Prentice Hall Lesson 11.5 EQ: How do you solve a radical equation? BOP:

Solution to BOP:

ALGEBRA 1 LESSON 11-4 pages 603–606  Exercises 1. 5 6 4. 2 7 7. yes 10. 4 2 11 – 4 7 26. 32 + 9 11 28. 2 7 + 2 3 30. –4 6 – 12 2 32. –5 11 – 5 3 34. 10 2 + 10; 24.1 12. 4 2 14. 7 16. 4 – 4 2 18. 6 – 2 3 20. 3 2 + 6 22. –9 – 14 6 11-4

52. Answers may vary. Sample: 8 2 + 4 3, 2 7 + 9 3, 6 5 + 3 7 ALGEBRA 1 LESSON 11-4 47. 8 2 units 50. 6 10 units 52. Answers may vary. Sample: 8 2 + 4 3, 2 7 + 9 3, 6 5 + 3 7 53. a. The student simplified 48 as 2 24 instead of 2 12 or 4 3. b. 2 6 + 4 3 54. a. 2 2 or 2.8 ft b. s 2 10 5 37. 7.4 ft 38. 5 10 41. 8 + 2 15 44. –24 11-4

61. They are unlike radicals. 62. a. 1, 0, 1, 1; 4, 1, 5, 17; ALGEBRA 1 LESSON 11-4 55. 9.1% 56. 12.8% 57. 15.5% 58. a. x b. x x 59. 60. about 251 years 61. They are unlike radicals. 62. a. 1, 0, 1, 1; 4, 1, 5, 17; 5, 3, 8, 34; 8, 6, 14, 10; 10, 9, 19, 181 b. No; the only values it worked for were 0 and 1. 63. a + b ≠≠ a + b 64. 67. 2 70 71. a. 2 6 b. 2 13 c. 2(p + q) 9 2 2 n 2 n – 1 2 ab b 11-4

Prentice Hall Lesson 11.5 EQ: How do you solve a radical equation? Toolbox: Radical equation – an equation with a variable in a radicand. To solve a radical equation, isolate the radical using inverse operations. Then square both sides and solve for the variable. To solve radical equations with radical expressions on both sides, square both sides of the equation. Then use inverse operations to solve for the variable.

Extraneous solution – a solution that does not satisfy the original equation. ALWAYS check your solutions to determine whether a solution is extraneous. If the only solution you get after checking your work is extraneous, then the original equation has no solution.

4 = 4 Solve each equation. Check your answers. a. x – 5 = 4 ALGEBRA 1 LESSON 11-5 Solve each equation. Check your answers. a. x – 5 = 4 x = 9 Isolate the radical on the left side of the equation. ( x)2 = 92 Square each side. x = 81 Check: x – 5 = 4 81 – 5 4 Substitute 81 for x. 9 – 5 4 4 = 4 11-5

b. x – 5 = 4 ( x – 5)2 = 42 Square each side. x – 5 = 16 Solve for x. ALGEBRA 1 LESSON 11-5 (continued) b. x – 5 = 4 ( x – 5)2 = 42 Square each side. x – 5 = 16 Solve for x. x = 21 Check: x – 5 = 4 21– 5 = 4 Substitute 21 for x. 16 = 4 4 = 4 11-5

of a car at the top of the loop. ALGEBRA 1 LESSON 11-5 On a roller coaster ride, your speed in a loop depends on the height of the hill you have just come down and the radius of the loop in feet. The equation v = 8 h – 2r gives the velocity v in feet per second of a car at the top of the loop. 11-5

The loop on a roller coaster ride has a radius of 18 ft. ALGEBRA 1 LESSON 11-5 (continued) The loop on a roller coaster ride has a radius of 18 ft. Your car has a velocity of 120 ft/s at the top of the loop. How high is the hill of the loop you have just come down before going into the loop? Solve v = 8 h – 2r for h when v = 120 and r = 18. 120 = 8 h – 2(18) Substitute 120 for v and 18 for r. = Divide each side by 8 to isolate the radical. 15 = h – 36 Simplify. 8 h – 2(18) 8 120 (15)2 = ( h – 36)2 Square both sides. 225 = h – 36 261 = h The hill is 261 ft high. 11-5

( 3x – 4)2 = ( 2x + 3)2 Square both sides. ALGEBRA 1 LESSON 11-5 Solve 3x – 4 = 2x + 3. ( 3x – 4)2 = ( 2x + 3)2 Square both sides. 3x – 4 = 2x + 3 Simplify. 3x = 2x + 7 Add 4 to each side. x = 7 Subtract 2x from each side. Check: 3x – 4 = 2x + 3 3(7) – 4 2(7) + 3 Substitute 7 for x. 17 = 17 The solution is 7. 11-5

(x)2 = ( x + 12)2 Square both sides. ALGEBRA 1 LESSON 11-5 Solve x = x + 12. (x)2 = ( x + 12)2 Square both sides. x2 = x + 12 x2 – x – 12 = 0 Simplify. (x – 4)(x + 3) = 0 Solve the quadratic equation by factoring. (x – 4) = 0 or (x + 3) = 0 Use the Zero–Product Property. x = 4  or   x = –3 Solve for x. Check: x = x + 12 4 4 + 12 –3 –3 + 12 4 = 4 –3 = 3 / The solution to the original equation is 4. The value –3 is an extraneous solution. 11-5

( 3x)2 = (–6)2 Square both sides. ALGEBRA 1 LESSON 11-5 Solve 3x + 8 = 2. 3x = –6 ( 3x)2 = (–6)2 Square both sides. 3x = 36 x = 12 Check: 3x + 8 = 2 3(12) + 8 2   Substitute 12 for x. 36 + 8 2 6 + 8 = 2   x = 12 does not solve the original equation. / 3x + 8 = 2 has no solution. 11-5

Solving Radical Equations ALGEBRA 1 LESSON 11-5 Solve each radical equation. 1. 7x – 3 = 4 2. 3x – 2 = x + 2 3. 2x + 7 = 5x – 8 4. x = 2x + 8 5. 3x + 4 + 5 = 3 2 5 7 2 5 4 no solution 11-5

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