# How do you solve sqrt(x+4 )= 1- sqrt(3x+13)?

Mar 11, 2018

$x = - 4$

#### Explanation:

Given:

$\sqrt{x + 4} = 1 - \sqrt{3 x + 13}$

We can square both sides to get:

$x + 4 = 1 - 2 \sqrt{3 x + 13} + \left(3 x + 13\right) = 3 x + 14 - 2 \sqrt{3 x + 13}$

Add $2 \sqrt{3 x + 13} - x - 4$ to both ends to get:

$2 \sqrt{3 x + 13} = 2 x + 10$

Divide both sides by $2$ to get:

$\sqrt{3 x + 13} = x + 5$

Square both sides to get:

$3 x + 13 = {x}^{2} + 10 x + 25$

Subtract $3 x + 13$ from both sides to get:

$0 = {x}^{2} + 7 x + 12 = \left(x + 3\right) \left(x + 4\right)$

So:

$x = - 3 \text{ }$ or $x = - 4$

We now need to check for extraneous roots, since these may have been introduced when we squared the equation:

$\sqrt{\left(\textcolor{b l u e}{- 3}\right) + 4} = 1 \ne - 1 = 1 - 2 = 1 - \sqrt{4} = 1 - \sqrt{3 \left(\textcolor{b l u e}{- 3}\right) + 4}$

$\sqrt{\left(\textcolor{b l u e}{- 4}\right) + 4} = 0 = 1 - 1 = 1 - \sqrt{1} = 1 - \sqrt{3 \left(\textcolor{b l u e}{- 4}\right) + 13}$

So $x = - 3$ is an extraneous solution caused by squaring $1 \ne - 1$, while $x = - 4$ is a solution of the original equation.