1. Solving Radical Equations
ALGEBRA 1 LESSON 11-5
pages 610–612  Exercises
1. 4
2. 49
3. 36
4. 137
5. 15
6. 16 ft
watts
9. 4.5
10. 3
11. 7
12. –2
13. 4
15. 2
16. –4
17. none
18. –
19. –7
20. none
21. 3
22. 5
23. no solution
24. 2
25. no solution
26. 4 or
, 1
29. a. 25 b
30. about 2.5 in.
31. An extraneous solution is
a solution of a new equation
that does not satisfy the
original equation.
32. Answers may vary. Sample:
x – 2 = – 2x , 3x = 3 5 4 1 4 1 2
11-5

2. Solving Radical Equations
ALGEBRA 1 LESSON 11-5 ft
34. 3
35. no solution
36. no solution
37. 1, 6
39. 11
40. 0, 12
41. 3, 6
42. 44
43. no solution
44. a. 68 ft
b mi/h
c. As radius increases,
velocity decreases.
As height decreases,
d. Velocity depends upon
the difference of the
height and the radius.
45. a.
b. approximately (6, 3.6)
c. 6; it is the x-coordinate of
the point of intersection.
46. a. V = 10x2
b. x =
c. 2, 3, 4, 5, 6, 7
47. a. – 7, 7
b. 49
c. In both cases 3 is
added to each side.
To solve the first
equation you find the
square roots of each
side, and in the second
square of each side.
48. –2, 8
49. 0 V 10
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3. Solving Radical Equations
ALGEBRA 1 LESSON 11-5
51. –1
52. Subtract 2x from each side.
Square both sides. Solve for x.
Check the solution if there is one.
53. The square of x – 1 will have
only 2 terms while x – 1 squared
will have 3 terms.
54. a. about 2.0 m
b. about 32.4 m
55. C
56. G
57. B
58. A
59. B
60. C
61. [2] – 5x = 4x – 3
15 – 5x = 4x – 3
–9x = –18
x = 2
Check: – 5(2) (2) – 3
5 = 5
The solution is 2.
[1] correct technique with a minor error
OR correct answer, no work shown
65. 32
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4. Solving Radical Equations
ALGEBRA 1 LESSON 11-5
66. 4( – 3)
, –4.5
, –0.4
70. –0.2, –4.8
71. –10.7, 0.7
72. –11.7, 1.7
73. –1.6, 3.1
74. (x + 12)(x – 2)
75. (m – 13)(m – 1)
76. (b + 18)(b – 2)
77. (2p + 1)(p + 7)
78. 3(d – 1)(d + 5)
79. (4v – 5)(v – 5)
11-5