# How do you simplify (sqrt3+3sqrt5)/(2sqrt8)?

Jul 24, 2017

See a solution process below:

#### Explanation:

First, we need to rationalize the denominator by multiplying the expression by the appropriate form of $1$ to eliminate the radical in the denominator:

$\frac{\sqrt{8}}{\sqrt{8}} \cdot \frac{\sqrt{3} + 3 \sqrt{5}}{2 \sqrt{8}} \implies \frac{\sqrt{8} \left(\sqrt{3} + 3 \sqrt{5}\right)}{2 \sqrt{8} \sqrt{8}} \implies$

$\frac{\sqrt{8} \sqrt{3} + 3 \sqrt{8} \sqrt{5}}{2 \cdot 8} \implies \frac{\sqrt{8} \sqrt{3} + 3 \sqrt{8} \sqrt{5}}{16}$

Next, we can use this rule for radicals to simplify the numerator:

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

$\frac{\sqrt{8} \sqrt{3} + 3 \sqrt{8} \sqrt{5}}{16} \implies \frac{\sqrt{8 \cdot 3} + 3 \sqrt{8 \cdot 5}}{16} \implies$

$\frac{\sqrt{24} + 3 \sqrt{40}}{16}$

We can rewrite this expression as:

$\frac{\sqrt{4 \cdot 6} + 3 \sqrt{4 \cdot 10}}{16}$

And use this rule for radicals (the opposite of the rule above) to further simplify the expression:

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

$\frac{\sqrt{4 \cdot 6} + 3 \sqrt{4 \cdot 10}}{16} \implies \frac{\sqrt{4} \sqrt{6} + 3 \sqrt{4} \sqrt{10}}{16} \implies$

$\frac{2 \sqrt{6} + \left(3 \cdot 2 \sqrt{10}\right)}{16} \implies \frac{2 \left(\sqrt{6} + 3 \sqrt{10}\right)}{16} \implies$

$\frac{\sqrt{6} + 3 \sqrt{10}}{8}$