# How do you rationalize the denominator and simplify sqrt(9xy)/(sqrt(3x^2y)?

Mar 31, 2016

$\pm \frac{\sqrt{3 x}}{x}$

#### Explanation:

Disregarding the possibilities of $\pm$

Consider the example
$\frac{\sqrt{16}}{\sqrt{4}} = \frac{4}{2} = 2$

Now view it as

$\sqrt{\frac{16}{4}} = \sqrt{4} = 2$
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Given: $\frac{\sqrt{9 x y}}{\sqrt{3 {x}^{2} y}}$

Write as

$\sqrt{\frac{9 x y}{3 {x}^{2} y}} \text{ "=" } \sqrt{\frac{3}{x}}$

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This is format is frowned upon so we need to 'get rid' of the root in the denominator.

Multiply by 1 but in the form of $1 = \frac{\sqrt{x}}{\sqrt{x}}$

$\frac{\sqrt{3}}{\sqrt{x}} \times \frac{\sqrt{x}}{\sqrt{x}} = \frac{\sqrt{3 x}}{x}$

Now think about $\left(- 2\right) \times \left(- 2\right) = + 4 = \left(+ 2\right) \times \left(+ 2\right)$

So our answer needs to be $\pm \frac{\sqrt{3 x}}{x}$