# How do you rationalize the denominator and simplify sqrt(12x^3)/sqrt(7y^5)?

Mar 23, 2016

$\frac{\sqrt{12 {x}^{3}}}{\sqrt{7 {y}^{5}}} = 2 \frac{\sqrt{21 {x}^{3} {y}^{5}}}{7 {y}^{5}}$

Or more simplified:

$\frac{\sqrt{12 {x}^{3}}}{\sqrt{7 {y}^{5}}} = \frac{2 x}{7 {y}^{3}} \cdot \sqrt{21 x y}$

#### Explanation:

To rationalize your expression, the idea is to multiply it by a therm that doesn't change it but cancels the $\sqrt{}$ at the denominator.

Therefore:

$\frac{\sqrt{12 {x}^{3}}}{\sqrt{7 {y}^{5}}} = \frac{\sqrt{12 {x}^{3}}}{\sqrt{7 {y}^{5}}} \cdot \frac{\sqrt{7 {y}^{5}}}{\sqrt{7 {y}^{5}}} =$

$= \frac{\sqrt{12 {x}^{3}} \cdot \sqrt{7 {y}^{5}}}{\sqrt{7 {y}^{5} \cdot 7 {y}^{5}}} =$

$= \frac{\sqrt{12 {x}^{3} \cdot 7 {y}^{5}}}{\sqrt{{\left(7 {y}^{5}\right)}^{2}}} = \frac{\sqrt{3 \cdot {2}^{2} \cdot 7 {x}^{3} {y}^{5}}}{7 {y}^{5}} = 2 \frac{\sqrt{21 {x}^{3} {y}^{5}}}{7 {y}^{5}}$

Now you can more simplify the results:

$2 \frac{\sqrt{21 {x}^{3} {y}^{5}}}{7 {y}^{5}} = 2 \frac{\sqrt{21 {x}^{2} \cdot x \cdot {y}^{4} \cdot y}}{y} ^ 5 = 2 x \cancel{{y}^{2}} \frac{\sqrt{21 x y}}{7 {y}^{{\cancel{5}}^{3}}} =$
$= \frac{2 x}{7 {y}^{3}} \cdot \sqrt{21 x y}$