# How do you simplify (3root4(4))/(2root4(8))?

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

#### Explanation:

In looking at this, the thing we want to address first is the roots in both the numerator and denominator. We can work with the denominator to make it cancel the numerator:

$\frac{3 \sqrt[4]{4}}{2 \sqrt[4]{8}} = \frac{3 \sqrt[4]{4}}{2 \sqrt[4]{4 \times 2}} = \frac{3 \sqrt[4]{4}}{2 \sqrt[4]{4} \sqrt[4]{2}} = \frac{3}{2 \sqrt[4]{2}}$

From here, we need to rationalize the denominator. To help with this, I'll express $\sqrt[4]{2} = {2}^{\frac{1}{4}}$:

$\frac{3}{2 \times {2}^{\frac{1}{4}}} \left(1\right) = \frac{3}{2 \times {2}^{\frac{1}{4}}} \left(\frac{{2}^{\frac{3}{4}}}{{2}^{\frac{3}{4}}}\right) = \frac{3 \times {2}^{\frac{3}{4}}}{2 \times {2}^{\frac{1}{4} + \frac{3}{4}}} = \frac{3 \times {2}^{\frac{3}{4}}}{2 \times {2}^{\frac{4}{4}}} = \frac{3 \times {2}^{\frac{3}{4}}}{2 \times 2} = \frac{3}{4} \times {2}^{\frac{3}{4}} = \frac{3}{4} \sqrt[4]{8}$