# How do you simplify square root of 128 + the square root of 32?

May 2, 2018

$\sqrt{128} + \sqrt{32}$

$\sqrt{\left(4 \times 4\right) \times \left(2 \times 2\right) \times 2} + \sqrt{\left(4 \times 4\right) \times 2}$

$4 \times 2 \sqrt{2} + 4 \sqrt{2}$

$8 \sqrt{2} + 4 \sqrt{2}$

$12 \sqrt{2}$

May 2, 2018

$12 \sqrt{2}$

#### Explanation:

Simplify:

$\sqrt{128} = \sqrt{16} \times \sqrt{8} \to 4 \times \sqrt{4} \times \sqrt{2} = 8 \sqrt{2}$.

$\sqrt{32} = \sqrt{16} \times \sqrt{2} = 4 \sqrt{2}$

$8 \sqrt{2} + 4 \sqrt{2} = 12 \sqrt{2}$

May 2, 2018

$\setminus \sqrt{128} + \setminus \sqrt{32} = 2 \cdot \setminus \sqrt{12}$

#### Explanation:

$\setminus \sqrt{128} + \setminus \sqrt{32}$
Write both numbers as a product of prime numbers.
(unique factorisation ).

For example:

$128 = 2 \cdot 64 = 2 \cdot 2 \cdot 32$
$32 = 2 \cdot 16 = 2 \cdot 2 \cdot 8 = 2 \cdot 2 \cdot 2 \cdot 4 = {2}^{5} = {2}^{4} \cdot 2$

$\setminus \sqrt{128} + \setminus \sqrt{32} = \setminus \sqrt{2 \cdot 2 \cdot 32} + \setminus \sqrt{32} =$
$= \setminus \sqrt{2 \cdot 2} \cdot \setminus \sqrt{32} + \setminus \sqrt{32} = 2 \cdot \setminus \sqrt{32} + \setminus \sqrt{32} =$
=3*\sqrt(32)=3* \sqrt(2^4*2)= 3*\sqrt(2^4)*\sqrt(2)= 3*2^2*\sqrt(2)=
$= 3 \cdot 4 \cdot \setminus \sqrt{2} = 12 \setminus \sqrt{2}$

May 2, 2018

$12 \sqrt{2}$

#### Explanation:

Given: $\sqrt{128} + \sqrt{32}$

You are looking for squared values that you can 'extract' from the roots.

If you are ever not sure do a quick sketch of a prime factor tree to help:

$\sqrt{128} + \sqrt{32}$

$\sqrt{{2}^{2} \times {2}^{2} \times {2}^{2} \times 2} + \sqrt{{2}^{2} \times {2}^{2} \times 2}$

$\textcolor{w h i t e}{\text{ddddd")8sqrt(2)color(white)("dddddd")+color(white)("dddd")4sqrt(2)color(white)("dddd")=color(white)("d}} 12 \sqrt{2}$