# How do you solve sqrt(6x+5) = sqrt53 and find any extraneous solutions?

Sep 25, 2016

$x = 8$

#### Explanation:

First, we need to square both sides of the equation so that we can easily work with it.

${\sqrt{6 x + 5}}^{2} = {\sqrt{53}}^{2}$

Doing this will turn the equation into something much more simple.

$6 x + 5 = 53$

Next, we subtract $5$ from both sides.

$6 x = 48$

Then, we divide both sides of the equation by $6$.

$x = 8$

In order to determine whether or not $x = 8$ is extraneous, we need to plug it into the original equation.

$\sqrt{6 x + 5} = \sqrt{53}$
$\sqrt{6 \left(8\right) + 5} = \sqrt{53}$
$\sqrt{48 + 5} = \sqrt{53}$
$\sqrt{53} = \sqrt{53}$

Since the statement that resulted in plugging in $x = 8$ is true, we can confirm that $x = 8$ is not an extraneous solution.