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

1 Answer
George C.
Mar 15, 2018

#x=5#

Explanation:

Given:

#sqrt(x+4)-sqrt(x-4) = 2#

Squaring both sides, we get:

#(x+4)-2sqrt(x+4) sqrt(x-4)+(x-4) = 4#

That is:

#2x-2sqrt(x^2-16) = 4#

Divide both sides by #2# to get:

#x-sqrt(x^2-16) = 2#

Add #sqrt(x^2-16)-2# to both sides to get:

#x-2 = sqrt(x^2-16)#

Square both sides to get:

#x^2-4x+4 = x^2-16#

Add #-x^2+4x+16# to both sides to get:

#20 = 4x#

Transpose and divide both sides by #4# to get:

#x = 5#

Since we have squared both sides of the equation - which not a reversible operation - we need to check that this solution we have reached is a solution of the original equation.

We find:

#sqrt((color(blue)(5))+4)-sqrt((color(blue)(5))-4) = sqrt(9)-sqrt(1) = 3-1 = 2#

So #x=5# is a valid solution.

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