# How do you simplify sqrt14/(3sqrt35)?

Sep 6, 2016

$\frac{\sqrt{10}}{15}$

#### Explanation:

Write as $\text{ } \frac{\sqrt{2 \times 7}}{3 \sqrt{5 \times 7}}$

$\frac{\sqrt{2} \times \cancel{\sqrt{7}}}{3 \times \sqrt{5} \times \cancel{\sqrt{7}}}$

$\frac{\sqrt{2}}{3 \sqrt{5}}$

If at all possible mathematicians like to 'get rid' of any roots in the denominator (bottom number).

Multiply by 1 but in the form $1 = \frac{\sqrt{5}}{\sqrt{5}}$

$\frac{\sqrt{2}}{3 \sqrt{5}} \times 1 \text{ " ->" } \frac{\sqrt{2}}{3 \sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}}$

but $\text{ } \sqrt{5} \times \sqrt{5} = 5$

$\frac{\sqrt{2} \sqrt{5}}{3 \times 5}$

$\frac{\sqrt{10}}{15}$

Sep 6, 2016

$\frac{\sqrt{10}}{15}$

#### Explanation:

We have: $\frac{\sqrt{14}}{3 \sqrt{35}}$

Let's express the radicals as products:

$= \frac{\sqrt{2 \cdot 7}}{3 \cdot \sqrt{5 \cdot 7}}$

$= \frac{\sqrt{2} \cdot \sqrt{7}}{3 \cdot \sqrt{5} \cdot \sqrt{7}}$

We can cancel the $\sqrt{7}$ terms:

$= \frac{\sqrt{2}}{3 \sqrt{5}}$

Finally, let's rationalise the denominator:

$= \frac{\sqrt{2}}{3 \sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}$

$= \frac{\sqrt{10}}{3 \cdot 5}$

$= \frac{\sqrt{10}}{15}$