How do you solve and check for extraneous solutions in sqrt(x+2)=x-4?

Aug 1, 2015

$\textcolor{red}{x = 7}$ is a solution.
$\textcolor{red}{x = 2}$ is an extraneous solution.

Explanation:

SOLVE:

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

Square each side.

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

Remove parentheses.

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

Move all terms to the right hand side.

$0 = {x}^{2} - 8 x + 16 - x - 2$

Combine like terms.

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

Factor:

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

$x - 7 = 0$ and $x - 2 = 0$

$x = 7$ and $x = 2$

CHECK FOR EXTRANEOUS SOLUTIONS

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

If $x = 7$,

$\sqrt{7 + 2} = 7 - 4$

$\sqrt{9} = 3$

$3 = 3$

$x = 7$ is a solution.

If $x = 2$,

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

$\sqrt{4} = - 2$

$2 = - 2$

This is impossible, so $x = 2$ is an extraneous solution.