# How do you simplify sqrt3(sqrt12-sqrt6)?

Nov 9, 2015

First, expand the product: multiply $\sqrt{3}$with $\sqrt{12}$ and with $\sqrt{6}$, respectively:

$\sqrt{3} \left(\sqrt{12} - \sqrt{6}\right) = \sqrt{3} \cdot \sqrt{12} - \sqrt{3} \cdot \sqrt{6}$

Now, use the rule $\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$:

$\sqrt{3} \cdot \sqrt{12} - \sqrt{3} \cdot \sqrt{6}$
$= \sqrt{3 \cdot 12} - \sqrt{3 \cdot 6}$
$= \sqrt{36} - \sqrt{18}$
$= 6 - \sqrt{18}$

The last one, $\sqrt{18}$, can be simplifyed further since $18 = 2 \cdot 9$ and $9 = {3}^{2}$:

$6 - \sqrt{18}$
$= 6 - \sqrt{2 \cdot 9}$
$= 6 - \sqrt{2} \cdot \sqrt{9}$
$= 6 - \sqrt{2} \cdot 3$
$= 6 - 3 \sqrt{2}$

I hope that this helped!