Rationalising the denominator: $\frac{11}{3\sqrt{3}+7}$ For my homework, I have been asked to rationalise and simplify this surd;
$$\frac{11}{3\sqrt{3}+7}$$
Each time I do this I get the wrong answer. The method I am using is;
$$ \frac{11}{3\sqrt3+7} \times \frac{3\sqrt3-7}{3\sqrt3-7} $$
I ended up with $$\frac{33+11\sqrt3-77}{9+3+21+7\sqrt3-21-7\sqrt3}$$
This ends up no where near the right answer, even once it is simplified. Can someone tell me where I'm going wrong? 
Many thanks!
 A: You're mistakenly multiplying $\rm\; a * b\sqrt{3} \ =\ ab + a\sqrt{3}\:\,\;$ but $\rm\; ab\:\sqrt{3}\;$ is correct.  
In other words $\rm\; b\:\sqrt{3}\;$ means $\rm b * \sqrt{3}\:,\;$ not $\rm\; b + \sqrt{3}\:.$
Also, to rationalize the denominator use $\rm\; (a+b\sqrt 3)\:(a-b\sqrt 3)\ =\ a^2 - 3 b^2$
A: Do you mean rationalise
$$
\frac{11}{3\sqrt{3}-7} \qquad \text{?}
$$
And are you sure you're trying
$$
\frac{11}{3\sqrt{3}-7} \cdot \frac{3\sqrt{3} + 7}{3\sqrt{3} + 7}\qquad \text{?}
$$
In general, Wikipedia is almost always of great help too.
A: HINT: The general trick is
$$
\frac{1}{\sqrt{a}+b}=\frac{\sqrt{a}-b}{(\sqrt{a}+b)(\sqrt{a}-b)}=\frac{\sqrt{a}-b}{a-b^2}.
$$
A: After multiplying the numerator and denominator by $3\sqrt3-7$
the new denominator is 
$$(3\sqrt 3+7)(3\sqrt3-7)=(3\sqrt3)^2-7^2=27-49=-22$$
a nice integer to divide by.
A: Apparently you made a minor operational mistake when interpreting $ 3\sqrt{3} + 7 $ as its conjugate, $ 3\sqrt{3} - 7 $. The proper way to rationalize the denominator containing a radical is to remove that radical by multiplying the numerator and denominator by the original radical's denominator's conjugate, $ 3\sqrt{3} - 7 $:
$$ \frac{11}{3\sqrt{3} + 7} \times \frac{3\sqrt{3} - 7}{3\sqrt{3} - 7} $$
Multiplying the above expression yields:
$$ \frac{33\sqrt{3} - 77}{9\cdot3 - 21\sqrt{3} + 21\sqrt{3} - 49} $$
Canceling the $ 21\sqrt{3} $ in the denominator and simplifying,
$$ \frac{33\sqrt{3} - 77}{9\cdot3 - 21\sqrt{3} + 21\sqrt{3} - 49} = \frac{33\sqrt{3} - 77}{27 - 49} = \frac{33\sqrt{3} - 77}{-22} = -\frac{33\sqrt{3} - 77}{22}$$
A: the answer is
$$-\frac{33\sqrt3-77}{22}$$
