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Section 10.3 Solving Radical Equations 559
Essential QuestionEssential Question How can you solve an equation that
contains square roots?
Analyzing a Free-Falling Object
Work with a partner. The table shows the time t (in seconds) that it takes a free-falling object (with no air resistance) to fall d feet.
a. Use the data in the table to sketch the graph of t as a function of d. Use the coordinate plane below. b. Use your graph to estimate the time it takes the object to fall 240 feet. c. The relationship between d and t is given by the function
t = (^) √
— —^ d 16
Use this function to check your estimate in part (b). d. It takes 5 seconds for the object to hit the ground. How far did it fall? Explain your reasoning.
32
2
− 2
4
d
t
64 96 128 160 192 224 256 288 320
Solving a Square Root Equation
Work with a partner. The speed s (in feet per second) of the free-falling object in Exploration 1 is given by the function
s = √
— 64 d.
Find the distance the object has fallen when it reaches each speed. a. s = 8 ft/sec b. s = 16 ft/sec c. s = 24 ft/sec
Communicate Your AnswerCommunicate Your Answer
3. How can you solve an equation that contains square roots? 4. Use your answer to Question 3 to solve each equation. a. 5 = √
— x + 20 b. 4 = √
— x − 18 c. √— x + 2 = 3 d. − 3 = − 2 √— x
MODELING WITH
MATHEMATICS
To be proficient in math, you need to routinely interpret your mathematical results in the context of the situation and reflect on whether the results make sense.
10.3 Solving Radical Equations
d (feet) t (seconds)
0 0. 32 1. 64 2. 96 2. 128 2. 160 3. 192 3. 224 3. 256 4. 288 4. 320 4.
560 Chapter 10 Radical Functions and Equations
10.3 Lesson^ What You Will LearnWhat You Will Learn
Solve radical equations. Identify extraneous solutions. Solve real-life problems involving radical equations.
Solving Radical Equations A radical equation is an equation that contains a radical expression with a variable in the radicand. To solve a radical equation involving a square root, first use properties of equality to isolate the radical on one side of the equation. Then use the following property to eliminate the radical and solve for the variable.
Solving Radical Equations
Solve each equation. a. √— x + 5 = 13 b. 3 − √ — x = 0
SOLUTION
a. √— x + 5 = 13 Write the equation. √— x = 8 Subtract 5 from each side. (√— x )^2 = 82 Square each side of the equation. x = 64 Simplify.
The solution is x = 64.
b. 3 − √— x = 0 Write the equation. 3 = √— x Add √ — x to each side. 32 = (√— x )^2 Square each side of the equation. 9 = x Simplify.
The solution is x = 9.
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Solve the equation. Check your solution.
1. √— x = 6 2. √— x − 7 = 3 3. (^) √— y + 15 = 22 4. 1 − √— c = − 2
Check √— x + 5 = 13
√
— 64 + 5 =
Check 3 − √— x = 0
3 − √
— 9 =
radical equation, p. 560 Previous radical radical expression extraneous solution
Core VocabularyCore Vocabullarry
CoreCore ConceptConcept
Squaring Each Side of an Equation
Words If two expressions are equal, then their squares are also equal. Algebra If a = b , then a^2 = b^2.
562 Chapter 10 Radical Functions and Equations
Identifying Extraneous Solutions Squaring each side of an equation can sometimes introduce an extraneous solution.
Identifying an Extraneous Solution
Solve x = √
— x + 6.
SOLUTION
x = √
— x + 6 Write the equation.
x^2 = (√
— x + 6 )
2 Square each side of the equation. x^2 = x + 6 Simplify. x^2 − x − 6 = 0 Subtract x and 6 from each side. ( x − 3)( x + 2) = 0 Factor. x − 3 = 0 or x + 2 = 0 Zero-Product Property x = 3 or x = − 2 Solve for x.
Check Check each solution in the original equation.
3 =
— 3 + 6 Substitute for x. − 2 =
— − 2 + 6
3 =
— 9 Simplify. − 2 =
— 4
3 = 3 ✓ Simplify. − 2 ≠ 2 ✗
Because x = −2 does not satisfy the original equation, it is an extraneous solution. The only solution is x = 3.
Identifying an Extraneous Solution
Solve 13 + √
— 5 n = 3.
SOLUTION
— 5 n = 3 Write the equation.
√
— 5 n = − 10 Subtract 13 from each side.
(√— 5 n )^2 = (−10)^2 Square each side of the equation.
5 n = 100 Simplify. n = 20 Divide each side by 5.
Because n = 20 does not satisfy the original equation, it is an extraneous solution. So, the equation has no solution.
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Solve the equation. Check your solution(s).
11. √
— 4 − 3 x = x 12. √
— 3 m + 10 = 1 13. p + 1 = (^) √
— 7 p + 15
ATTEND TO PRECISION To understand how extraneous solutions can be introduced, consider the equation √ — x = −2. This equation has no real solution, however, you obtain x = 4 after squaring each side.
STUDY TIP Be sure to always substitute your solutions into the original equation to check for extraneous solutions.
Check 13 + √
— 5 n = 3 13 + (^) √
— 5(20) =
— 100 =
Section 10.3 Solving Radical Equations 563
Solving Real-Life Problems
Modeling with Mathematics
The period P (in seconds) of a pendulum is given by the
function P = 2 π √
— L — 32
, where L is the pendulum length (in feet). A pendulum has a period of 4 seconds. Is this pendulum twice as long as a pendulum with a period of 2 seconds? Explain your reasoning.
SOLUTION
1. Understand the Problem You are given a function that represents the period P of a pendulum based on its length L. You need to fi nd and compare the values of L for two values of P. 2. Make a Plan Substitute P = 2 and P = 4 into the function and solve for L. Then compare the values. 3. Solve the Problem
P = 2 π√
— L — 32
Write the function. P = 2 π√
— L — 32
2 = 2 π√
— L — 32
Substitute for P. 4 = 2 π√
— L — 32 2 — 2 π
= (^) √
— L — 32
Divide each side by 2 π. (^) —^4 2 π
= (^) √
— L — 32 1 — π
= (^) √
— L — 32
Simplify. (^) —^2 π
= (^) √
— L — 32 1 — π 2
= — L
Square each side and simplify. (^) —^4 π 2
= — L
— π 2
= L Multiply each side by 32. (^128) — π 2
= L
3.24 ≈ L Use a calculator. 12.97 ≈ L
No, the length of the pendulum with a period of 4 seconds is 128 — π^2
÷ —^32
π^2
= 4 times longer than the length of a pendulum with a
period of 2 seconds.
4. Look Back Use the trace feature of a graphing calculator to check your solutions.
0 0
6
X=3.24 Y=1.9992973 16
y = 2 π 323232 x
0 0
6
X=12.97 Y=4.000137 16
y = 2 π 323232 x
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14. What is the length of a pendulum that has a period of 2.5 seconds?
STUDY TIP
The period of a pendulum is the amount of time it takes for the pendulum to swing back and forth.
Section 10.3 Solving Radical Equations 565
MATHEMATICAL CONNECTIONS In Exercises 35 and 36, fi nd the value of x****.
35. Perimeter = 22 cm 36. Area = √
— 5 x − 4 ft^2
6 x − 5 cm
4 cm 2 ft
3 x + 12 ft
In Exercises 37–44, solve the equation. Check your solution. (See Example 4.)
37. √^3 — x = 4 38. (^) √^3 — y = 2 39. 6 = 3 √
— 8 g 40. 3
— r + 19 = 3
41. 3
— 2 s + 9 = − 3 42. − 5 = 3
— 10 x + 15
43. 3 √
— y + 6 = 3 √
— 5 y − 2 44. 3 √
— 7 j − 2 = 3 √
— j + 4
In Exercises 45–48, determine which solution, if any, is an extraneous solution.
45. √
— 6 x − 5 = x ; x = 5, x = 1
46. (^) √
— 2 y + 3 = y ; y = −1, y = 3
47. (^) √
— 12 p + 16 = − 2 p ; p = −1, p = 4
48. − 3 g = (^) √
— − 18 − 27 g ; g = −2, g = − 1
In Exercises 49–58, solve the equation. Check your solution(s). (See Examples 5 and 6.)
49. y = (^) √
— 5 y − 4 50. √
— − 14 − 9 x = x
51. √
— 1 − 3 a = 2 a 52. 2 q = (^) √
— 10 q − 6
53. 9 + (^) √
— 5 p = 4 54. √
— 3 n − 11 = − 5
55. √
— 2 m + 2 − 3 = 1 56. 15 + √
— 4 b − 8 = 13
57. r + 4 = √
— − 4 r − 19 58. √
— 3 − s = s − 1
ERROR ANALYSIS In Exercises 59 and 60, describe and correct the error in solving the equation.
59. (^2) + 5 √— x = 12
5 √— x = 10 5 x = 100 x = 20
x = √
— 12 − 4 x x^2 = 12 − 4 x x^2 + 4 x − 12 = 0 (x − 2)(x + 6) = 0 x = 2 or x = − 6 The solutions are x = 2 and x = − 6.
61. REASONING Explain how to use mental math to solve √
— 2 x + 5 = 1.
62. WRITING Explain how you would solve 4 √
— m + 4 − 4 √
— 3 m = 0.
63. MODELING WITH MATHEMATICS The formula V = √
— PR relates the voltage V (in volts), power P (in watts), and resistance R (in ohms) of an electrical circuit. The hair dryer shown is on a 120-volt circuit. Is the resistance of the hair dryer half as much as the resistance of the same hair dryer on a 240-volt circuit? Explain your reasoning. (See Example 7.)
64. MODELING WITH MATHEMATICS The time t (in seconds) it takes a trapeze artist to swing back and
forth is represented by the function t = 2 π√
— r — 32
where r is the rope length (in feet). It takes the trapeze artist 6 seconds to swing back and forth. Is this rope 3 — 2 as long as the rope used when it takes the trapeze artist 4 seconds to swing back and forth? Explain your reasoning.
REASONING In Exercises 65– 68, determine whether the statement is true or false****. If it is false, explain why.
65. If √ — a = b , then (√^ — a )^2 = b^2. 66. If √ — a = √
— b , then a = b.
67. If a^2 = b^2 , then a = b. 68. If a^2 = √^
— b , then a^4 = (√
— b )
2 .
566 Chapter 10 Radical Functions and Equations
69. COMPARING METHODS Consider the equation x + 2 = √
— 2 x − 3. a. Solve the equation by graphing. Describe the process. b. Solve the equation algebraically. Describe the process. c. Which method do you prefer? Explain your reasoning.
70. HOW DO YOU SEE IT? The graph shows two radical functions.
y = 2 x + 3
y = 4 x − 3
4
2
− 2 2 4 x
y
a. Write an equation whose solution is the x -coordinate of the point of intersection of the graphs. b. Use the graph to solve the equation.
71. MATHEMATICAL CONNECTIONS The slant height s of a cone with a radius of r and a height of h is given by s = √
— r^2 + h^2. The slant heights of the two cones are equal. Find the radius of each cone.
s
s
2 r 2 r
4
72. CRITICAL THINKING How is squaring √
— x + 2 different from squaring √— x + 2?
USING STRUCTURE In Exercises 73–78, solve the equation. Check your solution.
73. √
— m + 15 = √— m + √
— 5 74. 2 − √
— x + 1 = √
— x + 2
75. (^) √
— 5 y + 9 + (^) √
— 5 y = 9
76. √
— 2 c − 8 − √
— 2 c − 4 = 0
77. 2 √
— 1 + 4 h − 4 √
— h − 2 = 0
— 20 − 4 z + (^2) √ —− z = 10
79. OPEN-ENDED Write a radical equation that has a solution of x = 5. 80. OPEN-ENDED Write a radical equation that has x = 3 and x = 4 as solutions. 81. MAKING AN ARGUMENT Your friend says the equation (^) √
— (2 x + 5)^2 = 2 x + 5 is always true, because after simplifying the left side of the equation, the result is an equation with infinitely many solutions. Is your friend correct? Explain.
82. THOUGHT PROVOKING Solve the equation 3 √
— x + 1 = √
— x − 3. Show your work and explain your steps.
83. MODELING WITH MATHEMATICS The frequency f (in cycles per second) of a string of an electric guitar is given by the equation f = —^1 2 √
— —^ T m
, where is the length of the string (in meters), T is the string’s tension (in newtons), and m is the string’s mass per unit length (in kilograms per meter). The high E string of an electric guitar is 0.64 meter long with a mass per unit length of 0.000401 kilogram per meter. a. How much tension is required to produce a frequency of about 330 cycles per second? b. Would you need more or less tension to create the same frequency on a string with greater mass per unit length? Explain.
Maintaining Mathematical ProficiencyMaintaining Mathematical Proficiency
Find the product. (Section 7.2)
84. ( x + 8)( x − 2) 85. (3 p − 1)(4 p + 5) 86. ( s + 2)( s^2 + 3 s − 4)
Graph the function. Compare the graph to the graph of f ( x ) = x^2. (Section 8.1)
87. r ( x ) = 3 x^2 88. g ( x ) = 3 — 4 x^2 89. h ( x ) = − 5 x^2
Reviewing what you learned in previous grades and lessons
