# How do you solve sqrt(x-7) + 5 = 11 and find any extraneous solutions?

##### 1 Answer
Sep 25, 2016

$x = 43$

#### Explanation:

In the equation $\sqrt{x - 7} + 5 = 11$, we cannot have $x < 7$.

Now $\sqrt{x - 7} + 5 = 11$

$\Leftrightarrow \sqrt{x - 7} = 11 - 5 = 6$

or squaring both sides $x - 7 = {6}^{2} = 36$

and $x = 36 + 7 = 43$

(as it is not less than $7$, it is not extraneous).