# How do you solve sqrt(3x + 4) + x = 8 and find any extraneous solutions?

Apr 20, 2017

#### Explanation:

Given: $\sqrt{3 x + 4} + x = 8$

Subtract x from both sides:

$\sqrt{3 x + 4} = 8 - x$

Square both sides:

$3 x + 4 = 64 - 16 x + {x}^{2}$

Subtract 3x + 4 from both sides:

$0 = {x}^{2} - 19 x + 60$

This factors:

$\left(x - 4\right) \left(x - 15\right) = 0$

$x = 4 \mathmr{and} x = 15$

Check:

$\sqrt{3 \left(4\right) + 4} + 4 = 8$
$\sqrt{3 \left(15\right) + 4} + 15 = 8$

$8 = 8$
$22 = 8$

$x = 4$ is the solution.