# How do you solve and check for extraneous solutions in sqrt(5x) - x = 0?

Aug 1, 2015

$\textcolor{red}{x = 0}$ and $\textcolor{red}{x = 5}$ are solutions.
There are $\textcolor{red}{\text{no}}$ extraneous solutions.

#### Explanation:

SOLVE:

$\sqrt{5 x} - x = 0$

Add $x$ to each side.

$\sqrt{5 x} = x$

Square each side.

$5 x = {x}^{2}$

Subtract $5 x$from each side.

$0 = {x}^{2} - 5 x$

Factor.

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

$x = 0$ and $x - 5 = 0$

$x = 0$ and $x = 5$

CHECK FOR EXTRANEOUS SOLUTIONS

$\sqrt{5 x} - x = 0$

If $x = 0$,

$\sqrt{5 \left(0\right)} - 0 = 0$

$\sqrt{0} - 0 = 0$

$0 - 0 = 0$

$0 = 0$

$x = 0$ is a solution.

If $x = 5$

$\sqrt{5 \left(5\right)} - 5 = 0$

$\sqrt{25} - 5 = 0$

$5 - 5 = 0$

$0 = 0$

$x = 5$ is a solution.

There are no extraneous solutions.