Solve each equation. Remember to check for extraneous solutions

Question

11) (20 − r)^1/2 = r

12) (6b)^1/2 = (8 − 2b)^1/2

Answer 1

11)

The equationis

 

(20 − r)^1/2  =  r --------------------> (1)

 

Squaring on both sides

 

(20 − r)  =  r^2

 

r^2 - (20 − r)  =  0

 

r^2 - 20 + r  =  0

 

r^2 + r - 20  =  0

 

r^2 + 5r - 4r - 20  =  0

 

r(r + 5) - 4(r + 5)  =  0

 

(r + 5)(r - 4)  =  0

 

(r + 5)  =  0     ;       (r - 4)  =  0

 

r  =  - 5           ;       r  =  4

 

Substitute r = -5 in Eq (1)

 

(20 + 5)^1/2  =  - 5

 

(25)^1/2  =  - 5

 

5  is not equal to - 5 

 

Hence, it is an extraneous solution

 

Substitute r = 4 in Eq (1)

 

(20 - 4)^1/2  =  4

 

(16)^1/2  =  4

 

4  =  4

 

Hence, The solution is r = 4.

 

12)

The equation is

 

(6b)^1/2 = (8 − 2b)^1/2 ------------------> (2)

 

Squaring on both sides

 

6b  =  8 − 2b

 

6b + 2b  =  8

 

8b  =  8

 

b  =  8/8

 

b  =  1

 

Substitute b = 1 in Eq (2)

 

(6(1))^1/2 = (8 − 2(1))^1/2

 

6^1/2  =  (8 - 2)^1/2

 

6^1/2  =  6^1/2

 

Hence, The Solution is b = 1.

 

1)  The solution is r = 4

2)  The Solution is b = 1.