# How do you simplify (square root of 6/square root of 7) * (square root of 14/square root of 3)?

Apr 2, 2018

Write the roots as their factors..

#### Explanation:

$\left(\frac{\sqrt{6}}{\sqrt{7}}\right) \cdot \left(\frac{\sqrt{14}}{\sqrt{3}}\right)$
Because we are multiplying the two fractions, we know that we can simplify by cancelling right?
Write each of the fractions, if possible, in their factors . For example : $\sqrt{6} = \sqrt{2} \cdot \sqrt{3}$

You should get this : $\left(\frac{\sqrt{2} \cdot \sqrt{3}}{\sqrt{7}}\right) \cdot \left(\frac{\sqrt{7} \cdot \sqrt{2}}{\sqrt{3}}\right)$
Cancel out any factor that is both in the numerator and the denominator.
You should be left with root 2 times root 2, which is 2 :)

Apr 2, 2018

$2$

#### Explanation:

$\frac{\sqrt{6}}{\sqrt{7}} \cdot \frac{\sqrt{14}}{\sqrt{3}}$

$\therefore = \frac{\sqrt{84}}{\sqrt{21}}$

$\therefore = \frac{\sqrt{2 \cdot 2 \cdot 3 \cdot 7}}{\sqrt{21}}$

$\therefore \sqrt{2} \cdot \sqrt{2} = 2$

$\therefore = \frac{2 \sqrt{3 \cdot 7}}{\sqrt{21}}$

$\therefore = \frac{2 {\cancel{\sqrt{21}}}^{1}}{\cancel{\sqrt{21}}} ^ 1$

$\therefore = 2$