How do you solve #sqrt(x+4)-2=sqrt(x-12)#?

1 Answer
Sep 9, 2016

#x=21#

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 #x# from both sides and simplify to get;

#8-4sqrt(x+4) = -12#

Subtract #8# from both sides to get:

#-4sqrt(x+4) = -20#

DIvide both sides by #-4# to get:

#sqrt(x+4) = 5#

Square both sides to get:

#x+4 = 25#

Subtract #4# from both sides to get:

#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 #x=21# is a solution of the original equation.