# How do you solve sqrt(x+7) = x-5 and find any extraneous solutions?

Jan 22, 2017

$x = 9$

#### Explanation:

$\sqrt{x + 7} = x - 5$

$\implies {\left(\sqrt{x + 7}\right)}^{2} = {\left(x - 5\right)}^{2}$

$\implies x + 7 = {x}^{2} - 10 x + 25$

$\implies {x}^{2} - 11 x + 18 = 0$

$\implies \left(x - 2\right) \left(x - 9\right) = 0$

$\implies x - 2 = 0 \mathmr{and} x - 9 = 0$

$\implies x = 2 \mathmr{and} x = 9$

Test for extraneous solutions:

If $x = 2$, we have

$\sqrt{2 + 7} = 3$
$2 - 5 = - 3$

so $x = 2$ is not a solution.

If $x - 9$, we have

$\sqrt{9 + 7} = 4$
$9 - 5 = 4$

so $x = 9$ is a solution.

Thus the only solution is $x = 9$.