# What is the square root 48 - 8 square root 12 - 4 square root 27?

Jul 19, 2015

You try to get 'squares out of the root'

#### Explanation:

$48 = 16 \cdot 3 = {4}^{2} \cdot 3$ so:
$\sqrt{48} = \sqrt{16 \cdot 3} = \sqrt{{4}^{2}} \cdot \sqrt{3} = 4 \sqrt{3}$

Same goes for:
$12 = 4 \cdot 3 = {2}^{2} \cdot 3$ so $\sqrt{12} = 2 \sqrt{3}$

$27 = 9 \cdot 3 = {3}^{2} \cdot 3$ so $\sqrt{27} = 3 \sqrt{3}$

The original question then changes into:

$= 4 \sqrt{3} - 8 \cdot 2 \sqrt{3} - 4 \cdot 3 \sqrt{3}$

$= \left(4 - 16 - 12\right) \cdot \sqrt{3} = - 24 \sqrt{3}$