# How do you solve sqrt(3x-8) = 2 and find any extraneous solutions?

May 29, 2016

$x = 4$

#### Explanation:

As we have $\sqrt{3 x - 8}$ in the equation, we will require that $3 x - 8 \ge 0$, that is, $x \ge \frac{8}{3}$.

$\sqrt{3 x - 8} = 2$

$\implies {\left(\sqrt{3 x - 8}\right)}^{2} = {2}^{2}$

$\implies 3 x - 8 = 4$

(Note that typically ${\left(\sqrt{a}\right)}^{2} = \sqrt{{a}^{2}} = | a |$, but we already required $3 x - 8 \ge 0$, meaning $| 3 x - 8 | = 3 x - 8$)

$\implies 3 x = 12$

$\therefore x = 4$

Checking our answer, we find that

$\sqrt{3 \left(4\right) - 8} = \sqrt{4} = 2$

as desired.