# How do you evaluate √2/7 - √7/2?

Nov 7, 2015

There are two ways to go about this. In some problems the first works best, in others the second.

#### Explanation:

We first get rid of the radicals in de fractions:
$= \frac{\sqrt{2}}{\sqrt{7}} - \frac{\sqrt{7}}{\sqrt{2}} = \frac{\sqrt{2}}{\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} - \frac{\sqrt{7}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$
$= \frac{\sqrt{2 \cdot 7}}{\sqrt{{7}^{2}}} - \frac{\sqrt{7 \cdot 2}}{\sqrt{{2}^{2}}} = \frac{\sqrt{14}}{7} - \frac{\sqrt{14}}{2}$

Now we make the denominators equal:
$\frac{2 \sqrt{14}}{14} - \frac{7 \sqrt{14}}{14} = \frac{- 5 \sqrt{14}}{14} = - \frac{5}{14} \sqrt{14}$
You may want to write this as $- \frac{5}{\sqrt{14}}$

Another solution:
First we make the denominators equal:
$= \frac{\sqrt{2}}{\sqrt{7}} - \frac{\sqrt{7}}{\sqrt{2}} = \frac{\sqrt{2}}{\sqrt{7}} \cdot \frac{\sqrt{2}}{\sqrt{2}} - \frac{\sqrt{7}}{\sqrt{2}} \cdot \frac{\sqrt{7}}{\sqrt{7}}$
$= \frac{\sqrt{2 \cdot 2}}{\sqrt{7 \cdot 2}} - \frac{\sqrt{7 \cdot 7}}{\sqrt{7 \cdot 2}} = \frac{2}{\sqrt{14}} - \frac{7}{\sqrt{14}}$
$= - \frac{5}{\sqrt{14}}$