# How do you solve -x = sqrt(2x+15) and find any extraneous solutions?

May 7, 2017

Solve for x then substitute the values in to verify that they are not extraneous solutions; x=5 is extraneous.

#### Explanation:

First, we solve for x:
${\left(- x\right)}^{2} = 2 x + 15$
${x}^{2} - 2 x - 15 = 0$
$\left(x - 5\right) \left(x + 3\right) = 0$
$x = - 3 , 5$

If $x = - 3$
$- \left(- 3\right) = \sqrt{2 \left(- 3\right) + 15}$
$3 = \sqrt{- 6 + 15} = \sqrt{9} = 3$
therefore $x = - 3$ is not extraneous

If $x = 5$
$- 5 = \sqrt{2 \left(5\right) + 15}$
Since the square root of a number cannot be negative, $x = 5$ is an extraneous solution.