# How do you simplify sqrt12-sqrt3?

Mar 10, 2018

See a solution process below:

#### Explanation:

First rewrite the radical on the left as:

$\sqrt{4 \cdot 3} - \sqrt{3}$

Now, use this rule of radicals to simplify the radical on the left:

$\sqrt{\textcolor{red}{a} \cdot \textcolor{b l u e}{b}} = \sqrt{\textcolor{red}{a}} \cdot \sqrt{\textcolor{b l u e}{b}}$

$\sqrt{\textcolor{red}{4} \cdot \textcolor{b l u e}{3}} - \sqrt{3} \implies$

$\sqrt{\textcolor{red}{4}} \sqrt{\textcolor{b l u e}{3}} - \sqrt{3} \implies$

$2 \sqrt{\textcolor{b l u e}{3}} - \sqrt{3}$

We can now factor out the common term:

$2 \sqrt{\textcolor{b l u e}{3}} - \sqrt{\textcolor{b l u e}{3}} \implies$

$2 \sqrt{\textcolor{b l u e}{3}} - 1 \sqrt{\textcolor{b l u e}{3}} \implies$

$\left(2 - 1\right) \sqrt{\textcolor{b l u e}{3}} \implies$

$1 \sqrt{\textcolor{b l u e}{3}} \implies$

$\sqrt{\textcolor{b l u e}{3}}$

Mar 10, 2018

$\sqrt{3}$

#### Explanation:

Simplify $\sqrt{12}$

$\sqrt{4} \sqrt{3}$

And further, simplify to get

$2 \sqrt{3}$

Subtract $\sqrt{3}$ from $2 \sqrt{3}$

$\sqrt{3}$