# How do you simplify 4 * the square root of 7 + 8 * the square root of 63?

Apr 1, 2018

$28 \sqrt{7}$

#### Explanation:

$\text{using the "color(blue)"law of radicals}$

•color(white)(x)sqrtaxxsqrtbhArrsqrt(ab)

$\Rightarrow \sqrt{63} = \sqrt{9 \times 7} = \sqrt{9} \times \sqrt{7} = 3 \sqrt{7}$

$\Rightarrow 4 \sqrt{7} + \left(8 \times 3 \sqrt{7}\right)$

$= 4 \sqrt{7} + 24 \sqrt{7}$

$= 28 \sqrt{7}$

Apr 1, 2018

$28 \sqrt{7}$

#### Explanation:

Here,

4sqrt7+8sqrt63=4sqrt7+8sqrt(9xx7

$= 4 \sqrt{7} + 8 \cdot 3 \sqrt{7}$

$= 4 \sqrt{7} + 24 \sqrt{7}$

$= 28 \sqrt{7}$

Apr 1, 2018

$28 \sqrt{7}$

#### Explanation:

Assumption: You mean $\textcolor{w h i t e}{\text{d}} 4 \sqrt{7} + 8 \sqrt{63}$

7 is a prime number so lets leave that for the moment.

Consider $\sqrt{63}$

Useful trick follows.
note that for 63 $\to 6 + 3 = 9$ and as 9 is divisible by 3 then so is 63

$\sqrt{63} = \sqrt{3 \times 21} = \sqrt{3 \times 3 \times 7} \leftarrow \text{ The 7 is useful!}$
$\textcolor{w h i t e}{\text{dd...}} = 3 \sqrt{7}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Putting it all back together we have:

$4 \sqrt{7} + \left(8 \times 3 \sqrt{7}\right)$

$4 \sqrt{7} + 24 \sqrt{7}$

Factor out the $\sqrt{7}$

$\sqrt{7} \left(4 + 24\right)$

$28 \sqrt{7}$