# How do you simplify sqrt6 - sqrt 3?

May 29, 2016

$\sqrt{6} - \sqrt{3} = \sqrt{3} \left(\sqrt{2} - 1\right)$

#### Explanation:

This is already about as simplified as it could be. However, if you wanted to, this can be factored.

Recall that (for $a , b > 0$)

$\sqrt{a b} = \sqrt{a} \cdot \sqrt{b}$

Therefore we can say that

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

We can then rewrite the original expression:

$\sqrt{6} - \sqrt{3} = \left(\sqrt{2} \cdot \sqrt{3}\right) - \sqrt{3}$

Factor $\sqrt{3}$ from both terms:

$\left(\sqrt{2} \cdot \sqrt{3}\right) - \sqrt{3} = \sqrt{3} \left(\sqrt{2} - 1\right)$

I wouldn't go so far as to say that $\sqrt{3} \left(\sqrt{2} - 1\right)$ is a simplification of $\sqrt{6} - \sqrt{3}$, but they are equivalent statements. Using $\sqrt{3} \left(\sqrt{2} - 1\right)$ may be helpful in simplifying more complicated expressions, such as:

$\frac{\sqrt{6} - \sqrt{3}}{\sqrt{10} - \sqrt{5}} = \frac{\sqrt{3} \left(\sqrt{2} - 1\right)}{\sqrt{5} \left(\sqrt{2} - 1\right)} = \frac{\sqrt{3}}{\sqrt{5}} = \frac{\sqrt{3}}{\sqrt{5}} \left(\frac{\sqrt{5}}{\sqrt{5}}\right) = \frac{\sqrt{15}}{5}$