# How do you solve sqrt(x-4) – sqrt(x+8)=2?

Feb 8, 2016

Square both sides of the equation, and isolate one radical to one side of the equation (see solution below).

#### Explanation:

$\sqrt{x - 4} = 2 + \sqrt{x + 8}$

${\left(\sqrt{x - 4}\right)}^{2} = {\left(2 + \sqrt{x + 8}\right)}^{2}$

$x - 4 = 4 + 4 \sqrt{x + 8} + x + 8$

$- 4 - 4 - 8 = 4 \sqrt{x + 8}$

$- 16 = 4 \sqrt{x + 8}$

$- 4 = \sqrt{x + 8}$

Square both sides again to get rid of the remaining square root.

${\left(- 4\right)}^{2} = {\left(\sqrt{x + 8}\right)}^{2}$

16 = x + 8

8 = x

Check in the original equation to make sure the solution is not an extraneous root.

$\sqrt{8 - 4} - \sqrt{8 + 8} \ne 2$

So, this equation has no solution ($\emptyset$)

Practice exercises:

1. Solve for x. Watch out for extraneous solutions.

a) $\sqrt{2 x + 6} - \sqrt{x + 1} = 2$

b) $\sqrt{2 x - 1} + \sqrt{9 x + 4} = 10$

Good luck!