# What is Square root of 24 minus square root of 54 plus square root of 96?

Oct 8, 2015

$3 \sqrt{6}$

#### Explanation:

Your starting expression looks like this

$\sqrt{24} - \sqrt{54} + \sqrt{96}$

To try and simplify this expression, write out each value you have under a square root as a product of its prime factors.

This will get you

$24 = {2}^{3} \cdot 3 = {2}^{2} \cdot 2 \cdot 3$

$54 = 2 \cdot {3}^{3} = 2 \cdot {3}^{2} \cdot 3 = {3}^{2} \cdot 2 \cdot 3$

$96 = {2}^{5} \cdot 3 = {2}^{4} \cdot 2 \cdot 3$

Notice that each number can be written as the product between a perfect square and $6$. This means that you can write

$\sqrt{24} = \sqrt{{2}^{2} \cdot 6} = \sqrt{{2}^{2}} \cdot \sqrt{6} = 2 \sqrt{6}$

$\sqrt{54} = \sqrt{{3}^{2} \cdot 6} = \sqrt{{3}^{2}} \cdot \sqrt{6} = 3 \sqrt{6}$

$\sqrt{96} = \sqrt{{2}^{4} \cdot 6} = \sqrt{{2}^{4}} \cdot \sqrt{6} = {2}^{2} \sqrt{6} = 4 \sqrt{6}$

The expression can thus be written as

$2 \sqrt{6} - 3 \sqrt{6} + 4 \sqrt{6}$

which is equal to

$\sqrt{6} \cdot \left(2 - 3 + 4\right) = \textcolor{g r e e n}{3 \sqrt{6}}$