Question 62331
Given sqrt(3x-5) - sqrt(x+7)= 2

==>   sqrt(3x -5) = 2 + sqrt( x+7)

Square on both sides, we get

( 3x -5) = [ 2 + sqrt(x + 7)] ^2
 
3x - 5 = 4 + ( x + 7) + 4sqrt( x + 7)

3x - 5 - 4 -x - 7 = 4sqrt ( x + 7)

2x -   16 = 4 sqrt(x +7)

Divide throughout by 2, we get

x - 8 =  2sqrt (x + 7)

Square on both sides, we get

( x -8) ^ 2 = 4 ( x +  7)

x ^2 - 16x + 64 =  4x + 28

x ^2 - 16x -4x + 64 - 28 = 0

x^2 - 20x  +  + 36 = 0

x^2 - 18x - 2x + 36 = 0

x (x - 18) - 2 (x - 18) = 0

( x - 2) ( x- 18) = 0 

==> x - 2 = 0  or x - 18 = 0

==> x = 2 or x = 18