How do you solve #sqrt(x+4)-2=sqrt(x-12)#?
1 Answer
Sep 9, 2016
Explanation:
Given:
#sqrt(x+4)-2 = sqrt(x-12)#
Square both sides of the equation (noting that this may introduce extraneous solutions) to get:
#(x+4)-4sqrt(x+4)+4=(x-12)#
Subtract
#8-4sqrt(x+4) = -12#
Subtract
#-4sqrt(x+4) = -20#
DIvide both sides by
#sqrt(x+4) = 5#
Square both sides to get:
#x+4 = 25#
Subtract
#x=21#
Finally we need to check that this is a solution of the original equation, since squaring both sides of an equation squashes the distinction between positive and negative quantities.
#sqrt(color(blue)(21)+4) - 2 = sqrt(25)-2 = 5-2 = 3 = sqrt(9) = sqrt(color(blue)(21)-12)#
So