# What is the square root of 3 + the square root of 72 - the square root of 128 +the square root of 108?

Oct 1, 2015

$7 \sqrt{3} - 2 \sqrt{2}$

#### Explanation:

$\sqrt{3} + \sqrt{72} - \sqrt{128} + \sqrt{108}$

We know that $108 = 9 \cdot 12 = {3}^{3} \cdot {2}^{2}$, so $\sqrt{108} = \sqrt{{3}^{3} \cdot {2}^{2}} = 6 \sqrt{3}$

$\sqrt{3} + \sqrt{72} - \sqrt{128} + 6 \sqrt{3}$

We know that $72 = 9 \cdot 8 = {3}^{2} \cdot {2}^{3}$, so $\sqrt{72} = \sqrt{{3}^{2} \cdot {2}^{3}} = 6 \sqrt{2}$

$\sqrt{3} + 6 \sqrt{2} - \sqrt{128} + 6 \sqrt{3}$

We know that $128 = {2}^{7}$, so $\sqrt{128} = \sqrt{{2}^{6} \cdot 2} = 8 \sqrt{2}$

$\sqrt{3} + 6 \sqrt{2} - 8 \sqrt{2} + 6 \sqrt{3}$

Simplifying

$7 \sqrt{3} - 2 \sqrt{2}$