# How do you solve (x+9)^"2/3"=4?

Apr 3, 2018

See below.

#### Explanation:

To solve this problem, we need to do something with that $\frac{2}{3}$ power first.

Since the $\frac{2}{3}$ encompasses an expression using addition, the power does not distribute. So we have to do a little manipulation:

${\left(x + 9\right)}^{\frac{2}{3}} = 4 \implies \sqrt[3]{{\left(x + 9\right)}^{2}} = 4$

Now that we have that, we can start to simplify this a little.

Since the cube root encompasses the entirety of the left side, we can cube the whole thing to get rid of it.

color(blue)(Note:"What you do to one side, you must do to the other".

We should now have this:

$\sqrt[3]{{\left(x + 9\right)}^{2}} = 4$

$\implies {\left(\sqrt[3]{{\left(x + 9\right)}^{2}}\right)}^{3} = {4}^{3}$

$\implies {\left(x + 9\right)}^{2} = 64$

From here we need to put the ${\left(x + 9\right)}^{2}$ into standard form, and solve from there:

${\left(x + 9\right)}^{2} = 64$

$\implies {x}^{2} + 18 x + 81 = 64$

$\implies {x}^{2} + 18 x + 15 = 0$

To solve this, we can use the quadratic formula, which is defined as:

$- \textcolor{red}{b} \pm \frac{\sqrt{{\textcolor{red}{b}}^{2} - 4 \textcolor{b l u e}{a} \textcolor{g r e e n}{c}}}{2 \textcolor{b l u e}{a}}$

$\textcolor{b l u e}{a = 1}$ $\textcolor{red}{b = 18}$ $\textcolor{g r e e n}{c = 15}$

=(-color(red)(18)+-sqrt(color(red)(18)^2-4(color(blue)(1))(color(green)(15))))/(2(color(blue)(1))

$= - \frac{18 \pm \sqrt{324 - 60}}{2}$

$= \frac{- 18 \pm 2 \sqrt{66}}{2}$

$= - 9 \pm \sqrt{66}$

Hope this helped!