# How do you rationalize the denominator and simplify 2/root4(2)?

Apr 15, 2016

$\sqrt[4]{{2}^{3}}$

#### Explanation:

We have to multiply both sides of the division by $\sqrt{{2}^{3}}$:

$\frac{2}{\sqrt[4]{2}} \times \frac{\sqrt[4]{{2}^{3}}}{\sqrt[4]{{2}^{3}}} = \frac{2 \sqrt[4]{{2}^{3}}}{\sqrt[4]{2 \times {2}^{3}}}$

$= \frac{2 \sqrt[4]{{2}^{3}}}{\sqrt[4]{{2}^{4}}} = \frac{2 \sqrt[4]{{2}^{3}}}{2} = \sqrt[4]{{2}^{3}}$

Apr 15, 2016

$\sqrt[4]{8}$

#### Explanation:

Given:$\text{ } \frac{2}{\sqrt[4]{2}}$

'~~~~~~~~~~~~~~~~~~~~~~~~~
If you have $3 \sqrt{2}$ you may take the 3 inside the squar root as long as you square. So $3 \sqrt{2} \equiv \sqrt{{3}^{2} \times 2}$. In the same way if we had $3 \sqrt[4]{2} \equiv \sqrt[4]{{3}^{4} \times 2} \text{ }$ The three dashes instead of = means equivalent to.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~

Applying this idea we have

$\sqrt[4]{{2}^{4} / 2} = \sqrt[4]{{2}^{3}} = \sqrt[4]{8}$