## $\frac{4}{\sqrt{2}} + \frac{2}{\sqrt{3}}$

Feb 10, 2018

$\frac{2 \left(3 \sqrt{2} + \sqrt{3}\right)}{3}$

#### Explanation:

The first thing that you need to do here is to get rid of the two radical terms from the denominators.

To do that, you must rationalize the denominator by multiplying each radical term by itself.

So what you do is you take the first fraction and multiply it by $1 = \frac{\sqrt{2}}{\sqrt{2}}$ in order to keep its value the same. This will get you

$\frac{4}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{4 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}}$

Since you know that

$\sqrt{2} \cdot \sqrt{2} = \sqrt{2 \cdot 2} = \sqrt{4} = \sqrt{{2}^{2}} = 2$

you can rewrite the fraction like this

$\frac{4 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{4 \cdot \sqrt{2}}{2} = 2 \sqrt{2}$

Now do the same for the second fraction, only this time, multiply it by $1 = \frac{\sqrt{3}}{\sqrt{3}}$. You will get

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

Since

$\sqrt{3} \cdot \sqrt{3} = \sqrt{{3}^{2}} = 3$

you will have

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

This means that the original expression is now equivalent to

$\frac{4}{\sqrt{2}} + \frac{2}{\sqrt{3}} = 2 \sqrt{2} + \frac{2 \sqrt{3}}{3}$

Next, multiply the first term by $1 = \frac{3}{3}$ to get

$2 \sqrt{2} \cdot \frac{3}{3} + \frac{2 \sqrt{3}}{3} = \frac{6 \sqrt{2}}{3} + \frac{2 \sqrt{3}}{3}$

The two fractions have the same denominator, so you can add their numerators to get

$\frac{6 \sqrt{2}}{3} + \frac{2 \sqrt{3}}{3} = \frac{6 \sqrt{2} + 2 \sqrt{3}}{3}$

Finally, you can use $2$ as a common factor here to rewrite the fraction as

$\frac{6 \sqrt{2} + 2 \sqrt{3}}{3} = \frac{2 \left(3 \sqrt{2} + \sqrt{3}\right)}{3}$

And there you have it

$\frac{4}{\sqrt{2}} + \frac{2}{\sqrt{3}} = \frac{2 \left(3 \sqrt{2} + \sqrt{3}\right)}{3}$