6 3 binomial radical expressions page 162

C h

a p

t e

r

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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

re

25

20

15

10

5

0

5

10

15

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

6

1

= = x y

7

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

4

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.

V3

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

20

15

10

5

0

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 )

3

2

1

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

1

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 .

4

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 =

1

8

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

3

1

/

/

4

4

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

= 272/3

= 9

b) 5 5

3

1

/

/

4

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

6

9

) ) -

-

1

1

/

/

3

3 Power of a quotient rule

= 2 3

-

-

2

3 Power of a power rule

= 3 2

3

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

3

.

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/

2

3

) 1/2

= y1

x

/6

1

3 .

1

2 =

1

6

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

1

9 6 93. �16�

-1/2 94. �11�

-1/4

8

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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.

D

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.

3

r

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

)

35

30

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

10

258 20

6 15

4

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

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