rationalizing the denominators

Why do I have to rationalize the denominators?

In AMC( American mathematics contest ) contests I see that they never leave an answers as 2/root3 it will always be 2root3/3, but in SAT math tests they have answers such as 2/root3. What is the rule and meaning for rationalizing the denominators?(Wondering)

Thanks.

Vicky.

Times both sides by the denominator

$\displaystyle \frac {2}{\sqrt3}$

If I times that fraction by

$\displaystyle \frac{\sqrt3}{\sqrt3}$, I am actually just multiplying by one. But look at what happens:

$\displaystyle \frac {2}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}$

$\displaystyle =\frac{2\times\sqrt3}{\sqrt3\times\sqrt3}$

$\displaystyle =\frac{2\sqrt3}{3}$

So the rule is:

For any fraction $\displaystyle \frac{a}{\sqrt{b}}$ multiply by $\displaystyle \frac{\sqrt{b}}{\sqrt{b}}$ to give $\displaystyle \frac{a\sqrt{b}}{b}$

**For the purpose of your question, this is how you rationalize the denominator.**, but for a harder example, you may read on.

However, it gets harder if you have something such as:

$\displaystyle \frac{3}{\sqrt3+4}$

In this scenario, you have to use the 'difference of two squares' to rationalize, by making the denominator a difference of two squares, which is where $\displaystyle (a+b)(a-b)=a^2-b^2$. So apply that to the question, where $\displaystyle a=\sqrt{3}$ and b=4:

$\displaystyle \frac{3}{\sqrt3+4}\times\frac{\sqrt3-4}{\sqrt3-4}$

$\displaystyle =\frac{3(\sqrt{3}-4)}{(\sqrt3+4)(\sqrt3-4)}$

Note the 'difference of two squares' in the denominator? Using the difference of two squares rule above, we can simplify this to:

$\displaystyle =\frac{3(\sqrt{3}-4)}{{\sqrt3}^2-(4^2)}$

$\displaystyle =\frac{3(\sqrt{3}-4)}{3-16}$

$\displaystyle =\frac{3\sqrt3-12}{-13}$