6 to rationalize the denominator: 3 3 33 2 2 3 2x3 Vx V6 Vy V6 Vxy66 x 6 xy.

32 3V6y y Now do Exercises 19–28

y V3 y6V3 y6V3 y6

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� �

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590 Chapter 9 Radicals and Rational Exponents 9-34

U3V Dividing Radicals In Section 9.3 you learned how to add, subtract, and multiply radical expressions. To divide two radical expressions, simply write the quotient as a ratio and then simplify. In general, we have

n n n V6a a nV6a Vb6 ,

n bVb6

provided that all expressions represent real numbers. Note that the quotient rule is applied only to radicals that have the same index.

E X A M P L E 4 Dividing radicals with the same index Divide and simplify. Assume the variables represent positive numbers.

a) V106 V56 b) (3V26) (2V36) c) V3 10x26 V3 5x6

Solution

a) V106 V56 V V

56 06

a b a b , provided that b * 0.

1 5 0

Quotient rule for radicals

V26 Reduce.

b) (3V26) (2V36)

. Rationalize the denominator.

3 2 V . 3

V 2 66

Note that V66 2 * V36.

c) V3 10x26 V3 5x6

3 1 5 0 x x2

Quotient rule for radicals

V3 2x6 Reduce.

V3 10x26 V3 5x6

V36 V36

3V26 2V36

Now do Exercises 29–36

Note that in Example 4(a) we applied the quotient rule to get V106 V65 V62. In Example 4(b) we did not use the quotient rule because 2 is not evenly divisible by 3. Instead, we rationalized the denominator to get the result in simplified form.

When working with radicals it is usually best to write them in simplified radical form before doing any operations with the radicals.

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� �

�

9-35 9.4 Quotients, Powers, and Rationalizing Denominators 591

E X A M P L E 5 Simplifying before dividing Divide and simplify. Assume the variables represent positive numbers.

a) V126 V72x6 b) V4 16a6 V4 a56 Solution

a) V126 V72x6 V V

46 66

V V

36 2x6

Factor out perfect squares.

V V

36 x6

Simplify.

V V

36 2x6

V V

x6 x6

Reduce 2 6

to 1 3

and rationalize.

V 6 6 x x6

Multiply the radicals.

b) V4 16a6 V4 a56 Factor out perfect fourth powers.

Reduce.

2 a

Simplify the radicals.

V4 166 V4 a46

V4 166 . V4 a6 V4 a46 . V4 a6

Now do Exercises 37–44

In Chapter 10 it will be necessary to simplify expressions of the type found in Example 6.

E X A M P L E 6 Simplifying radical expressions Simplify.

a) b)

Solution a) First write V126 in simplified form. Then simplify the expression.

4 � 4 V126 4 �

4 2V36

Simplify V126.

2� (2 2� �

V 2

36) Factor.

2 � 2 V36

Divide out the common factor.

b) �6 �

�2 V206 �6 �

�2 2V56

�2�(3 �

�

2� V56)

3 � V56

�6 � V206 �2

4 � V126 4

U Helpful Hint V

The expressions in Example 6 are the types of expressions that you must simplify when learning the quadratic formula in Chapter 10.

Now do Exercises 45–48

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592 Chapter 9 Radicals and Rational Exponents 9-36

E X A M P L E 7

E X A M P L E 8

CAUTION To simplify the expressions in Example 6, you must simplify the radical, factor the numerator, and then divide out the common factors. You

V12 V3cannot simply “cancel” the 4’s in 4 � 6 or the 2’s in 2 � 6 because they 4 2

are not common factors.

U4V Rationalizing Denominators Using Conjugates A simplified expression involving radicals does not have radicals in the denominator. If an expression such as 4 � V36 appears in a denominator, we can multiply both the numerator and denominator by its conjugate 4 � V36 to get a rational number in the denominator.

Rationalizing the denominator using conjugates Write in simplified form.

2 � V63 V56 a) b)

4 � V63 V66 � V62

Solution 2 � V36 (2 � V63)(4 � V36)

a) Multiply the numerator and denominator by 4 � V36 . 4 � V36 (4 � V63)(4 � V36)

8 � 6V36 � 3 (4 � V36 )(4 � V63) 16 � 3 13 13

11 � 6V36 Simplify.

V65 V65(V66 � V26 ) b) Multiply the numerator and denominator by

V66 � V62 (V66 � V62)(V66 � V62) V66 � V26. V306 � V106 (V66 � V26 )(V66 � V62) 6 � 2 4

Now do Exercises 49–58

U5V Powers of Radical Expressions In Example 8, we use the power of a product rule [(ab)n anbn] and the power of a power rule [(am)n amn] with radical expressions. We also use the fact that a root and

n na power can be found in either order. That is, (Va6 )m V6am.

Finding powers of rational expressions Simplify. Assume the variables represent positive numbers.

63 )4 3 4a) (5V26)3 b) (2Vx c) (3wV26w)3 d) (2tV36t)3

Solution a) (5V26)3 53(V26)3 Power of a product rule

125V86 (V62)3 V623 V86 125 . 2V26 V68 V64V26 2V26 250V26

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9-37 9.4 Quotients, Powers, and Rationalizing Denominators 593

b) (2Vx3 )4 = 24(Vx3 )4 Power of a product rule = 24V(x3)4 (Vn a )m = Vn am = 16Vx12 (am)n = amn

= 16x6

c) (3wV3 2w )3 = 33w3(V3 2w )3

= 27w3(2w)

= 54w4

d) (2tV4 3t )3 = 23t3(V4 3t )3 = 8t3V4 27t3 Now do Exercises 59–70

Warm-Ups ▼

Fill in the blank. 1. The numbers V2 , V3 , and V5 are numbers. 2. Writing a fraction with an irrational denominator as an

equivalent fraction with a rational denominator is the denominator.

3. A simplified square root expression has no perfect as factors of the radicand.

4. A simplified radical expression has no fractions inside the .

5. A simplified radical expression has no radicals in the .

True or false?

6. V V

6 2

= V3

7. V 2 2

= V2

8. 4 -

2 V10 = 2 - V10

9. 4 -

2 V10 = 2 - V5

10. V 1 3

= V 3 3

11. (2V4 )2 = 16 12. (3V5 )3 = 27V125

9 .4Exercises

U Study Tips V • Personal issues can have a tremendous effect on your progress in any course. If you need help, get it. • Most schools have counseling centers that can help you to overcome personal issues that are affecting your studies.

V 3 V 6 U1V Rationalizing the Denominator 3. 4. V 7 V 5 All variables in the following exercises represent positive numbers.

Rewrite each expression with a rational denominator. See 1 7 5. 6.

Example 1. 3 3 V 4 V3 3 4 2 5 V 6 V 2 1. 2. 7. 8. 3 4 V 5 V 3 V 5 V27

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� �

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� � � � � �

� �

594 Chapter 9 Radicals and Rational Exponents 9-38

U2V Simplifying Radicals Simplify. See Example 6.

Write each radical expression in simplified radical form. 6 � 6 V50 V45 10 � 645. 46. See Example 2. 3 5

V65 V67 9. 10. �2 � V12 �6 � 66 V72V12 V18 47. 48.6 6

�2 �6 V63 V62

11. 12. V12 V186 6

1 3 U4V Rationalizing Denominators13. 14. 2 8 Using Conjugates

Simplify each expression by rationalizing the denominator.2 3 15. 3 16. 3 See Example 7.3 5

4 7 1 49.

17. 3 18. 4 2 � V684 5 6

50. 3 � V186Simplify. See Example 3.

x x2 51. 3

19. 20. V11 66 � V5y a 63 5a w 52.21. 7 22. 3 V56 � V146b y

1 � V26 a 5x 53.

23. 24. V36 � 13b 2y 2 � V36

a 4a 54. 3 325. 26. V62 � V66

b b V26

55.5 3 3 327. 28. V66 � V63

b2 22 4a 5

56. V76 � V56

U3V Dividing Radicals 2V36 Divide and simplify. See Examples 4 and 5. 57.

3V26 � V56 6 629. V156 V5 30. V14 V67 3V56

58. 5V26 � V6631. V36 V56 32. V56 V67

33. (3V36) (5V66) 34. (2V2 (4V106) 6) U5V Powers of Radical Expressions

Simplify. See Example 8. 35. (2V36) (3V66) 36. (5V12 (4V66)6)

59. (2V26)5 60. (3V63)4

2 3 2 61. (Vx6)5 62. (2V6y )3 37. V246a 6 6 V486xV72a 38. V32x 3 363. (�3Vx6)3 64. (�2Vx6 )4

3 3 3 339. V26 V2 40. V6 V2x 320 6 8x7 6 3265. (2xVx6)3 66. (2yV46y)3 4 4

41. V48 V6 4a10 V62 3 4 6 4 3 42. V6 2a 367. (�2V56)2 68. (�3V46)2 4 4 3 44 5 4 2 36 w 6 b 69. (V6)6 y43. V16w V6 44. V81b5 V6 x 70. (2V6)3

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9-39

Miscellaneous

Simplify.

V36 2 71. �

V62 V26 2 5

72. � V67 V76

V36 3V66 73. �

V26 2

V66 1 75. .

2 V36

8 � V326 77.

5 � V756 79.

81. Va6(Va6 � 3)

83. 4Va6(a � Va6)

6)285. (2V3m

2 87. (�2Vxy6z)2

3 2 5 89. Vm 3 m 3 m6(V6 � V6) 3 3

91. V86x4 � V276x4

4 6)32 93. (2mV2m

x � 9 95.

V6x � 3 x � y

96. V6x � Vy6

3Vk6 97.

Vk6 � V67 V6hk

98. Vh6 � 3Vk6

V63 V56 74. �

2V62 3V26

V66 V146 76. .

V67 V36

4 � V286 78.

3 � V186 80.

82. 3Vm 6 � 6)6(2Vm

84. V3ab 6 � V36(V3a 6)

6)286. (�3V4y

6)288. (5aVab

4 4 4 3 7 90. V6(V6 � Vw6)w w 3 3 92. V166a4 � aV26a

6 94. (�2tV26t2)5

9.4 Quotients, Powers, and Rationalizing Denominators 595

5 3 99. �

V62 � 1 V26 � 1 V63 V36

100. � V66 � 1 V66 � 1

1 1 101. �

V62 V36

4 1 102. �

2V63 V65 3 4

103. � V26 � 1 V26 � 1

3 2 104. �

V56 � V63 V65 � V63

V6x 3Vx6 105. �

V6x � 2 Vx6 � 2

V56 V56y 106. �

3 � V6y 3 � Vy6

1 1 107. �

V6x 1 � Vx6

Vx6 5 108. �

V6x � 3 Vx6

Getting More Involved

109. Exploration

A polynomial is prime if it cannot be factored by using integers, but many prime polynomials can be factored if we use radicals.

3 3 3 a) Find the product (x � V26)(x2 � V62x � V46). b) Factor x3 � 5 using radicals. c) Find the product

3 3 3 3 3 (V5 2 25 10 46 � V6)(V6 � V6 � V6). d) Use radicals to factor a � b as a sum of two cubes

and a � b as a difference of two cubes.

110. Discussion

Which one of the following expressions is not equivalent to the others?

4 3 3 3 4 a) (V6x)4 b) Vx6 c) Vx6 4�3 1�3)4d) x e) (x

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596 Chapter 9 Radicals and Rational Exponents 9-40

In This Section

U1V The Odd-Root Property

U2V The Even-Root Property

U3V Equations Involving Radicals

U4V Equations Involving Rational

9.5 Solving Equations with Radicals and Exponents

One of our goals in algebra is to keep increasing our knowledge of solving equations because the solutions to equations can give us the answers to various applied questions. In this section, we will apply our knowledge of radicals and exponents to solving some new types of equations.

Exponents

U5V Applications U1V The Odd-Root Property

3Because (�2)3 � 8 and 23 8, the equation x 8 is equivalent to x 2. The equation x3 � 8 is equivalent to x � 2. Because there is only one real odd root of each real number, there is a simple rule for writing an equivalent equation in this situation.

Odd-Root Property

If n is an odd positive integer, n

xn k is equivalent to x V6k for any real number k.

nNote that xn k is equivalent to x Vk6 means that these two equations have the same 33real solutions. So x 1 and x V16 each have only one real solution.

E X A M P L E 1 Using the odd-root property Solve each equation.

a) x3 27 b) x5 � 32 0 c) (x � 2)3 24

Solution a) x3 27

x V3 276 Odd-root property x 3

Check 3 in the original equation. The solution set is {3}.

b) x5 � 32 0

x5 �32 Isolate the variable.

x V5 �326 Odd-root property x �2

Check �2 in the original equation. The solution set is {�2}.

c) (x � 2)3 24

x � 2 V3 246 Odd-root property x 2 � 2V3 36 V3 246 V3 86 . V3 36 2V3 36

Check. The solution set is {2 � 2V3 36 }. Now do Exercises 1–8

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9-41 9.5 Solving Equations with Radicals and Exponents 597

U2V The Even-Root Property In solving the equation x2 4, you might be tempted to write x 2 as an equivalent

2equation. But x 2 is not equivalent to x 4 because 22 4 and (�2)2 4. So the 2 2solution set to x 4 is {�2, 2}. The equation x 4 is equivalent to the compound

sentence x 2 or x �2, which we can abbreviate as x ±2. The equation x ±2 is read “x equals positive or negative 2.”

Equations involving other even powers are handled like the squares. Because 4 424 16 and (�2)4 16, the equation x 16 is equivalent to x ±2. So x 16

has two real solutions. Note that x4 �16 has no real solutions. The equation x6 5 6is equivalent to x ±V65. We can now state a general rule.

Even-Root Property

Suppose n is a positive even integer. If k � 0, then xn k is equivalent to x ±Vn k6. If k 0, then xn k is equivalent to x 0. If k � 0, then xn k has no real solution.

nNote that xn k for k � 0 is equivalent to x ±Vk6 means that these two equations have the same real solutions.

actually take an odd root of each side.

E X A M P L E 2 Using the even-root property Solve each equation.

a) x2 10 b) w8 0 c) x4 �4

Solution a) x2 10

x ±V106 Even-root property The solution set is {�V106, V106}, or {±V106}.

b) w8 0 w 0 Even-root property

The solution set is {0}. c) By the even-root property, x4 �4 has no real solution. (The fourth power of any

real number is nonnegative.)

U Helpful Hint V

We do not say, “take the square root of each side.” We are not doing the same thing to each side of x2 9 when we write x ±3. This is the third time that we have seen a rule for obtaining an equivalent equation without “doing the same thing to each side.” (What were the other two?) Because there is only one odd root of every real number, you can

Now do Exercises 9–14

Whether an equation has a solution depends on the domain of the variable. For example, 2x 5 has no solution in the set of integers and x2 �9 has no solution in the set of real numbers. We can say that the solution set to both of these equations is the empty set, �, as long as the domain of the variable is clear. In Section 9.6 we intro duce a new set of numbers, the imaginary numbers, in which x2 �9 will have two solutions. So in this section it is best to say that x2 �9 has no real solution, because in Section 9.6 its solution set will not be �. An equation such as x x � 1 never has a solution, and so saying that its solution set is � is clear.

In Example 3, the even-root property is used to solve some equations that are a bit more complicated than those of Example 2.

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� � �

598 Chapter 9 Radicals and Rational Exponents 9-42

E X A M P L E 3 Using the even-root property Solve each equation.

a) (x � 3)2 4 b) 2(x � 5)2 � 7 0

c) x4 � 1 80

Solution a) (x � 3)2 4

x � 3 2 or x � 3 �2 Even-root property

x 5 or x 1 Add 3 to each side.

The solution set is {1, 5}.

b) 2(x � 5)2 � 7 0

2(x � 5)2 7 Add 7 to each side.

(x � 5)2 7 2

Divide each side by 2.

x � 5 7 2

or x � 5 � 7 2

Even-root property

x 5 � V

2 1646

or x 5 � V

2 1646 7

2 V V

76 26

V V

26 26

V 2 146

x 10 �

2 V1646

or x 10 �

2 V1646

The solution set is {10 � 2 V1646 , 10 � 2 V1646}. c) x4 � 1 80

x4 81

x ±V4 816 ±3 The solution set is {�3, 3}.

Now do Exercises 15–24

In Chapter 5 we solved quadratic equations by factoring. The quadratic equations that we encounter in this chapter can be solved by using the even-root property as in parts (a) and (b) of Example 3. In Chapter 10 you will learn general methods for solv ing any quadratic equation.

U3V Equations Involving Radicals If we start with the equation x 3 and square both sides, we get x2 9, which has the solution set {�3, 3}. But the solution set to x 3 is {3}. Squaring both sides pro duced an equation with more solutions than the original. We call the extra solutions extraneous solutions. The same problem can occur when we raise each side to any even power. Note that we don’t always get extraneous solutions. We might get one or more of them.

Raising each side to an odd power does not cause extraneous solutions. For exam ple, if we cube each side of x 3 we get x3 27. The solution set to both equations is {3}. Likewise, x � 3 and x3 � 27 both have solution set {�3}.

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9-43 9.5 Solving Equations with Radicals and Exponents 599

E X A M P L E 4

U Calculator Close-Up V

If 14 satisfies the equation

V2x � 3 0, 6 � 5 then (14, 0) is an x-intercept for the graph of

y V2x � 3 6 � 5. So the calculator graph shown here provides visual support for the con clusion that 14 is the only solution to the equation.

�2 20

�10

Raising each side of an equation to a power

If n is odd, then a b and an bn are equivalent equations.

If n is even, then a b and an bn may not be equivalent. However, the solution set to an bn contains all of the solutions to a b.

It has always been important to check solutions any time you solve an equation. When raising each side to a power, it is even more important. We use these ideas most often with equations involving radicals as shown in Example 4.

Raising each side to a power to eliminate radicals Solve each equation.

a) V2x � 3 0 b) V3x � 5 V6 68 x6 � 5 3 6 3 x � 1 c) V3x � 16

Solution a) Eliminate the square root by raising each side to the power 2:

V62x � 3 � 5 0 Original equation V62x � 3 5 Isolate the radical.

(V2x � 3 Square both sides.6)2 52

2x � 3 25

2x 28

x 14

Check by evaluating x 14 in the original equation:

V62(14) �63 � 5 0

V28 � 3 0 6 � 5

V25 0 6 � 5

0 0

The solution set is {14}. 3 3 b) V6 x � 1 Original equation3x � 5 V6

3 3 (V3x � 5 (Vx � 1 6)3 6)3 Cube each side. 3x � 5 x � 1

2x �6 x �3

Check x �3 in the original equation:

V3(�3) 6 V�3 � 63 6� 5 3 61 3 6 3 6 V�4 V�4

Note that V�43 6 is a real number. The solution set is {�3}. In this example, we checked for arithmetic mistakes. There was no possibility of extraneous solutions here because we raised each side to an odd power.

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600 Chapter 9 Radicals and Rational Exponents 9-44

c) V3x � 1686 x Original equation (V3x � 1686)2 x2 Square both sides.

3x � 18 x2 Simplify.

�x2 � 3x � 18 0 Subtract x2 from each side to get zero on one side.

x2 � 3x � 18 0 Multiply each side by �1 for easier factoring.

(x � 6)(x � 3) 0 Factor.

x � 6 0 or x � 3 0 Zero factor property

x 6 or x �3

Because we squared both sides, we must check for extraneous solutions. If

x �3 in the original equation V3x � 1686 x, we get

V3(�3)6� 186 �3 V96 �3

3 �3,

which is not correct. If x 6 in the original equation, we get

V3(6) �6 186 6, which is correct. The solution set is {6}.

U Calculator Close-Up V

The graphs of

y1 V3x � 1686 and y2 x provide visual support that 6 is the only value of x for which

x and V3x � 1686 are equal.

�10 10

Now do Exercises 25–44 �10

In Example 5, the radicals are not eliminated after squaring both sides of the equation. In this case, we must square both sides a second time. Note that we square the side with two terms the same way we square a binomial.

E X A M P L E 5 Squaring both sides twice Solve V5x � 1 66 � Vx � 2 1.

Solution It is easier to square both sides if the two radicals are not on the same side.

V5x � 1 6 Original equation6 � Vx � 2 1 V65x � 1 1 � Vx � 2 Add Vx � 26 6 to each side. 6)2 6)2

5x � 1 1 � 2Vx � 2 Square the right side like (V5x � 1 (1 � Vx � 2 Square both sides.

6 � x � 2 a binomial.

5x � 1 3 � x � 2Vx � 2 Combine like terms on6 the right side.

4x � 4 2Vx � 2 Isolate the square root.6 2x � 2 Vx � 2 Divide each side by 2.6

6)2 4x2 � 8x � 4 x � 2 Square the binomial on

the left side.

(2x � 2)2 (Vx � 2 Square both sides.

4x2 � 9x � 2 0 (4x � 1)(x � 2) 0

4x � 1 0 or x � 2 0 1

x or x 2 4

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� � � �

�

9-45 9.5 Solving Equations with Radicals and Exponents 601

Check to see whether V5x � 16 � Vx � 26 1 for x 1 4

and for x 2:

5 . 1 4

� 1 � 1 4

� 2 1 4

� 9 4

1 2 �

3 2 � 1

V5 . 2 �6 16 � V2 � 26 V96 � V46 3 � 2 1

So the original equation is not satisfied for x 1 4

but is satisfied for x 2. Since

2 is the only solution to the equation, the solution set is {2}.

Now do Exercises 45–60

U4V Equations Involving Rational Exponents Equations involving rational exponents can be solved by combining the methods that you just learned for eliminating radicals and integral exponents. For equa tions involving rational exponents, always eliminate the root first and the power second.

E X A M P L E 6 Eliminating the root, then the power Solve each equation.

a) x2�3 4

b) (w � 1)�2�5 4

Solution a) Because the exponent 2�3 indicates a cube root, raise each side to the power 3:

x2�3 4 Original equation

(x2�3)3 43 Cube each side. x2 64 Multiply the exponents: 2

3 . 3 2.

x 8 or x �8 Even-root property

All of the equations are equivalent. Check 8 and �8 in the original equation. The solution set is {�8, 8}.

b) (w � 1)�2�5 4 Original equation

[(w � 1)�2�5]�5 4�5 Raise each side to the power �5 to eliminate the negative exponent.

(w � 1)2 10

1 24

Multiply the exponents: � 2

5 (�5) 2.

w � 1 ± 10

1 24

Even-root property

w � 1 3 1 2

or w � 1 � 3 1 2

w 3 3 3 2

or w 3 3 1 2

Check the values in the original equation. The solution set is �3 3 1 2, 3 3 3 2�.

U Calculator Close-Up V

Check that 31�32 and 33�32 satisfy the original equation.

U Helpful Hint V

Note how we eliminate the root first by raising each side to an integer power, and then apply the even-root property to get two solutions in Example 6(a). A common mistake is to raise each side to the 3�2 power and get x 43�2 8. If you do not use the even-root property, you can easily miss the solution �8.

Now do Exercises 61–72

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602 Chapter 9 Radicals and Rational Exponents 9-46

An equation with a rational exponent might not have a real solution because all even powers of real numbers are nonnegative.

E X A M P L E 7 An equation with no solution Solve (2t � 3)�2�3 � 1.

Solution Raise each side to the power �3 to eliminate the root and the negative sign in the exponent:

(2t � 3)�2�3 � 1 Original equation

[(2t � 3)�2�3]�3 (�1)�3 Raise each side to the �3 power. (2t � 3)2 � 1 Multiply the exponents: �

3 (�3) 2.

By the even-root property this equation has no real solution. The square of every real number is nonnegative.

Now do Exercises 73–74

The three most important rules for solving equations with exponents and radicals are restated here.

Strategy for Solving Equations with Exponents and Radicals

1. In raising each side of an equation to an even power, we can create an equation that gives extraneous solutions. We must check all possible solu tions in the original equation.

2. When applying the even-root property, remember that there is a positive and a negative even root for any positive real number.

3. For equations with rational exponents, raise each side to a positive or negative integral power first and then apply the even- or odd-root property. (Positive fraction—raise to a positive power; negative fraction—raise to a negative power.)

U5V Applications The square of the hypotenuse of any right triangle is equal to the sum of the squares of the legs (the Pythagorean theorem). In Example 8 we use this fact and the even-root property to find a distance on a baseball diamond.

E X A M P L E 8 Diagonal of a baseball diamond A baseball diamond is actually a square, 90 feet on each side. What is the distance from third base to first base?

Solution First make a sketch as in Fig. 9.1. The distance x from third base to first base is the length of the diagonal of the square shown in Fig. 9.1. The Pythagorean theorem can be applied to

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9-47 9.5 Solving Equations with Radicals and Exponents 603

the right triangle formed from the diagonal and two sides of the square. The sum of the squares of the sides is equal to the diagonal squared:

x2 = 902 + 902

x2 = 8100 + 8100 x2 = 16,200 x = ±V16,200k = ±90V2k

The length of the diagonal of a square must be positive, so we disregard the negative solution. Checking the answer in the original equation verifies that the exact length of the diagonal is 90V2k feet.

Now do Exercises 95–110Figure 9.1

1st base3rd base

2nd base

Home plate

90 ft

x ft

Warm-Ups ▼

Fill in the blank. 1. If n is an positive integer, then xn = k is equivalent

to x = V n

kk for any real number k. 2. If n is an positive integer, then xn = k is equivalent

to x = ±V n

kk for k > 0. 3. An solution is a solution that appears when

solving an equation but does not satisfy the original equation.

4. Raising each side of an equation to an power can produce an extraneous solution.

True or false? 5. The equations x2 = 4 and x = 2 are equivalent.

6. The equation x2 = -25 has no real solution.

7. The equation x2 = 0 has no solution.

8. The equation x3 = 8 is equivalent to x = 2.

9. Squaring both sides of Vxk = - 7 yields an equation with an extraneous solution.

10. The equations x2 - 6 = 0 and x = ±V6k are equivalent.

9 .5Exercises

U Study Tips V • Try changing subjects or tasks every hour when you study. The brain does not easily assimilate the same material hour after hour. • You will learn more from working on a subject 1 hour per day than 7 hours on Saturday.

U1V The Odd-Root Property U2V The Even-Root Property

Solve each equation. See Example 1. Find all real solutions to each equation. See Examples 2

1. x3 = -1000 2. y3 = 125 and 3.

3. 32m5 - 1 = 0 4. 243a5 + 1 = 0 9. x2 = 25 10. x2 = 36

5. (y - 3)3 = -8 6. (x - 1)3 = -1 11. x2 - 20 = 0 12. a2 - 40 = 0 7.

1 -- x3 + 4 = 0 8. 3(x - 9)7 = 0 2

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604 Chapter 9 Radicals and Rational Exponents 9-48

13. x2 � 9 0 14. w2 � 49 0

15. (x � 3)2 16 16. (a � 2)2 25

17. (x � 1)2 � 8

18. (w � 3)2 � 12

1 12 219. x 5 20. x 6 2 3

21. (y � 3)4 0 22. (2x � 3)6 0

23. 2x6 128 24. 3y4 48

U3V Equations Involving Radicals

Solve each equation and check for extraneous solutions. See Example 4.

25. Vx6� 3 � 3 4 6 � 526. Va � 1 1

27. 2Vw6� 4 5 628. 3Vw � 1 6

3 3 3 329. V2x � 3 V6 30. Va � 3 V66 x � 12 6 2a � 7

31. V2t � 4 Vt � 1 32. Vw � 3 V4w � 66 6 6 615

33. V46x2 �6x � 3 2x 34. Vx2 � 566x � 2 x

35. Vx2 � 26 3 36. Vx6 � 4 46x � 6 2 � x6

37. V2x2 � 6 x 38. V2x2 �60 x61 63x � 16

39. V26x2 � 5x � 6 x 40. V2x2 �66 66x � 9 x

41. Vx6� 1 x � 1 642. V2x � 1 2x � 1

43. x � Vx � 9 9 6 � 3x 16 44. V3x � 1

Solve each equation and check for extraneous solutions. See Example 5.

45. Vx6 � Vx � 3 36

46. Vx6 � Vx � 3 36

47. Vx � 2 66 � Vx � 1 3

48. Vx6 � Vx � 5 56

49. Vx � 3 66 � Vx � 2 1

50. V62x � 1 � V6x 1

51. V3x � 1 66 � V2x � 1 1

52. V4x � 1 66 � V3x � 2 1

53. V2x � 2 66 � Vx � 3 2

54. V3x 66 � Vx � 2 4

55. V4 � x 66 � Vx � 6 2

56. V6 � x 66 � Vx � 2 2

57. V6x � 5 � V6x 3

58. V2x 62 66 � V2x � 16

59. V3x � 1 66 � V2x � 4 3

60. V2x � 5 66 � Vx � 2 1

U4V Equations Involving Rational Exponents

Solve each equation. See Examples 6 and 7. 2�3 2�361. x 3 62. a 2

�2�3 �2�363. y 9 64. w 4

1�3 1�365. w 8 66. a 27

1 t�1�2 �1�467. 9 68. w

69. (3a � 1)�2�5 1 70. (r � 1)�2�3 1

71. (t � 1)�2�3 2 72. (w � 3)�1�3 1

73. (x � 3)2�3 �4 74. (x � 2)3�2 �1

Miscellaneous

Solve each equation. See the Strategy for Solving Equations with Exponents and Radicals box on page 602.

75. 2x2 � 3 7

76. 3x2 � 5 16

b =

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9-49 9.5 Solving Equations with Radicals and Exponents 605

C ap

si ze

s cr

ee ni

ng v

al ue

w ith

13 .5

f t

3 3 477. Y42w + 43 = Yw - 2

78. Y w 22w - 43 42 - = Y3 428

79. (w + 1)213 = -3

80. (x - 2)413 = -2

81. (a + 1)113 = -2

82. (a - 1)113 = -3

83. (4y - 5)7 = 0

84. (5x)9 = 0

85. Y5x2 +4 -44x + 1 x = 0 86. 3 + Y4x2 - 84x = 0

87. Y4x x + 242 =

288. Y94x = x + 6

89. (t + 2)4 = 32

90. (w + 1)4 = 48

91. Yx2 - 34 =4x x 92. Y4x4 - 4 = -x 4 448

-93. x 3 = 8

-94. x 2 = 4

U5V Applications

Solve each problem by writing an equation and solving it. Find the exact answer and simplify it using the rules for radicals. See Example 8.

95. Side of a square. Find the length of the side of a square whose diagonal is 8 feet.

96. Diagonal of a patio. Find the length of the diagonal of a square patio with an area of 40 square meters.

97. Side of a sign. Find the length of the side of a square sign whose area is 50 square feet.

98. Side of a cube. Find the length of the side of a cubic box whose volume is 80 cubic feet.

99. Diagonal of a rectangle. If the sides of a rectangle are 30 feet and 40 feet in length, find the length of the diago nal of the rectangle.

100. Diagonal of a sign. What is the length of the diagonal of a rectangular billboard whose sides are 5 meters and 12 meters?

101. A 30-60-90 triangle. In a 30°-60°-90° triangle, the side opposite the 30° angle is half the length of the hypotenuse. See the accompanying figure. a) Find the length of the hypotenuse if the side opposite

the 30° angle is 1. b) Find the length of the side opposite 60° if the side

opposite 30° is 1. c) Find the length of the side opposite 60° if the length

of the hypotenuse is 1.

102. An isosceles right triangle. An isosceles right triangle has two 45° angles. The sides opposite those angles are equal in length. See the accompanying figure. a) Find the length of the hypotenuse if the length of each

of the equal sides is 1. b) Find the length of each of the equal sides if the length

of the hypotenuse is 1.

60A

30A

45A

Figure for Exercises 101 and 102

103. Sailboat stability. To be considered safe for ocean sail ing, the capsize screening value C should be less than 2 (www.sailing.com). For a boat with a beam (or width) b in feet and displacement d in pounds, C is determined by the function

C = 4d-113b.

C = 54d -113b.

0 10 20 30 Displacement

(thousands of pounds)

Figure for Exercise 103

http:www.sailing.com

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29.46 ears

11.86 years

5.2 AU

606 Chapter 9 Radicals and Rational Exponents 9-50

a) Find the capsize screening value for the Tartan 4100, which has a displacement of 23,245 pounds and a beam of 13.5 feet.

b) Solve this formula for d.

c) The accompanying graph shows C in terms of d for the Tartan 4100 (b 13.5). For what displacement is the Tartan 4100 safe for ocean sailing?

104. Sailboat speed. The sail area-displacement ratio S pro vides a measure of the sail power available to drive a boat. For a boat with a displacement of d pounds and a sail area of A square feet S is determined by the function

S 16Ad�2�3. a) Find S to the nearest tenth for the Tartan 4100, which

is the volume of a cube with surface area 12 square feet?

111. Kepler’s third law. According to Kepler’s third law of T 2planetary motion, the ratio 3 has the same value for every R

planet in our solar system. R is the average radius of the orbit of the planet measured in astronomical units (AU), and T is the number of years it takes for one complete orbit of the sun. Jupiter orbits the sun in 11.86 years with an average radius of 5.2 AU, whereas Saturn orbits the sun in 29.46 years. Find the average radius of the orbit of Saturn. (One AU is the distance from the earth to the sun.)

has a sail area of 810 square feet and a displacement of y 23,245 pounds.

b) Write d in terms of A and S. Jupiter

Sun Saturn

105. Diagonal of a side. Find the length of the diagonal of a side of a cubic packing crate whose volume is 2 cubic

meters.

106. Volume of a cube. Find the volume of a cube on which the diagonal of a side measures 2 feet.

107. Length of a road. An architect designs a public park in the shape of a trapezoid. Find the length of the diagonal road marked a in the figure.

108. Length of a boundary. Find the length of the border of the park marked b in the trapezoid shown in the figure.

6 km

5 km b 3 km a

12 km

Figure for Exercises 107 and 108

109. Average annual return. In Exercise 127 of Section 9.2, the function

S�1�n r � 1�P was used to find the average annual return for an investment. a) Write S in terms of r, P, and n. b) Write P in terms of r, S, and n.

110. Surface area of a cube. The function A 6V 2�3 gives the surface area of a cube in terms of its volume V. What

Figure for Exercise 111

112. Orbit of Venus. If the average radius of the orbit of Venus is 0.723 AU, then how many years does it take for Venus to complete one orbit of the sun? Use the information in Exercise 111.

Use a calculator to find approximate solutions to the follow ing equations. Round your answers to three decimal places.

2113. x 3.24

114. (x � 4)3 7.51

115. Vx � 2 1.736

116. V3 6x � 5 3.7 2�3117. x 8.86

118. (x � 1)�3�4 7.065

Getting More Involved

119. Cooperative learning

Work in a small group to write a formula that gives the side of a cube in terms of the volume of the cube and explain the formula to the other groups.

120. Cooperative learning

Work in a small group to write a formula that gives the side of a square in terms of the diagonal of the square and explain the formula to the other groups.

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9-51 9.6 Complex Numbers 607

9.6 Complex Numbers

In Chapter 1, we discussed the real numbers and the various subsets of the real numbers. In this section, we define a set of numbers that has the real numbers as a subset. This new set of numbers is the set of complex numbers. Although it is hard to imagine numbers beyond the real numbers, the complex numbers are used to model some very real phenomena in physics and electrical engineering. These applications are beyond the scope of this text, but if you want a better understanding of them, search the Internet for “applications of complex numbers.”

In This Section

U1V Definition

U2V Addition, Subtraction, and Multiplication

U3V Division of Complex Numbers

U4V Square Roots of Negative Numbers

U5V Imaginary Solutions to Equations

U1V Definition The equation 2x 1 has no solution in the set of integers, but in the set of rational numbers, 2x 1 has a solution. The situation is similar for the equation x2 �4. It has no solution in the set of real numbers because the square of every real number is nonnegative. However, in the set of complex numbers x2 �4 has two solutions. The complex numbers were developed so that equations such as x2 �4 would have solutions.

The complex numbers are based on the symbol V�16. In the real number system this symbol has no meaning. In the set of complex numbers this symbol is given mean ing. We call it i. We make the definition that

i V�1 and i26 � 1.

Complex Numbers

The set of complex numbers is the set of all numbers of the form

a � bi,

where a and b are real numbers, i V�1 �1.6, and i2

In the complex number a � bi, a is called the real part and b is called the imaginary part. If b * 0, the number a � bi is called an imaginary number. If b 0, then the complex number a � 0i is the real number a.

In dealing with complex numbers, we treat a � bi as if it were a binomial, with i being a variable. Thus, we would write 2 � (�3)i as 2 � 3i. We agree that 2 � i3, 3i � 2, and i3 � 2 are just different ways of writing 2 � 3i (the standard form). Some examples of complex numbers are

2 3 �3 � 5i, � i, 1 � i V62, 9 � 0i, and 0 � 7i.

3 4

For simplicity we write only 7i for 0 � 7i. The complex number 9 � 0i is the real number 9, and 0 � 0i is the real number 0. Any complex number with b 0 is a real number. For any real number a,

a � 0i a.

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608 Chapter 9 Radicals and Rational Exponents 9-52

U Helpful Hint V

Note that a complex number does not have to have an i in it. All real numbers are complex numbers. So 1, 2, and 3 are complex numbers.

The set of real numbers is a subset of the set of complex numbers. See Fig. 9.2.

Complex numbers

3, �, 0, �9,

Real numbers

i, 2 � 3i, �3 � 8i,

Imaginary numbers

, 5 — 2 2 �5� �

Figure 9.2

U2V Addition, Subtraction, and Multiplication Addition and subtraction of complex numbers are performed as if the complex numbers were algebraic expressions with i being a variable.

E X A M P L E 1 Addition and subtraction of complex numbers Find the sums and differences.

a) (2 � 3i) � (6 � i) b) (�2 � 3i) � (�2 � 5i)

c) (3 � 5i) � (1 � 2i) d) (�2 � 3i) � (1 � i)

Solution a) (2 � 3i) � (6 � i) 8 � 4i

b) (�2 � 3i) � (�2 � 5i) �4 � 2i

c) (3 � 5i) � (1 � 2i) 2 � 3i

d) (�2 � 3i) � (1 � i) �3 � 2i

Now do Exercises 1–8

Informally, we add and subtract complex numbers as in Example 1. Formally, we use the following symbolic definition. We include this definition for completeness, but you don’t need to memorize it. Just add or subtract as in Example 1.

Addition and Subtraction of Complex Numbers

The sum and difference of a � bi and c � di are defined as follows:

(a � bi) � (c � di) (a � c) � (b � d)i (a � bi) � (c � di) (a � c) � (b � d)i

Complex numbers are multiplied as if they were algebraic expressions. Whenever i2 appears, we replace it by �1.

Products of complex numbersE X A M P L E 2 Find each product.

a) 2i(1 � i) b) (2 � 3i)(4 � 5i) c) (3 � i)(3 � i)

Solution a) 2i(1 � i) 2i � 2i2 Distributive property

2i � 2(�1) � 1i2

�2 � 2i

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9-53 9.6 Complex Numbers 609

b) Use the FOIL method to find the product:

(2 � 3i)(4 � 5i) 8 � 10i � 12i � 15i2

8 � 22i � 15(�1) Replace i2 by �1.

8 � 22i � 15

�7 � 22i

c) This product is the product of a sum and a difference.

(3 � i)(3 � i) 9 � 3i � 3i � i2

9 � (�1) i2 �1

U Calculator Close-Up V

Many graphing calculators can per form operations with complex numbers.

Now do Exercises 9–26

For completeness we give the following symbolic definition of multiplication of complex numbers. However, it is simpler to find products as we did in Example 2 than to use this definition.

Multiplication of Complex Numbers

The complex numbers a � bi and c � di are multiplied as follows:

(a � bi)(c � di) (ac � bd) � (ad � bc)i

We can find powers of i using the fact that i2 �1. For example,

i3 i2 . i �1 . i �i.

The value of i4 is found from the value of i3:

i4 i3 . i �i . i �i2 1

Using i2 � 1, i3 � i, and i4 1, you can actually find any power of i by factoring out all of the fourth powers. For example,

i13 (i4)3 i18 (i4)4 . i2 . i (1)3 . i i and (1)4 . i2 � 1.

E X A M P L E 3 Powers of imaginary numbers Write each expression in the form a � bi.

a) (2i)2 b) (�2i)4 c) i6

d) i22 e) i19

Solution a) (2i)2 22 . i2 4(�1) �4

b) (�2i)4 (�2)4 . i4 16 . 1 16

c) i6 i2 . i 4 �1 . 1 �1

d) i22 (i4)5 . i2 (1)5 . i2 �1

e) i19 (i4)4 . i3 (1)4 . i3 �i

Now do Exercises 27–38

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610 Chapter 9 Radicals and Rational Exponents 9-54

U Helpful Hint V

Here is that word “conjugate” again. It is generally used to refer to two things that go together in some way.

U3V Division of Complex Numbers To divide a complex number by a real number, divide each term by the real number, just as we would divide a binomial by a number. For example,

4 � 6i 2(2 � 3i) 2 2

2 � 3i.

To understand division by a complex number, we first look at imaginary numbers that have a real product. The product of the two imaginary numbers in Example 2(c) is a real number:

(3 � i)(3 � i) 10

We say that 3 � i and 3 � i are complex conjugates of each other.

Complex Conjugates

The complex numbers a � bi and a � bi are called complex conjugates of one another. Their product is the real number a2 � b2.

E X A M P L E 4 Products of conjugates Find the product of the given complex number and its conjugate.

a) 2 � 3i b) 5 � 4i

Solution a) The conjugate of 2 � 3i is 2 � 3i.

(2 � 3i)(2 � 3i) 4 � 9i2

4 � 9

b) The conjugate of 5 � 4i is 5 � 4i.

(5 � 4i)(5 � 4i) 25 � 16

Now do Exercises 39–46

We use complex conjugates to divide complex numbers. The process is the same as rationalizing the denominator. We multiply the numerator and denominator of the quotient by the complex conjugate of the denominator. If we were to use V�16 instead of i, then Example 5 here would look just like Example 7 in Section 9.4.

E X A M P L E 5 Dividing complex numbers Find each quotient. Write the answer in the form a � bi.

5 3 � i 3 � 2i a) b) c)

3 � 4i 2 � i i

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9-55 9.6 Complex Numbers 611

Solution a) Multiply the numerator and denominator by 3 � 4i , the conjugate of 3 � 4i:

3 � 5

4i (3 � 5(

4 3 i) �

(3 4 �

i) 4i)

1 9 5 �

�

1 2 6 0 i2 i

15 � 25

20i 9 � 16i2 9 � 16(�1) 25

1 2 5 5

� 2 2 0 5

3 5

� 4 5

b) Multiply the numerator and denominator by 2 � i, the conjugate of 2 � i:

3 2

�

i i

( ( 3 2

�

i i ) ) ( ( 2 2

�

i i ) )

6 � 4 �

5i � i2

6 4 �

�

5 ( i �

�

1) 1

5 � 5

1 � i

c) Multiply the numerator and denominator by �i, the conjugate of i:

3 � i

2i (3 � i(�

2i i

) ) (�i)

�3i �

�

i2 2i2

�3i 1 � 2

2 � 3i

Now do Exercises 47–58

The symbolic definition of division of complex numbers follows.

Division of Complex Numbers

We divide the complex number a � bi by the complex number c � di as follows:

a � bi (a � bi)(c � di) c � di (c � di)(c � di)

U4V Square Roots of Negative Numbers In the complex number system, negative numbers have two square roots. Because i2 �1 and (�i)2 �1, both i and �i are square roots of �1. Because (2i)2 �4 and (�2i)2 � 4, both 2i and �2i are square roots of �4. We use the radical symbol only for the square root that has the positive coefficient, as in V�4 2i.6

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612 Chapter 9 Radicals and Rational Exponents 9-56

Square Root of a Negative Number

For any positive real number b,

V�b6 iVb6.

For example, V�9 iV9 3i and V�7 iV7 6i6 6 6 6. Note that the expression V7 could easily be mistaken for the expression V76i, where i is under the radical. For this reason, when the coefficient of i is a radical, we write i preceding the radical.

Note that the product rule (Va6 . Vb Vab6 6) does not apply to negative numbers. For example, V�2 6 * V66:6 . V�3

V�2 6 6 . iV36 i2V66 �V666 . V�3 iV2 Square roots of negative numbers must be written in terms of i before operations are performed.

E X A M P L E 6 Square roots of negative numbers Write each expression in the form a � bi, where a and b are real numbers.

a) 3 � V�96 b) V�126 � V�276

c) �1 �

3 V�186

d) V�46 . V�96

Solution a) 3 � V�96 3 � iV96

3 � 3i

b) V�126 � V�276 iV126 � iV276 2iV36 � 3iV36 5iV36

c) �1 �

3 V�186 �1 �

3 iV186

�1 � 3 3iV26

� 1 3

� iV26

d) V�46 . V�96 iV46 . iV96 2i . 3i 6i2 �6

V126 V46 V36 2V36 V276 V96 V36 3V36

Now do Exercises 59–78

U5V Imaginary Solutions to Equations In the complex number system the even-root property can be restated so that x2 k is

2equivalent to x ±Vk6 for any k * 0. So an equation such as x �9 that has no real solutions has two imaginary solutions in the complex numbers.

E X A M P L E 7 Imaginary solutions to equations Find the imaginary solutions to each equation.

2 2 � 2a) x �9 b) 3x 0

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� �

9-57 9.6 Complex Numbers 613

Solution a) First apply the even-root property:

x2 �9 x ±V�96 Even-root property

±iV96 ±3i

Check these solutions in the original equation:

(3i)2 9i2 9(�1) �9 (�3i)2 9i2 �9

The solution set is {±3i}. b) First solve the equation for x2:

3x2 � 2 0

x2 � 2 3

x ± � 2 3

±i 2 3

±i V 3 66

Check these solutions in the original equation. The solution set is {±iV 3 66}. Now do Exercises 79–86

The basic facts about complex numbers are listed in the following box.

Complex Numbers

1. Definition of i: i V�16, and i2 �1. 2. A complex number has the form a � bi, where a and b are real numbers. 3. The complex number a � 0i is the real number a. 4. If b is a positive real number, then V�b6 iVb6. 5. The numbers a � bi and a � bi are called complex conjugates of each

other. Their product is the real number a2 � b2 . 6. Add, subtract, and multiply complex numbers as if they were algebraic

expressions with i being the variable, and replace i2 by �1. 7. Divide complex numbers by multiplying the numerator and denominator by

the conjugate of the denominator. 8. In the complex number system, x2 k for any real number k is equivalent to

x ±Vk6.

Warm-Ups ▼

Fill in the blank. 1. A number is a number of the form a � bi

where a and b are real numbers.

2. An number is a complex number in which b � 0.

3. The union of the real numbers and the imaginary numbers is the set of numbers.

4. The of a � bi is a � bi.

5. The of a � bi and a � bi is a2 � b2 .

6. If b 0, then a � bi is a number.

7. The set of real numbers is a of the set of complex numbers.

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614 Chapter 9 Radicals and Rational Exponents 9-58 9

True or false? 12. i4 = 1 8. V-s1 = i 13. i3 = i 9. 2 - V-s6 = 2 - 6i 14. i48 = 1

10. V-s9 = ±3 15. (2 - i)(2 + i) = 5 11. 2 - 3i - (4 - 2i) = -2 - i 16. If x2 = -9, then x = ±3i.

Exercises

U Study Tips V • When studying for a midterm or final, review the material in the order it was originally presented. This strategy will help you to see

connections between the ideas. • Studying the oldest material first will give top priority to material that you might have forgotten.

U2V Addition, Subtraction, and Multiplication

Find the indicated sums and differences of complex numbers. See Example 1.

1. (2 + 3i) + (-4 + 5i) 2. (-1 + 6i) + (5 - 4i)

3. (2 - 3i) - (6 - 7i) 4. (2 - 3i) - (6 - 2i)

5. (-1 + i) + (-1 - i) 6. (-5 + i) + (-5 - i)

7. (-2 - 3i) - (6 - i) 8. (-6 + 4i) - (2 - i)

Find each product. Express each answer in the form a + bi. See Example 2.

9. 3(2 + 5i) 10. 4(1 - 3i)

11. 2i(i - 5) 12. 3i(2 - 6i)

13. -4i(3 - i) 14. -5i(2 + 3i)

15. (2 + 3i)(4 + 6i) 16. (2 + i)(3 + 4i)

17. (-1 + i)(2 - i) 18. (3 - 2i)(2 - 5i)

19. (-1 - 2i)(2 + i) 20. (1 - 3i)(1 + 3i)

21. (5 - 2i)(5 + 2i) 22. (4 + 3i)(4 + 3i)

23. (1 - i)(1 + i) 24. (2 + 6i)(2 - 6i)

25. (4 + 2i)(4 - 2i) 26. (4 - i)(4 + i)

Find the indicated powers of complex numbers. See Example 3.

27. (3i)2 28. (5i)2

29. (-5i)2 30. (-9i)2

31. (2i)4 32. (-2i)3

33. i9 34. i12

35. i18 36. i33

37. i25 38. i31

U3V Division of Complex Numbers

Find the product of the given complex number and its conjugate. See Example 4.

39. 3 + 5i 40. 3 + i

41. 1 - 2i 42. 4 - 6i

43. -2 + i 44. -3 - 2i

45. 2 - iV3s 46. V5s - 4i

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9-59 9.6 Complex Numbers 615

Find each quotient. Express each answer in the form a � bi. 83. 2x2 � 5 0 84. 3x2 � 4 0 See Example 5.

3 6 47. 48.

4 � i 7 � 2i 85. 3x2 � 6 0 86. x2 � 1 0

2 � i 3 � 5i 49. 50.

3 � 2i 2 � i

4 � 3i 5 � 6i Miscellaneous51. 52. i 3i

Write each expression in the form a � bi, where a and b are 2 � 6i 9 � 3i real numbers.53. 54.

2 �6 87. (2 � 3i)(3 � 4i) 88. (2 � 3i)(2 � 3i)1 � i 2 � i

55. 56. 3i � 2 i � 5 6 8 89. (2 � 3i) � (3 � 4i) 90. (3 � 5i) � (2 � 7i)

57. 58. 3i �2i

2 � 3i �3i 91. 92.

U4V Square Roots of Negative Numbers 3 � 4i 3 � 6i

Write each expression in the form a � bi, where a and b are real numbers. See Example 6.

93. i(2 � 3i) 94. �3i (4i � 1) 59. V�625 60. V�816 61. 2 � V�64 62. 3 � V�96

95. (�3i)2 96. (�2i)6

63. 2V�69 � 5 64. 3V�166 � 2 97. V�12 6 6 � V�256 � V�3 98. V�49 6

65. 7 � V�66 66. V�56 � 3 99. (2 � 3i)2 100. (5 � 3i)2

67. V�68 � V6�18 6 � V�4568. 2V�20 6

�4 � V�32 �2 � �276 V6 2 � V�12 �6 � 6 101. 102.6 V�18

69. 70. 2 �6 2 3

�4 � V�24 8 � �20 Getting More Involved6 V6 71. 72.

4 �4 103. Writing

Explain why 2 � i is a solution to

73. V�2 6 6 . V�15 x 0.6 . V�6 74. V�3 6 2 � 4x � 5

104. Cooperative learning 75. V�63 . V6�27 6 . V�776. V�3 6 Work with a group to verify that �1 � iV36 and

V68 V66 �1 � iV36 satisfy the equation 77. 78.

V�4 6 x 0. In the complex number system there are three cube roots

U5V Imaginary Solutions to Equations of 8. What are they? Find the imaginary solutions to each equation. See Example 7. 105. Discussion

6 V�2 3 � 8

279. x �36 80. x2 � 4 0 What is wrong with using the product rule for radicals to get

V�4 6 64) V16 4? What is the correct product?

81. x2 �12 82. x2 �25 6 . V�4 V(�4)(�6 6

C h

a p

t e

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616 Chapter 9 Radicals and Rational Exponents 9-60

Wrap-Up 9 Summary

Powers and Roots Examples

nth roots If a bn for a positive integer n, then b is an 2 and �2 are fourth roots of 16. nth root of a.

Principal root The positive even root of a positive number The principal fourth root of 16 is 2.

4 6Radical notation If n is a positive even integer and a is positive, V16 2

then the symbol Va6 denotes the principal nth 4n V616 * �2 root of a.

n 3 3 If n is a positive odd integer, then the symbol Va V�8 �2, V86 26 6 denotes the nth root of a.

n 5 6 If n is any positive integer, then V60 0. V60 0, V60 0 If x is any real number, then use absolute Vx �x �, Vx �x3�62 66 value when an even root of an exponential Vx4 x2 4 x4 �x �6 , V6 expression has an odd exponent.

Domain of a The set of all real numbers that can V6x, domain [0, �) radical expression be used in place of the variable in the 3 x � 1V6, domain (��, �)

radical expression Vx � 54 6, domain [5, �)

Domain of a The domain of the radical expression G(x) Vx6 radical function that defines the function Domain [0, �)

n 31�n 81�3Definition of If n is any positive integer, then a V6a, V68 2 n1�na provided that V6a is a real number. (�4)1�2 is not real.

1�n)m 82�3 (81�3)2Definition of If m and n are positive integers, then am�n (a , 22 4 am�n provided that a1�n is a real number. (�16)3�4 is not real.

1 1 8�2�3Definition of If m and n are positive integers and a * 0, �3

a�m�n 8 2 4�m�n 1then a

�n , provided that a1�n is a real

am number.

Rules for Radicals Examples

Product rule for Provided that all roots are real, V26 . V63 V66 radicals nV6 n n V46x 2Vx6ab Va6 . Vb6.

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� � �

�

9-61

Quotient rule for radicals

Provided that all roots are real and b * 0,

n a b

. Vn a6 Vn b6

Simplified radical form for radicals of index n

A simplified radical of index n has 1. no perfect nth powers as factors of the

radicand, 2. no fractions inside the radical, and 3. no radicals in the denominator.

Rules for Rational Exponents

If a and b are nonzero real numbers and r and s are rational numbers, then the following rules hold, provided all expressions represent real numbers.

Product rule ar . as ar�s

Quotient rule a a

s a r�s

Power of a (ar)s ars power rule

Power of a (ab)r arbr

product rule

Power of a quotient rule

�a b � r a

Equations

Equations with 1. In raising each side of an equation to an even radicals and power, we can create an equation that gives exponents extraneous solutions. We must check.

2. When applying the even-root property, remember that there is a positive and a negative root.

3. For equations with rational exponents, raise each side to a positive or a negative power first and then apply the even- or odd-root property.

Chapter 9 Summary 617

5 V56 9 3

V106 V65 V62

V206 V4 . 5 2V56 6

3 V63 2 V62

V63 V63 V62 V66 .

V26 V26 V62 2

Examples

31�4 . 31�2 33�4

3�4x 1�2x1�4x

(21�2)�1�2 2�1�4 3�4)4 3(x x

(a2b6)1�2 ab3

8 6 �

2�3 4 4 �x x

Examples

Vx6 �3 x 9

2x 36 x ±6

�2�3x 4 �2�3)�3 4�3(x

12x 64

1 x ±

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�

618 Chapter 9 Radicals and Rational Exponents 9-62

Complex Numbers Examples

Complex numbers Numbers of the form a � bi, where a and b 2 � 3i are real numbers: �6i i V�1 �1 V26 � i 6, i2

Complex conjugates Complex numbers of the form a � bi and a � bi: (2 � 3i)(2 � 3i) 22 � 32

Their product is the real number a2 � b2. 13

Complex number Add, subtract, and multiply as algebraic (2 � 5i) � (4 � 2i) 6 � 3i operations expressions with i being the variable. (2 � 5i) � (4 � 2i) �2 � 7i

Simplify using i2 �1. (2 � 5i)(4 � 2i) 18 � 16i

Divide complex numbers by multiplying (2 � 5i) (4 � 2i) the numerator and denominator by the (2 � 5i)(4 � 2i) conjugate of the denominator.

(4 � 2i)(4 � 2i)

�2 � 24i 1 6 � � i

20 10 5

Square root of a For any positive real number b, V�b iVb V�9 iV9 3i6 6. 6 6 negative number

Imaginary solutions In the complex number system, x2 k for any x2 �25 to equations real k is equivalent to x 6. x ±V�25 ±5i ± Vk 6

Enriching Your Mathematical Word Power

Fill in the blank. 9. The set of real numbers that can be used in place of the variable in an algebraic expression is the of the1. A number b such that bn a is the nth of a. expression.

2. The expression a2 is the of a. 10. If an exponent is one of the numbers in the set

3. A number b such that b3 a is the root of a. n

{. . . , �3, �2, �1, 0, 1, 2, 3, . . .}, then it is an 4. If a � 0 and n is even, then Va6 is the nth root exponent.

of a. 11. A number has the form a � bi where a and b are 5. If n is an odd number and bn a, then b is an root real numbers.

of a. 12. In a � bi, i is the unit. n

6. The number n in Va6 is the . 13. The complex numbers a � bi and a � bi are complex n

7. The number a in V6a is the . . 8. Radicals with the same radicand and the same index are 14. A complex number in which b � 0 is an

radicals. number.

Review Exercises

9.1 Radicals 3 55. V�27 6. V�326 6 Find each root. Variables represent any real numbers. Use absolute value when necessary. 7. V100 8. V66

3 1000

2 121. V816 2. V16 6 10. Vy4 6 9. Vy 4 6 3 5 3 53 b103. V27 4. V32 11. Va 12. V66 6 6

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� �

� � � �

9-63 Chapter 9 Review Exercises 619

4 6 6 49. (26)1�3 50. (52)1�2 30 51. 100�3�2 52. 1000�2�3

13. Vn24 14. Vm32

15. Vn20 16. Vm4 6 6 �1�2 2 �33x (x y z)1�2 Simplify each radical expression. Assume all variables 53. �1 54.3�2x 1�2 �1�2x yzrepresent positive real numbers.

17. V72 18. V48 55. (a1�2 b)3(ab1�4)2 56. (t�1�2)�2(t�2v2)6 6 12 1019. Vx6 20. Va6 1�2 1�457. (x y1�4)(x y) 58. (a1�3b1�6)2(a1�3b2�3)

21. Vx6 22. Va93 6 3 6 9.3 Adding, Subtracting, and Multiplying Radicals

23. V26x9 24. V63a7 Perform the operations and simplify. Assume the variables 25. V8w 26. V20n represent positive real numbers.65 625

3 4 59. V136 . V613 27. V16x 28. V54b56 3 6 3 3 3

9b5 11 14 14 1460. V6 . V6 . V629. Va 30. V32m4 6 4 6 61. V27 6 � V756 � V45 63 3x 12a

31. 32. 62. V12 6 � V726 � V50 616 25 63. 3V26 (5V26 � 7V63)

Find the domain of each radical expression. Use interval notation. 64. �2Va6 (Va 6)6 � Vab6

33. V2x � 56 65. (2 � V36)(3 � V26)

34. V3x � 1662 66. (2Vx6 � V6y)(Vx6 � V6y) 3 635. V7x � 1

9.4 Quotients, Powers, and Rationalizing Denominators3 636. V9 � 2x Perform the operations and simplify. Assume the variables

4 6 1 represent positive real numbers.37. V�3x �6

67. 5 V62 68. (10V66) (2V26) 4 6 138. V�5x �6 2 1

69. 70. 5 61

39. x � 1 2 2 1

71. 3 72. 3 3 92

40. x � 2 3 2 3

73. 74. V63x V2y6Find the domain of each radical function.

41. T(x) Vx � 5 V6y 66 10 3 V5x575. 76. V66 V8642. W(x) V6 � 2x6

a4 343. y V20 � x 77. 78. 33 6 Va2 4

6 V2a 6

44. y V3x6 5 3 79. 4 80. b 45. S(x) V17x – 16 V6262 3x 4

6Va2b3 5 6 6)446. T(x) V9 � 5x 81. (V3 82. (�2V6x)9

2 � V8 �3 � 66 V189.2 Rational Exponents 83. 84. 2 �6Simplify the expressions involving rational exponents. Assume

all variables represent positive real numbers. Write your V66 V156 85. 86.

answers with positive exponents. 1 � V63 2 � V56

47. (�27)�2�3 48. �253�2

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�

620 Chapter 9 Radicals and Rational Exponents 9-64

2V36 87.

3V66 � V126

�Vxy6 88.

3Vx6 � Vxy6

3 4 2 3 6)6 6)889. (2wV2w 90. (mVm

9.5 Solving Equations with Radicals and Exponents Find all real solutions to each equation.

2 91. x 16 2 92. w 100

93. (a � 5)2 4

94. (m � 7)2 25

95. (a � 1)2 5

96. (x � 5)2 3

97. (m � 1)2 �8

98. (w � 4)2 16

99. Vm � 1 3 6 100. 3Vx � 5 12 6 101. V63 2x � 9 3

102. V64 2x � 1 2

2�3103. w 4

�4�3104. m 16

105. (m � 1)1�3 5

106. (w � 3)�2�3 4

107. Vx � 3 Vx � 2 6 6 � 1

108. Vx6x � 6 4 2 � 36

62 109. V5x � x6 V66

110. Vx � 4 6 6 � 2Vx � 1 �1

111. Vx � 7 6 6 � 2Vx �2

112. Vx 66 � Vx � 1 1

113. 2Vx 66 � Vx � 3 3

114. 1 � Vx � 7 V2x � 7 6 6

9.6 Complex Numbers Perform the indicated operations. Write answers in the form a � bi.

115. (2 � 3i)(�5 � 5i)

116. (2 � i)(5 � 2i)

117. (2 � i) � (5 � 4i)

118. (2 � i) � (3 � 6i)

119. (1 � i) � (2 � 3i)

120. (3 � 2i) � (1 � i)

6 � 3i 121.

3 8 � 12i

122. 4

4 � V�126 123.

2 6 � V�18 6

124. 3

2 � 3i 125.

4 � i 3 � i

126. 2 � 3i

127. (�2i)4

128. (�2i)5

i14129.

i21130.

Find the imaginary solutions to each equation.

131. x2 � 100 0

132. 25a2 � 3 0

133. 2b2 � 9 0

134. 3y2 � 8 0

Miscellaneous

Determine whether each equation is true or false and explain your answer. An equation involving variables should be marked true only if it is an identity. Do not use a calculator.

135. 23 . 32 65

41�2 136. 161�4

137. (V62)3 2V62 3 138. V96 3

8200 . 8200 64200 139.

140. V295 66 . V295 295

41�2 141. V26

142. Va62 � a � 143. 52 . 52 254

144. V66 V26 V36

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�

9-65 Chapter 9 Review Exercises 621

10 5 165. Dropping a rock. The time in seconds t that it takes for a145. Vw6 w rock to fall to the earth from a height of h feet is given by

146. Va616 a4 the function 147. Vx66 x3

6 16

3 6 t(h) 0.25h 1�2.148. V6 V4

149. Vx68 x4 a) Find t(100). 9

150. V266 22�3 b) If it takes 4 seconds for a rock to reach the ground 6151. V16 2 when dropped from the top of a tall building, then

152. 21�2 . 21�4 23�4 what is the height of the building?

2600 4300153. 4 6154. V26 . V26 V26 166. Skid marks. Under certain conditions the speed S in

miles per hour prior to an accident is determined from the2 � V66 155.

2 1 � V66 length L in feet of the skid marks by the function

4 � 2V36 156. 2 � V36 S(L) (20L)1�2.

2 4 2 a) Find S(80).157. 6 3

b) Find the length of the skid marks for a car traveling 8200 . 8200 8400158. 70 mph.

159. 812�4 811�2 167. Guy wire. If a guy wire of length 40 feet is attached to

160. (�64)2�6 (�64)1�3 an antenna at a height of 30 feet, then how far from the base of the antenna is the wire attached to the ground?

161. (a4b2)1�2 � a2b � 2 1�2 � a �

162. �ab b6 � 3 Solve each problem.

163. Falling objects. Neglecting air resistance, the number of feet s that an object falls from rest during t seconds is 40 ft given by the formula s 16t2. How long would it take 30 ft

an object to reach the earth if it is dropped from 12,000 feet?

164. Timber. Anne is pulling on a 60-foot rope attached to the top of a 48-foot tree while Walter is cutting the tree at its base. How far from the base of the tree is Anne standing?

x ft

Figure for Exercise 167

168. Touchdown. Suppose at the kickoff of a football game, the receiver catches the football at the left side of the goal line and runs for a touchdown diagonally across the field. How many yards would he run? (A football field is

60 ft 48 ft 100 yards long and 160 feet wide.)

169. Long guy wires. The manufacturer of an antenna recommends that guy wires from the top of the antenna to the ground be attached to the ground at a distance from the base equal to the height of the antenna. How long

x ft would the guy wires be for a 200-foot antenna? Figure for Exercise 164

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�

622 Chapter 9 Radicals and Rational Exponents 9-66

170. Height of a post. Betty observed that the lamp post in front of her house casts a shadow of length 8 feet when the angle of inclination of the sun is 60 degrees. How tall is the lamp post? (In a 30-60-90 right triangle, the side opposite 30 is one-half the length of the hypotenuse.)

30�

x ft

60�

8 ft

Figure for Exercise 170

171. Manufacturing a box. A cubic box has a volume of 40 cubic feet. The amount of recycled cardboard that it takes to make the six-sided box is 10% larger than the surface area of the box. Find the exact amount of recycled cardboard used in manufacturing the box.

172. Shipping parts. A cubic box with a volume of 32 cubic feet is to be used to ship some machine parts. All of the parts are small except for a long, straight steel connecting rod. What is the maximum length of a connecting rod that will fit into this box?

173. Health care costs. The total annual cost of health care in the United States grew from $993.3 billion in 1995 to $2151.1 billion in 2009 (Statistical Abstract of the United States, www.census.gov).

4000

3000

2000

1000

0 0 5 10 15 20

Years since 1995

Figure for Exercise 173

C os

t ( bi

lli on

s of

d ol

la rs

) a) Find the average annual rate of growth r for that

period by solving 2151.1 993.3(1 � r)14.

b) Estimate the total annual cost of health care in 2015 by reading the accompanying graph.

174. Population growth. The formula P P0(1 � r)n gives the population P at the end of an n-year time period, where P0 is the initial population and r is the average annual growth rate. The U.S. population grew from 248.7 million in 1990 to 307.8 million in 2009 (U.S. Census Bureau). Find r for that period.

175. Landing speed. Aircraft engineers determine the proper landing speed V (in feet per second) for an airplane from the formula

841L V ,

where L is the gross weight of the aircraft in pounds, C is the coefficient of lift, and S is the wing surface area in square feet. Rewrite the formula so that the expression on the right-hand side is in simplified radical form.

176. Spillway capacity. Civil engineers use the formula

Q 3.32LH3�2

to find the maximum discharge that the dam (a broad- crested weir) shown in the figure can pass before the water breaches its abutments (Standard Handbook for Civil Engineers, 1968). In the formula, Q is the discharge in cubic feet per second, L is the length of the spillway in feet, and H is the depth of the spillway. Find Q given that L 60 feet and H 5 feet. Find H given that Q 3000 cubic feet per second and L 70 feet.

Figure for Exercise 176

http:www.census.gov

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9-67 Chapter 9 Test 623

Chapter 9 Test

Find each root. Variables represent any real numbers. Use Write each expression in the form a � bi. absolute value when necessary. 31. (3 � 2i)(4 � 5i)

3 61. V366 2. V�125

32. i4 � i5 33. Vt62 4. V3 6p

4 4 3 � i

5. Vw8 6. Vw12 33.6 6 1 � 2i

Simplify each expression. Assume all variables represent �6 � V�12 positive numbers. 34. 8

7. 82�3

Find all real or imaginary solutions to each equation. 8. 4�3�2

35. (x � 2)2 49 9. V216 V76

10. 2V56 . 3V5 36. 2Vx � 4 36 6

11. V20 66 � V5 2�337. w 4 1

12. V56 � V56 38. 9y2 � 16 0

. 21�213. 21�2

14. V72 39. V26x x � 12 x6 2 � 6 5 40. Vx � 1 66 � Vx � 4 515. 12

Show a complete solution to each problem.6 � V186 16. 41. Find the exact length of the side of a square whose diagonal6 17. (2V36 � 1)(V36 � 2) is 3 feet.

4 5 818. V326a y 42. Two positive numbers differ by 11, and their square roots differ by 1. Find the numbers.

119. 3 2V62x

43. If the perimeter of a rectangle is 20 feet and the diagonal is

20. 8a 6 feet, then what are the length and width?9 2V13

3 b

921. V3 6�27x 44. The average radius R of the orbit of a planet around the sun 3 is determined by R T2�3, where T is the number of years22. V20m6

for one orbit and R is measured in astronomical units (AU). 1�2 1�423. x . x If it takes Pluto 248.530 years to make one orbit of the sun,

4 1�2)1�224. (9y x then what is the average radius of the orbit of Pluto? If the average radius of the orbit of Neptune is 30.08 AU, then3 625. V40x7 how many years does it take Neptune to complete one orbit

26. (4 � V36)2 of the sun?

Find the domain of each radical expression. Use interval notation. 45. The maximum speed for a sailboat in knots M is a function27. V4 � x6

3 6 of the length of the waterline in feet w, given by M(w)28. V5x � 3

1.3Vw6. Rationalize the denominator and simplify. a) Find M(16) and M(25).

2 V66 b) Find the length of the waterline if the maximum speed29. 30. 5 � V63 4V63 � V26 is 9.1 knots.

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624 Chapter 9 Radicals and Rational Exponents 9-68

MakingConnections A Review of Chapters 1–9

Evaluate each expression

1. 3 � 2V14 � 2 . 5 2. 4 � 3⏐5 � 2 . 4⏐66 3. 5 � 2(6 � 2 . 42) 4. V132 � 6 � 66122

62 36 � 25 6362) � 235. V62 � 3 6. V4(7 � 7. (4 � 32) ⏐5 � 2 . 9⏐ 6 � ⏐9 � 16⏐8. V9 � 16

9. V(6�30)26� 4 . 9 . 25 10. V(�23)2612 . 56 6 � 4 . 6

Which elements of 3 1 33{�5, �V62, � , 0, 1, V64, 2.99, �, , 2 � 3i}9 4

are members of these sets?

11. Whole numbers 12. Natural numbers

13. Integers 14. Rational numbers

15. Irrational numbers 16. Real numbers

17. Imaginary numbers 18. Complex numbers

Fill in the blank.

19. Zero is the identity.

20. One is the identity.

21. According to the property of addition, a � b b � a for all real numbers a and b.

22. According to the property of multiplication, ab ba for all real numbers a and b.

23. According to the property of addition, a � (b � c) (a � b) � c for all real numbers a, b, and c.

24. According to the property, a(b � c) ab � ac for all real numbers a, b, and c.

25. Every real number a has a(n) inverse �a such that a � (�a) 0.

26. Every nonzero real number a has a(n) 1 1

inverse such that a . 1. a a

Find all real solutions to each equation or inequality. For the inequalities, also sketch the graph of the solution set.

27. 3(x � 2) � 5 7 � 4(x � 3)

28. V6x � 7 46

29. � 2x � 5 � � 1

30. 8x3 � 27 0

31. 2x � 3 � 3x � 4

32. V2x � 3 66 � V3x � 4 0 w w � 4 11

33. � 3 2 2

34. 2(x � 7) � 4 x � (10 � x)

35. (x � 7)2 25

�1�236. a 4

37. x � 3 � 2 or x � 2x � 6

�2�3 2 � 138. a 16 39. 3x 0

40. 5 � 2(x � 2) 3x � 5(x � 2) � 1

41. �3x � 4 � � 5

42. 3x � 1 0 643. Vy � 1 9

44. �5(x � 2) � 1� 3

45. 0.06x � 0.04(x � 20) 2.8

3V62 V63 46. �3x � 1� � �2 47.

x 4V65 Vx6 � 4 1

48. x Vx6 � 5

3V2 � 4 xV18 6 6 49.

V26 3V62 � 2

x 2V56 � V62 50.

2V56 � V26 x

V2x � 5 36 � 51.

x V2x � 56

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9-69

V66 � 2 2 52.

x V66 � 4

x � 1 V66 53.

V66 x

x � 3 V10 6 54.

V10 x6

1 1 1 55. � �

x x � 1 6

1 1 2 56. �

x2 � 2x x 3

Chapter 9 Making Connections 625

popped corn v (in cubic centimeters) that results is modeled by the formula

2v �94.8 � 21.4x � 0.761x .

a) Use the formula to find the volume that results when 1 gram of popcorn with moisture content 11% is popped.

b) Use the accompanying graph to estimate the moisture content that will produce the maximum volume of popped corn.

c) Use the graph to estimate the maximum possible volume for popping 1 gram of popcorn in a hot-air popper.

b2 � 46The expression �b � V6ac will be used in Chapter 10 to solve 2a

quadratic equations. Evaluate it for the given values of a, b, and c.

57. a 1, b 2, c �15

58. a 1, b 8, c 12

59. a 2, b 5, c �3

60. a 6, b 7, c �3 Vo lu

m e

of p

op pe

d co

rn (

cm )

8 10 12 14 16 18 0

Solve each problem. Moisture content of popcorn (%)

61. Popping corn. If 1 gram of popcorn with moisture content x% is popped in a hot-air popper, then the volume of Figure for Exercise 61

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Figure for Exercise 7

626 Chapter 9 Radicals and Rational Exponents 9-70

CriticalThinking For Individual or Group Work Chapter 9

These exercises can be solved by a variety of techniques, which may or may not require algebra. So be creative and think critically. Explain all answers. Answers are in the Instructor’s Edition of this text.

1. Wagon wheel. A wagon wheel is placed against a wall as shown in the accompanying figure. One point on the edge of the wheel is 5 inches from the ground and 10 inches from the wall. What is the radius of the wheel?

Figure for Exercise 1

2. Comparing jobs. Bob has two job offers with a starting salary of $100,000 per year and monthly paychecks. The Atlanta employer will raise his annual salary by $2000 at the end of every year, while the Chicago employer will raise his annual salary by $1000 at the end of every six months.

a) Which job is the better deal? b) How much more will Bob have made at the end of

10 years with the better deal?

3. Floor tiles. A square floor is tiled using 121 square floor tiles. Only whole tiles are used. How many tiles are neither diagonal tiles nor edge tiles?

4. Counting days. If the first day of this century was January 1, 2000, then how many days are there in this century? (A year is a leap year if it is divisible by 4, unless it’s divisible by 100, in which case it isn’t, unless it’s divisible by 400, in which case it is.)

5. Planting trees. How can you plant 10 trees in five rows with four trees in each row?

6. Counting rectangles. How many rectangles of any size are there on an 8 by 8 checker board?

7. Chime time. The clock in the bell tower at Webster College chimes every hour on the hour: once at 1 o’clock, twice at 2 o’clock, and so on. The clock takes 5 seconds to chime at 4 o’clock and 15 seconds to chime at 10 o’clock. The time needed to chime 1 o’clock is negligible. What is the total number of seconds needed for the clock to do all of its chiming in a 24-hour period starting at 1 P.M.?

Photo for Exercise 7

8. Arranging digits. In how many ways can you arrange the digits 8, 7, 6, and 3 to form a four-digit number divisible by 9, using each digit once and only once?