# How do you solve sqrt(x + 6) = x and find any extraneous solutions?

Oct 9, 2017

$x = 3 , x = - 2 \text{ is extraneous}$

#### Explanation:

$\textcolor{b l u e}{\text{squaring both sides}}$

${\left(\sqrt{x + 6}\right)}^{2} = {x}^{2}$

$\Rightarrow x + 6 = {x}^{2}$

$\text{rearrange in standard form } a {x}^{2} + b x + c = 0$

$\Rightarrow {x}^{2} - x - 6 = 0$

$\text{the factors of - 6 which sum to - 1 are - 3 and + 2}$

$\Rightarrow \left(x - 3\right) \left(x + 2\right) = 0$

$\text{equate each factor to zero and solve for x}$

$x - 3 = 0 \Rightarrow x = 3$

$x + 2 = 0 \Rightarrow x = - 2$

$\textcolor{b l u e}{\text{As a check}}$

$\text{substitute each of the possible solutions into the}$
$\text{original equation to test their validity}$

$x = 3 \to \sqrt{3 + 6} = \sqrt{9} = 3 = x \Rightarrow \text{ valid solution}$

$x = - 2 \to \sqrt{- 2 + 6} = \sqrt{4} = 2 \ne x \Rightarrow \text{extraneous}$