# How do you simplify sqrt48+sqrt147?

May 29, 2018

See a solution process below:

#### Explanation:

First, we can rewrite the terms under the radicals as:

$\sqrt{16 \cdot 3} + \sqrt{49 \cdot 3}$

Now, we can use this rule of exponents to do the simplification:

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

$\left(\sqrt{16} \cdot \sqrt{3}\right) + \left(\sqrt{49} \cdot \sqrt{3}\right) \implies$

$4 \sqrt{3} + 7 \sqrt{3} \implies$

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

$11 \sqrt{3}$

May 29, 2018

$11 \sqrt{3}$

#### Explanation:

$\sqrt{48} + \sqrt{147}$

$\sqrt{16} \cdot \sqrt{3} + \sqrt{49} \cdot \sqrt{3}$

$4 \sqrt{3} + 7 \sqrt{3}$

$11 \sqrt{3}$

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You must find first number which should be a square number.

Like $16$ and $49$ are square numbers.

May 29, 2018

$11 \sqrt{3}$

#### Explanation:

$\sqrt{48} + \sqrt{147}$

$\sqrt{2 \cdot 2 \cdot 2 \cdot 2 \cdot 3} + \sqrt{3 \cdot 7 \cdot 7}$

$\sqrt{{2}^{2} \cdot {2}^{2} \cdot 3} + \sqrt{3 \cdot {7}^{2}}$

$2 \cdot 2 \cdot \sqrt{3} + 7 \cdot \sqrt{3}$

$4 \sqrt{3} + 7 \sqrt{3}$

$11 \sqrt{3}$