Simplify each expression. See Example 8.

79. 80.

5�3� 5�2� �3�6� �

43�2�

�

81. 82.

��1�5�

2 3� �

�2

3�1�0��

83. 84.

�13

227�3��

85. 86.

�19 �

2171�7��

Simplify.

87. �5�a� �2�0�a� 88. a�6� � a�1�2�3�5�a� 6a2�2�

89. �7�5� � �6� 90. �2�4� � �1�5�0�

�5�

22�

� �3�6�

91. (5 3�5�)2 92. (�6� � �5�)(�6� �5�)70 30�5� 1

93. �5� ��

32�0�� 94.

�5�

35�

� 2�2� �3�

Use a calculator to find the approximate value of eachexpression to three decimal places.

95. 1.866 96. 0.045�2 �3���

�6

2 �3��

2

5���8� � �3�

3�1�0� 12�2� 5 4�5����

13

�5� 4��3�2� � �5�

�7� � 5�2�7� 1

�6� � 3 �2� � �3����

2

�2� � �3����3� � 1

2 �3��5 � �3�

2���2� �5�

�3����5� � �3�

3���6� �2�

5���3� � �2�

432 (8-20) Chapter 8 Powers and Roots

97. �1.974 98. �1.465

Solve each problem.

99. Find the exact area and perimeter of the given rectangle.12 ft2, 10�2� ft

F I G U R E F O R E X E R C I S E 9 9

100. Find the exact area and perimeter of the given triangle.9, 5�3� �39�

F I G U R E F O R E X E R C I S E 1 0 0

101. Find the exact volume in cubic meters of the given rec-tangular box.6 m3

F I G U R E F O R E X E R C I S E 1 0 1

102. Find the exact surface area of the box in Exercise 101.2�6� 4�3� 6�2� m2

√––

3 m

√––

6 m√––

2 m

√—–

12

√—–

39

√—–

27

√––

8 ft

√—–

18 ft

�5 � �2����3� �7�

�4 � �6���5 � �3�

S O L V I N G E Q U A T I O N S W I T H R A D I C A L S A N D E X P O N E N T S

Equations involving radicals and exponents occur in many applications. In this sec-tion you will learn to solve equations of this type, and you will see how these equa-tions occur in applications.

The Square Root Property

An equation of the form x2 � k can have two solutions, one solution, or no solu-tions, depending on the value of k. For example,

x2 � 4

8.4I n t h i ss e c t i o n

● The Square Root Property

● Obtaining EquivalentEquations

● Squaring Each Side of anEquation

● Solving for the IndicatedVariable

● Applications

has two solutions because (2)2 � 4 and (�2)2 � 4. So x2 � 4 is equivalent to thecompound equation

x � 2 or x � �2,

which is also written as x � �2 and is read as “x equals positive or negative 2.” Theonly solution to x2 � 0 is 0. The equation

x2 � �4

has no real solution because the square of every real number is nonnegative.These examples illustrate the square root property.

Square Root Property (Solving x2 � k)

For k � 0 the equation x2 � k is equivalent to the compound equation

x � �k� or x � ��k�. (Also written x � ��k�.)

For k � 0 the equation x2 � k is equivalent to x � 0.For k � 0 the equation x2 � k has no real solution.

The expression �9� has a value of 3 only, but the equationx2 � 9 has two solutions: 3 and �3.

E X A M P L E 1 Using the square root propertySolve each equation.

a) x2 � 12 b) 2(x � 1)2 � 18 � 0

c) x2 � �9 d) (x � 16)2 � 0

Solutiona) x2 � 12

x � ��1�2� Square root property

x � �2�3�Check: (2�3�)2 � 4 � 3 � 12, and (�2�3�)2 � 4 � 3 � 12.

b) 2(x � 1)2 � 18 � 0

2(x � 1)2 � 18 Add 18 to each side.

(x � 1)2 � 9 Divide each side by 2.

x � 1 � ��9� Square root property

x � 1 � 3 or x � 1 � �3

x � 2 or x � �4

Check 2 and �4 in the original equation. Both �4 and 2 are solutions to theequation.

c) The equation x2 � �9 has no real solution because no real number has a squarethat is negative.

d) (x � 16)2 � 0

x � 16 � 0 Square root property

x � 16

Check: (16 � 16)2 � 0. The equation has only one solution, 16. ■

8.4 Solving Equations with Radicals and Exponents (8-21) 433

h e l p f u l h i n t

C A U T I O N

We do not say “take the squareroot of each side.” We are notdoing the same thing to eachside of x 2 � 12 when we writex � ��12�. This is the secondtime that we have seen a rulefor obtaining an equivalentequation without “doing thesame thing to each side.”(What was the first?)

Obtaining Equivalent Equations

When solving equations, we use a sequence of equivalent equations with each onesimpler than the last. To get an equivalent equation, we can

1. add the same number to each side,

2. subtract the same number from each side,

3. multiply each side by the same nonzero number, or

4. divide each side by the same nonzero number.

However, “doing the same thing to each side” is not the only way to obtain anequivalent equation. In Chapter 5 we used the zero factor property to obtain equiv-alent equations. For example, by the zero factor property the equation

(x � 3)(x � 2) � 0

is equivalent to the compound equation

x � 3 � 0 or x � 2 � 0.

In this section you just learned how to obtain equivalent equations by the squareroot property. This property tells us how to write an equation that is equivalent tothe equation x2 � k. Note that the square root property does not tell us to “take thesquare root of each side.” To become proficient at solving equations, we must un-derstand these methods. One of our main goals in algebra is to keep expanding ourskills for solving equations.

Squaring Each Side of an Equation

Some equations involving radicals can be solved by squaring each side:

�x� � 5

(�x�)2 � 52 Square each side.

x � 25

All three of these equations are equivalent. Because �2�5� � 5 is correct, 25 satis-fies the original equation.

However, squaring each side does not necessarily produce an equivalent equa-tion. For example, consider the equation

x � 3.

Squaring each side, we get

x2 � 9.

Both 3 and �3 satisfy x2 � 9, but only 3 satisfies the original equation x � 3. Sox2 � 9 is not equivalent to x � 3. The extra solution to x2 � 9 is called an extra-neous solution.

These two examples illustrate the squaring property of equality.

Squaring Property of Equality

When we square each side of an equation, the solutions to the new equationinclude all of the solutions to the original equation. However, the new equa-tion might have extraneous solutions.

434 (8-22) Chapter 8 Powers and Roots

s t u d y t i p

Make sure that you knowwhat your instructor expectsfrom you. You can deter-mine what your instructorfeels is important by lookingat the examples that yourinstructor works in class andthe homework assigned.When in doubt, ask yourinstructor what you will beresponsible for and writedown the answer.

This property means that we may square each side of an equation, but we mustcheck all of our answers to eliminate extraneous solutions.

E X A M P L E 2 Using the squaring property of equalitySolve each equation.

a) �x2� �� 1�6� � 3 b) x � �2�x��� 3� c) �x2� �� 4�x� � �2� �� 3�x�

Solutiona) �x2� �� 1�6� � 3

(�x2� �� 1�6�)2 � 32 Square each side.

x2 � 16 � 9

x2 � 25

x � �5 Square root property

Check each solution:

�5�2��� 1�6� � �2�5� �� 1�6� � �9� � 3

�(��5�)2� �� 1�6� � �2�5� �� 1�6� � �9� � 3

So both 5 and �5 are solutions to the equation.

b) x � �2�x��� 3�x2 � (�2�x��� 3�)2 Square each side.

x2 � 2x � 3

x2 � 2x � 3 � 0 Solve by factoring.

(x � 3)(x � 1) � 0 Factor.

x � 3 � 0 or x � 1 � 0 Zero factor property

x � 3 or x � �1

Check in the original equation:

Check x � 3: Check x � �1:

3 � �2� �� 3� �� 3� �1 � �2�(��1�)��� 3�3 � �9� Correct �1 � �1� Incorrect

Because �1 does not satisfy the original equation, �1 is an extraneous solu-tion. The only solution is 3.

c) �x2� �� 4�x� � �2� �� 3�x�x2 � 4x � 2 � 3x Square each side.

x2 � x � 2 � 0

(x � 2)(x � 1) � 0

x � 2 � 0 or x � 1 � 0 Zero factor property

x � 2 or x � �1

Check each solution in the original equation:

Check x � 2: Check x � �1:

�2�2��� 4� �� 2� � �2� �� 3� �� 2� �(��1�)2� �� 4�(��1�)� � �2� �� 3�(��1�)����4� � ���4� �5� � �5�

Because ���4� is not a real number, 2 is an extraneous solution. The onlysolution to the original equation is �1. ■

8.4 Solving Equations with Radicals and Exponents (8-23) 435

s t u d y t i p

Most instructors believe thatwhat they do in class is impor-tant. If you miss class, you misswhat is important to your in-structor.You miss what is mostlikely to appear on the test.

In the next example, one of the sides of the equation is a binomial. When wesquare each side, we must be sure to square the binomial properly.

E X A M P L E 3 Squaring each side of an equationSolve the equation x � 2 � ���2� �� 3�x�.

Solutionx � 2 � ���2� �� 3�x�

(x � 2)2 � (���2� �� 3�x�)2 Square each side.

x2 � 4x � 4 � �2 � 3x Square the binomial on the

x2 � 7x � 6 � 0left side.

(x � 6)(x � 1) � 0 Factor.

x � 6 � 0 or x � 1 � 0

x � �6 or x � �1

Check these solutions in the original equation:

Check x � �6: Check x � �1:

�6 � 2 � ���2� �� 3�(��6�)� �1 � 2 � ���2� �� 3�(��1�)��4 � �1�6� Incorrect 1 � �1� Correct

The solution �6 does not check. The only solution to the original equation is �1. ■

Solving for the Indicated Variable

In the next example we use the square root property to solve a formula for an indi-cated variable.

E X A M P L E 4 Solving for a variableSolve the formula A � �r2 for r.

SolutionA � �r2

�

A � r2 Divide each side by �.

���

A� � r Square root property

The formula solved for r is

r � ���

A�.

If r is the radius of a circle with area A, then r is positive and

r � ��

A�.

■

Applications

Equations involving exponents occur in many applications. If the exact answer to aproblem is an irrational number in radical notation, it is usually helpful to find adecimal approximation for the answer.

436 (8-24) Chapter 8 Powers and Roots

E X A M P L E 5 Finding the side of a square with a given diagonalIf the diagonal of a square window is 10 feet long, then what are the exact and ap-proximate lengths of a side? Round the approximate answer to two decimal places.

SolutionFirst make a sketch as in Fig. 8.1. Let x be the length of a side. The Pythagoreantheorem tells us that the sum of the squares of the sides is equal to the diagonalsquared:

x2 � x2 � 102

2x2 � 100x2 � 50x � ��5�0�

� �5�2�Because the length of a side must be positive, we disregard the negative solution.The exact length of a side is 5�2� feet. Use a calculator to get 5�2� � 7.07. Thesymbol � means “is approximately equal to.” The approximate length of a side is7.07 feet. ■

True or false? Explain your answer.

1. The equation x2 � 9 is equivalent to the equation x � 3. False

2. The equation x2 � �16 has no real solution. True

3. The equation a2 � 0 has no solution. False

4. Both ��5� and �5� are solutions to x2 � 5 � 0. False

5. The equation �x2 � 9 has no real solution. True

6. To solve �x��� 4� � �2�x��� 9�, first take the square root of each side.False

7. All extraneous solutions give us a denominator of zero. False

8. Squaring both sides of �x� � �1 will produce an extraneous solution.True

9. The equation x2 � 3 � 0 is equivalent to x � ��3�. True

10. The equation �2 � �6�x2� �� x� �� 8� has no solution. True

8.4 Solving Equations with Radicals and Exponents (8-25) 437

x ft

x ft

10 ft

F I G U R E 8 . 1

W A R M - U P S

Reading and Writing After reading this section, write out theanswers to these questions. Use complete sentences.

1. What is the square root property?The square root property says that x2 � k for k � 0 isequivalent to x � ��k�.

2. When do we take the square root of each side of anequation?We do not take the square root of each side of an equation.

3. What new techniques were introduced in this section forsolving equations?The square root property and squaring each side are twonew techniques used for solving equations.

4. Is there any way to obtain an equivalent equation other thandoing the same thing to each side?The square root property and the zero factor property pro-duce equivalent equations without doing the same thing toeach side.

5. Which property for solving equations can give extraneousroots?Squaring each side can produce extraneous roots.

6. Which property of equality does not always give you anequivalent equation?Squaring each side does not always give equivalentequations.

E X E R C I S E S8 . 4

438 (8-26) Chapter 8 Powers and Roots

Solve each equation. See Example 1.

7. x2 � 16 8. x2 � 49�4, 4 �7, 7

9. x2 � 40 � 0 10. x2 � 24 � 0

�2�1�0�, 2�1�0� �2�6�, 2�6�

11. 3x2 � 2 12. 2x2 � 3

��

3

6�,

�3

6� �

�2

6�,

�2

6�

13. 9x2 � �4 14. 25x2 � 1 � 0No solution No solution

15. (x � 1)2 � 4 16. (x � 3)2 � 9�1, 3 �6, 0

17. 2(x � 5)2 � 1 � 7 18. 3(x � 6)2 � 4 � 11

5 � �3�, 5 � �3� 6 � �5�, 6 � �5�

19. (x � 19)2 � 0 20. 5x2 � 5 � 5�19 0

Solve each equation. See Example 2.

21. �x��� 9� � 9 22. �x��� 3� � 4

90 13

23. �2�x��� 3� � �4 24. �3�x��� 5� � �9

No solution No solution

25. 4 � �x2� �� 9� 26. 1 � �x2� �� 1��5, 5 ��2�, �2�

27. x � �1�8� �� 3�x� 28. x � �6�x��� 2�7�3 9

29. x � �x� 30. x � �2�x�0, 1 0, 2

31. �x��� 1� � �2�x��� 5� 32. �1� �� 3�x� � �x��� 5�6 �1

33. 3�2�x��� 1� � 3 � 5 34. 4�x��� 5� � 3 � 9

1

1

3

8 4

Solve each equation. See Example 3.

35. x � 3 � �2�x��� 6� 36. x � 1 � �3�x��� 5�3, 5 2, 3

37. �x��� 1�3� � x � 1 38. x � 1 � �2�2� �� 2�x�3 3

39. �1�0�x��� 4�4� � x � 2 40. �8�x��� 7� � x � 16, 8 2, 4

Solve each formula for the indicated variable. See Example 4.

41. V � �r2h for r 42. V � 4

3�r2h for r

r � ���

V�h� r � ��

4

3��

V�h�

43. a2 � b2 � c2 for b 44. y � ax2 � c for x

b � ��c2� �� a�2� x � ��y��

a�c

�

45. b2 � 4ac � 0 for b 46. s � 1

2gt2 � v for t

b � �2�a�c� t � ��2�s��

g�2�v

�47. v � �2�p�t� for t 48. y � �2�x� for x

t � 2

v

p

2

x � y

2

2

Solve each equation.

49. 3x2 � 6 � 0 50. 5x2 � 3 � 0

��2�, �2� No solution

51. �2�x��� 3� � �3�x��� 1� 52. �2�x��� 4� � �x��� 9�No solution No solution

53. (2x � 1)2 � 8 54. (3x � 2)2 � 18

1 �

2

2�2�,

1 �

2

2�2�

2 �

3

3�2�,

2 �

3

3�2�

55. �2�x��� 9� � 0 56. �5� �� 3�x� � 0

9

2

5

3

57. x � 1 � �2�x��� 1�0� 58. x � 3 � �2�x��� 1�8�3 9

59. 3(x � 1)2 � 27 � 0 60. 2(x � 3)2 � 50 � 0�4, 2 �2, 8

61. (2x � 5)2 � 0 62. (3x � 1)2 � 0

5

2

1

3

Use a calculator to find approximate solutions to eachequation. Round your answers to three decimal places.

63. x2 � 3.25 64. (x � 1)2 � 20.3�1.803, 1.803 �5.506, 3.506

65. �x��� 2� � 1.73 66. �2�.3�x��� 1�.4� � 3.30.993 5.343

67. 1.3(x � 2.4)2 � 5.4 68. �2.4x2 � �9.550.362, 4.438 �1.995, 1.995

Find the exact answer to each problem. If the answer is irra-tional, then find an approximation to three decimal places. SeeExample 5.

69. Side of a square. Find the length of the side of a squarewhose area is 18 square feet. 3�2� or 4.243 ft

70. Side of a field. Find the length of the side of a square wheatfield whose area is 75 square miles. 5�3� or 8.660 mi

F I G U R E F O R E X E R C I S E 7 0

x miles

x miles

71. Side of a table. Find the length of the side of a square cof-fee table whose diagonal is 6 feet.

3�2� or 4.243 ft

72. Side of a square. Find the length of the side of a squarewhose diagonal measures 1 yard.

�2

2� or 0.707 yd

73. Diagonal of a tile. Find the length of the diagonal of asquare floor tile whose sides measure 1 foot each.

�2� or 1.414 ft

74. Diagonal of a sandbox. The sandbox at Totland is shapedlike a square with an area of 20 square meters. Find thelength of the diagonal of the square.

2�1�0� or 6.325 meters

75. Diagonal of a tub. Find the length of the diagonal of a rec-tangular bathtub with sides of 3 feet and 4 feet.5 ft

F I G U R E F O R E X E R C I S E 7 5

76. Diagonal of a rectangle. What is the length of the diago-nal of a rectangular office whose sides are 6 feet and 8 feet?10 ft

77. Falling bodies. If we neglect air resistance, then the num-ber of feet that a body falls from rest during t seconds is

3 ft

4 ft

8.4 Solving Equations with Radicals and Exponents (8-27) 439

given by s � 16t2. How long does it take a pine cone to fallfrom the top of a 100-foot pine tree?2.5 sec

78. America’s favorite pastime. A baseball diamond is actu-ally a square, 90 feet on each side. How far is it from homeplate to second base?

90�2� or 127.279 ft

F I G U R E F O R E X E R C I S E 7 8

79. Length of a guy wire. Aguy wire from the top of a 200-foottower is to be attached to a point on the ground whose

distance from the base of the tower is 23

of the height of the

tower. Find the length of the guy wire.

200�

3

1�3� or 240.37 ft

80. America’s favorite pastime. The size of a rectangular tele-vision screen is commonly given by the manufacturer asthe length of the diagonal of the rectangle. If a televisionscreen measures 10 inches wide and 8 inches high, thenwhat is the exact length of the diagonal of the screen? Whatis the approximate size of this television screen to thenearest inch?

2�4�1� or 13 inches

F I G U R E F O R E X E R C I S E 8 0

10 in.

8 in.x in.

1st base3rd base

2nd base

Home plate

90 ft

90 ft

100 ft

F I G U R E F O R E X E R C I S E 7 7