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

Mar 15, 2018

$x = 5$

#### Explanation:

Given:

$\sqrt{x + 4} - \sqrt{x - 4} = 2$

Squaring both sides, we get:

$\left(x + 4\right) - 2 \sqrt{x + 4} \sqrt{x - 4} + \left(x - 4\right) = 4$

That is:

$2 x - 2 \sqrt{{x}^{2} - 16} = 4$

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:

${x}^{2} - 4 x + 4 = {x}^{2} - 16$

Add $- {x}^{2} + 4 x + 16$ to both sides to get:

$20 = 4 x$

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:

$\sqrt{\left(\textcolor{b l u e}{5}\right) + 4} - \sqrt{\left(\textcolor{b l u e}{5}\right) - 4} = \sqrt{9} - \sqrt{1} = 3 - 1 = 2$

So $x = 5$ is a valid solution.