# How do you rationalize the denominator and simplify sqrt49/sqrt53?

Sep 30, 2015

The answer is $\frac{7 \sqrt{53}}{53}$ .

#### Explanation:

$\frac{\sqrt{49}}{\sqrt{53}}$

Rationalize the denominator.

$\frac{\sqrt{49} \sqrt{53}}{\sqrt{53} \sqrt{53}}$

Apply the rule $\sqrt{a} \sqrt{a} = a$ .

$\frac{\sqrt{49} \sqrt{53}}{53}$

Write the prime factors for $\sqrt{49}$ and $\sqrt{53}$.

(sqrt(7xx7)sqrt(53) ($53$ is a prime number)

Rewrite $7 \times 7$ as ${7}^{2}$.

$\frac{\sqrt{{7}^{2}} \sqrt{53}}{53} =$

Apply the rule $\sqrt{{a}^{2}} = \left\mid a \right\mid$ .

$\frac{7 \sqrt{53}}{53}$