# How do you simplify sqrt13+sqrt52?

Mar 18, 2018

$3 \sqrt{13}$

#### Explanation:

Given: $\sqrt{13} + \sqrt{52}$

We have:

$\sqrt{52}$

$= \sqrt{13 \cdot 4}$

Using $\sqrt{a b} = \sqrt{a} \sqrt{b} \setminus \text{for} \setminus a , b \ge 0$.

$= \sqrt{4} \cdot \sqrt{13}$

$= 2 \sqrt{13}$ (only use principal square root)

So, the expression becomes

$\iff \sqrt{13} + 2 \sqrt{13}$

Factoring,

$\sqrt{13} \left(1 + 2\right)$

$= 3 \sqrt{13}$

Mar 18, 2018

$3 \sqrt{13}$

#### Explanation:

With any surd, we can use the law:

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

Simplifying $\sqrt{13}$

$\to$ Since it is a prime number it cannot be simplified any further

Simplifying $\sqrt{52}$

We can figure out that we can get $52$ from $13 \times 4$

$\therefore$ $\to \sqrt{13} \times \sqrt{4}$

As said earlier, $\sqrt{13}$ cannot be simplified, but $\sqrt{4}$ can, using our squared numbers of $1 , 4 , 9 , 12 , 25. . .$, $\sqrt{4} = 2$

Therefore this turns to:

$2 \sqrt{13}$ as we always put the value in front of the square root#

$\sqrt{13} + 2 \sqrt{13}$
Since $\sqrt{13}$ means $1 \sqrt{13}$ we can just add both $1$ and $2$ to get us $3 \sqrt{13}$ as the root of $\sqrt{13}$ is the same in both expressions.
$\therefore$ $\sqrt{13} + 2 \sqrt{13} \to 3 \sqrt{13}$