# How do you solve sqrt(x+2)=x-4 and find any extraneous solutions?

Sep 2, 2016

$\left\{7\right\}$ is the solution set.

#### Explanation:

Square both sides.

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

$x + 2 = {x}^{2} - 8 x + 16$

$0 = {x}^{2} - 9 x + 14$

$0 = \left(x - 7\right) \left(x - 2\right)$

$x = 7 \mathmr{and} 2$

Checking in the original equation:

1. $x = 7$

sqrt(7 + 2) =^? 7 - 4

$\sqrt{9} = 3$

1. $x = 2$

sqrt(2 + 2) =^? 2 - 4

$2 \ne - 2$

Hence, the only actual solution is $x = 7$.

Hopefully this helps!