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

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

BOP:

Solution to BOP:

ALGEBRA 1 LESSON 11-4ALGEBRA 1 LESSON 11-4

37.7.4 ft

38.5 10

41.8 + 2 15

44.–24

11-4

10 547.

48.8 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 48as 2 24 instead of 2 12 or 4 3.

b. 2 6 + 4 354.a. 2 2 or 2.8 ft

b. s 2

63. a + b ≠≠ a + b

64.

67. 2

69. 70

71. a. 2 6b. 2 13c. 2(p + q)

55. 9.1%

56. 12.8%

57. 15.5%

58. a. xb. 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.


11-4

n

2n – 1

2

ab b

9 22

Prentice Hall Lesson 11.5EQ: 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.

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 = 16 Solve for x.

x = 21

( x – 5)2 = 42 Square each side.

(continued)

Check: x – 5 = 4 21– 5 = 4 Substitute 21 for x. 16 = 4

4 = 4

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. 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 8

(15)2 = ( h – 36)2 Square both sides.225 = h – 36261 = h

The hill is 261 ft high.

(continued)

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.

The solution is 7.

Check: 3x – 4 = 2x + 3

3(7) – 4 2(7) + 3 Substitute 7 for x.

17 = 17

11-5


(x)2 = ( x + 12)2 Square both sides.

x2 = x + 12

x2 – x – 12 = 0 Simplify.

The solution to the original equation is 4. The value –3 is an extraneous solution.

Solve x = x + 12.

(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 /

11-5


Solve 3x + 8 = 2.

3x = –6

( 3x)2 = (–6)2 Square both sides.

3x = 36

x = 12

3x + 8 = 2 has no solution.

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./

11-5

Solving Radical EquationsALGEBRA 1 LESSON 11-5ALGEBRA 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 4

no solution

2 57

11-5

Summary:

Don’t forget to write your minimum three sentence summary answering today’s essential question on your summary paper!

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

Documents

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