How do I find the extraneous solution of sqrt(x+4)=x-2?

Mar 8, 2018

$x = 0 \text{ is an extraneous solution}$

Explanation:

$\text{square both sides to 'undo' the radical}$

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

$\Rightarrow x + 4 = {x}^{2} - 4 x + 4$

$\text{rearrange into standard form}$

$\Rightarrow {x}^{2} - 5 x = 0 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\Rightarrow x \left(x - 5\right) = 0$

$\Rightarrow x = 0 \text{ or } x = 5$

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

$\text{Substitute these values into the original equation}$

$x = 0 \to \sqrt{4} = 2 \text{ and } 0 - 2 = - 2$

$2 \ne - 2 \Rightarrow x = 0 \textcolor{red}{\text{ is an extraneous solution}}$

$x = 5 \to \sqrt{9} = 3 \text{ and "5-2=3larr" True}$

$\Rightarrow x = 5 \textcolor{red}{\text{ is the solution}}$