# Question #fde5d

Oct 16, 2016

$\frac{\sqrt{21 {x}^{2} y}}{\sqrt{75 x {y}^{5}}} = \frac{\sqrt{7 x}}{5 {y}^{2}}$

#### Explanation:

We will use the property that $\sqrt{a} \sqrt{b} = \sqrt{a b}$ to eliminate the radical from the denominator. To do so, we will multiply the numerator and denominator by such a value as to make the denominator the square root of a perfect square. With that, we will apply $\sqrt{{a}^{2}} = a$ for $a > 0$.

First, let's figure out what that value will be. Note that ${\left({x}^{a}\right)}^{b} = {x}^{a b}$.

$75 x {y}^{5} = 3 \left({5}^{2}\right) x {y}^{5}$

$= {5}^{2} {\left({y}^{2}\right)}^{2} \cdot 3 x y$

$= {\left(5 {y}^{2}\right)}^{2} \cdot 3 x y$

Thus, to make it a perfect square, that is, to have all of the powers be even, we need to multiply by $3 x y$:

${\left(5 {y}^{2}\right)}^{2} \cdot 3 x y \cdot 3 x y = {\left(5 {y}^{2}\right)}^{2} \cdot {\left(3 x y\right)}^{2} = {\left(15 x {y}^{3}\right)}^{2}$

So, using the property we mentioned initially, we can proceed:

$\frac{\sqrt{21 {x}^{2} y}}{\sqrt{75 x {y}^{5}}} = \frac{\sqrt{21 {x}^{2} y} \cdot \sqrt{3 x y}}{\sqrt{75 x {y}^{5}} \cdot \sqrt{3 x y}}$

$= \frac{\sqrt{21 {x}^{2} y \cdot 3 x y}}{\sqrt{75 x {y}^{5} \cdot 3 x y}}$

$= \frac{\sqrt{63 {x}^{3} {y}^{2}}}{\sqrt{{\left(15 x {y}^{3}\right)}^{2}}}$

$= \frac{\sqrt{63 {x}^{3} {y}^{2}}}{15 x {y}^{3}}$

The denominator is now rationalized, but we can do some further simplification by pulling out any squares from the remaining square root.

$= \frac{\sqrt{\left({3}^{2} \cdot 7\right) \left({x}^{2} \cdot x\right) {y}^{2}}}{15 x {y}^{3}}$

$= \frac{\sqrt{{\left(3 x y\right)}^{2} \cdot 7 x}}{15 x {y}^{3}}$

$= \frac{\sqrt{{\left(3 x y\right)}^{2}} \cdot \sqrt{7 x}}{15 x {y}^{3}}$

$= \frac{3 x y \sqrt{7 x}}{15 x {y}^{3}}$

$= \frac{\sqrt{7 x}}{5 {y}^{2}}$