# How do you simplify 2√12 - 7√3?

Oct 1, 2015

$- 3 \sqrt{3}$

#### Explanation:

First of all, recall the following rule for non-negative real numbers $a$ and $b$:
(1) $\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$
For those who need a proof of this, just raise the right side to the power of $2$ getting the expression under a square root on the left:
${\left(\sqrt{a} \cdot \sqrt{b}\right)}^{2} = \sqrt{a} \cdot \sqrt{b} \cdot \sqrt{a} \cdot \sqrt{b} =$
$= \sqrt{a} \cdot \sqrt{a} \cdot \sqrt{b} \cdot \sqrt{b} = a \cdot b$,
which proves the property (1) above.

Using this property (1),
$2 \sqrt{12} - 7 \sqrt{3} = 2 \sqrt{4 \cdot 3} - 7 \sqrt{3} =$
$= 2 \sqrt{4} \cdot \sqrt{3} - 7 \sqrt{3} = 4 \sqrt{3} - 7 \sqrt{3} = - 3 \sqrt{3}$