# How do you simplify sqrt6(sqrt3+6)?

Oct 18, 2015

$3 \sqrt{2} \cdot \left(1 + 2 \sqrt{3}\right)$

#### Explanation:

Yourfirst step is to expand the paranthesis by using the distributive property of multiplication.

That is, you can distribute $\sqrt{6}$ to to both the terms that are currently in the paranthesis

$\textcolor{red}{\sqrt{6}} \cdot \left(\sqrt{3} + 6\right) = \textcolor{red}{\sqrt{6}} \cdot \sqrt{3} + \textcolor{red}{\sqrt{6}} \cdot 6$

Now, use the product property of radicals to write

$\sqrt{6} \cdot \sqrt{3} = \sqrt{6 \cdot 3} = \sqrt{18}$

The trick now is to realize that $18$ can be written as a product between a perfect square and another number

$18 = 9 \cdot 2 = {3}^{2} \cdot 2$

This means that the expression can be written as

$\sqrt{18} + 6 \sqrt{6} = \sqrt{{3}^{2} \cdot 2} + 6 \sqrt{6}$

$= \sqrt{{3}^{2}} \cdot \sqrt{2} + 6 \sqrt{6}$

$= 3 \sqrt{2} + 6 \sqrt{6}$

We're not done yet. Notice that you can write

$\sqrt{6} = \sqrt{2 \cdot 3} = \sqrt{2} \cdot \sqrt{3}$

This means that the expression becomes

$3 \sqrt{2} + 6 \cdot \sqrt{2} \cdot \sqrt{3}$

Use $3 \sqrt{2}$ as a common factor to get

$3 \sqrt{2} + 6 \cdot \sqrt{2} \cdot \sqrt{3} = \textcolor{g r e e n}{3 \sqrt{2} \cdot \left(1 + 2 \sqrt{3}\right)}$