# How do you simplify 3sqrt98-4sqrt28?

Jun 25, 2018

See a solution process below:

#### Explanation:

First, rewrite the expression as:

$3 \sqrt{49 \cdot 2} - 4 \sqrt{4 \cdot 7}$

Next, use this rule for radicals to rewrite each radical:

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

$3 \sqrt{\textcolor{red}{49} \cdot \textcolor{b l u e}{2}} - 4 \sqrt{\textcolor{red}{4} \cdot \textcolor{b l u e}{7}} \implies$

$3 \sqrt{\textcolor{red}{49}} \sqrt{\textcolor{b l u e}{2}} - 4 \sqrt{\textcolor{red}{4}} \sqrt{\textcolor{b l u e}{7}} \implies$

(3 * 7)sqrt(color(blue)(2)) - (4 * 2))sqrt(color(blue)(7)) =>

$21 \sqrt{2} - 8 \sqrt{7}$

Jun 25, 2018

$21 \sqrt{2} - 8 \sqrt{7}$

#### Explanation:

Let's see if we can factor a perfect square out of the radicals. The key here is that we can leverage the radical property

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

We can rewrite $98$ as $49 \cdot 2$ and $28$ as $4 \cdot 7$. We now have

$3 \sqrt{49 \cdot 2} - 4 \sqrt{4 \cdot 7}$

$\implies 3 \cdot 7 \sqrt{2} - 4 \cdot 2 \sqrt{7}$

We can simplify this to

$21 \sqrt{2} - 8 \sqrt{7}$

Hope this helps!