# How do you solve and check for extraneous solutions in 2(x + 8) ^ (4/5) - 12 = 150?

Aug 26, 2015

$x = 235 \text{ }$ or $\text{ } x = - 251$

#### Explanation:

Start by rewriting your equation using the radical form for the term that has the fractional exponent

$2 \sqrt[5]{{\left(x + 8\right)}^{4}} - 12 = 150$

Next, isolate the radical term on the left-hand side of the equation by adding $12$ to both sides and dividing all the terms by $2$

$2 \sqrt[5]{{\left(x + 8\right)}^{4}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{12}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{12}}} = 150 + 12$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \sqrt[5]{{\left(x + 8\right)}^{4}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} = \frac{162}{2}$

$\sqrt[5]{{\left(x + 8\right)}^{4}} = 81$

This can be rewritten as

${\left(\sqrt[5]{x + 8}\right)}^{4} = 81$

Take the fourth root from both sides of the equation - do not forget that you have positive and negative roots for the fourth root of $81$!

$\sqrt[4]{{\left(\sqrt[5]{x + 8}\right)}^{4}} = \sqrt[4]{81}$

$\sqrt[5]{x + 8} = \pm 3$

Raise both sides of the equation to the fifth power

${\left(\sqrt[5]{x + 8}\right)}^{5} = {\left(\pm 3\right)}^{5}$

This equation will now produce two solutions

$x + 8 = {3}^{5}$

$x = 243 - 8 = \textcolor{g r e e n}{235}$

and

$x + 8 = {\left(- 3\right)}^{5}$

$x = - 243 - 8 = \textcolor{g r e e n}{- 251}$

Your original equation will thus have two valid solutions and no extraneous solutions.