# How do you simplify sqrt(250/252)?

Mar 14, 2017

$\frac{5 \cdot \sqrt{5}}{3 \cdot \sqrt{14}}$

#### Explanation:

First, simplify the fraction in the square root:
$\sqrt{\frac{125}{126}}$
Then, use the rule$\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$:
$\frac{\sqrt{125}}{\sqrt{126}}$
Now, onto simplifying the radicals on the top and bottom: split the square roots using the rule $\sqrt{a b} = \sqrt{a} \cdot \sqrt{b}$:
$\frac{\sqrt{25} \cdot \sqrt{5}}{\sqrt{9} \cdot \sqrt{14}}$ (see notes below if this doesn't make sense)
Simplify $\sqrt{25}$ and $\sqrt{9}$:
$\frac{5 \cdot \sqrt{5}}{3 \cdot \sqrt{14}}$
And that's as far as you can go!

Note on simplification of $\sqrt{125}$ & $\sqrt{126}$:
The way I used to simplify these is to find all the prime factors:
$\sqrt{125} = \sqrt{5} \cdot \sqrt{5} \cdot \sqrt{5}$
$\sqrt{126} = \sqrt{2} \cdot \sqrt{3} \cdot \sqrt{3} \cdot \sqrt{7}$
Since $\sqrt{5} \cdot \sqrt{5} = {\sqrt{5}}^{2} = 5$ and $\sqrt{3} \cdot \sqrt{3} = {\sqrt{3}}^{2} = 3$, you can simplify these to:
$\sqrt{125} = 5 \cdot \sqrt{5}$
$\sqrt{126} = \sqrt{2} \cdot 3 \cdot \sqrt{7}$
Using the rule $\sqrt{a b} = \sqrt{a} \cdot \sqrt{b}$ on the second expression:
$\sqrt{125} = 5 \cdot \sqrt{5}$
$\sqrt{126} = 3 \cdot \sqrt{14}$
And you end up in the same place as the first method