# How do you evaluate 2root5{ 2} + 2root5{ 64}?

Aug 4, 2017

See a solution process below:

#### Explanation:

First, we can rewrite the expression as:

$2 \sqrt[5]{2} + 2 \sqrt[5]{32 \cdot 2} \implies$

$2 \sqrt[5]{2} + 2 \sqrt[5]{{2}^{5} \cdot 2}$

Now we can use this rule of radicals to simplify the radical on the right side of the expression:

$\sqrt[n]{\textcolor{red}{a} \cdot \textcolor{b l u e}{b}} = \sqrt[n]{\textcolor{red}{a}} \cdot \sqrt[n]{\textcolor{b l u e}{b}}$

$2 \sqrt[5]{2} + 2 \sqrt[5]{\textcolor{red}{{2}^{5}} \cdot \textcolor{b l u e}{2}} \implies$

$2 \sqrt[5]{2} + 2 \sqrt[5]{\textcolor{red}{{2}^{5}}} \sqrt[5]{\textcolor{b l u e}{2}} \implies$

$2 \sqrt[5]{2} + \left(2 \cdot 2 \sqrt[5]{\textcolor{b l u e}{2}}\right) \implies$

$2 \sqrt[5]{2} + 4 \sqrt[5]{\textcolor{b l u e}{2}}$

We can now factor out the common term of $\sqrt[5]{3}$ and combine the remaining terms:

$2 \sqrt[5]{2} + 4 \sqrt[5]{2} \implies$

$\left(2 + 4\right) \sqrt[5]{2} \implies$

$6 \sqrt[5]{2}$