# How do you rationalize the denominator and simplify sqrt15/sqrt35?

Apr 7, 2016

Very slightly different approach

$\frac{\pm \sqrt{21}}{7}$

#### Explanation:

Given:$\text{ } \frac{\sqrt{15}}{\sqrt{35}}$

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Consider the example: $\sqrt{16} = 4$
(actually it is $\pm 4$)

Write as $\sqrt{4} \times \sqrt{4} = 2 \times 2 = 4$
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Using the principle demonstrated in the example:

Write as:$\text{ } \frac{\sqrt{3} \times \sqrt{5}}{\sqrt{5} \times \sqrt{7}}$

$\implies \frac{\sqrt{3}}{\sqrt{7}} \times \frac{\sqrt{5}}{\sqrt{5}} \text{ " =" } \frac{\sqrt{3}}{\sqrt{7}}$

Multiply by 1 but in the form of $1 = \frac{\sqrt{7}}{\sqrt{7}}$

$\implies \frac{\sqrt{3}}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}}$

Both 3 and 7 are prime so we have

$\frac{\pm \sqrt{3 \times 7}}{7} = \frac{\pm \sqrt{21}}{7}$