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Solve each equation. Remember to check for extraneous solutions

0 votes
11) (20 − r)^1/2 = r
12) (6b)^1/2 = (8 − 2b)^1/2
asked Oct 29, 2018 in ALGEBRA 1 by anonymous
reshown Oct 29, 2018 by bradely

1 Answer

0 votes
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.
 
Answer :

1)  The solution is r = 4

2)  The Solution is b = 1.
answered Oct 31, 2018 by homeworkhelp Mentor

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