# How do you multiply sqrt(2x^5y^2)/sqrt(14x^3y^8)?

Jan 11, 2018

$\frac{x \sqrt{7}}{7 {y}^{3}}$

#### Explanation:

First, remember that:
$\sqrt{{a}^{2} b} = a \sqrt{b}$

Using our rule, we simplify the radicals.
$\frac{\sqrt{2 {x}^{5} {y}^{2}}}{\sqrt{14 {x}^{3} {y}^{8}}}$ becomes $\frac{y {x}^{2} \sqrt{2 x}}{x {y}^{4} \sqrt{14 x}}$

We now try to rationalize the denominator.

$\frac{y {x}^{2} \sqrt{2 x}}{x {y}^{4} \sqrt{14 x}} \times \frac{\sqrt{14 x}}{\sqrt{14 x}} \implies \frac{y {x}^{2} \sqrt{2 x} \cdot \sqrt{14 x}}{x {y}^{4} \left(14 x\right)}$

Now, remember that:
$\sqrt{a} \cdot \sqrt{b} = \sqrt{a b}$

$\frac{y {x}^{2} \sqrt{2 x} \cdot \sqrt{14 x}}{x {y}^{4} \left(14 x\right)}$ becomes
$\frac{y {x}^{2} \sqrt{28 {x}^{2}}}{x {y}^{4} \left(14 x\right)}$ Using our first rule, we can simplify this further.

$\frac{y {x}^{2} \sqrt{28 {x}^{2}}}{x {y}^{4} \left(14 x\right)} \implies \frac{y {x}^{2} \left(2 x \sqrt{7}\right)}{x {y}^{4} \left(14 x\right)}$

We multiply this out to get:
$\frac{2 {x}^{3} y \sqrt{7}}{14 {x}^{2} {y}^{4}}$

Lastly, remember that:
$\frac{{a}^{n}}{{a}^{m}} = {a}^{n - m}$

Using this rule, we now have:
$\frac{x \sqrt{7}}{7 {y}^{3}}$