# How do you solve then check for extraneous solutions: (x+5)^(2/3)=4?

Mar 25, 2018

$x = 3$

#### Explanation:

Raise both sides to the power of $\frac{3}{2}$

${\left({\left(x + 5\right)}^{\frac{2}{3}}\right)}^{\frac{3}{2}} = {4}^{\frac{3}{2}}$

$x + 5 = {\left(\pm \sqrt{4}\right)}^{3}$

There are two possible solutions for $\sqrt{4}$

$x + 5 = {\left(+ 2\right)}^{3} \text{ "or" } x + 5 = {\left(- 2\right)}^{3}$

$\text{ "x+5 = 8" "or" } x + 5 = - 8$

$x = 8 - 5 \text{ "or" } x = - 8 - 5$

$x = 3 \mathmr{and} x = - 13$

Check both in the original equation.

${\left(3 + 5\right)}^{\frac{2}{3}} = {8}^{\frac{2}{3}} = 4$

${\left(13 + 5\right)}^{\frac{2}{3}} = {13}^{\frac{2}{3}} = 1.768 \text{ } \leftarrow$ this does not give $4$