<s>[INST] 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!
 [/INST] 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,
$$ \frac{33\sqrt{3} - 77}{9\cdot3 - 21\sqrt{3} + 21\sqrt{3} - 49} = \frac{33\sqrt{3} - 77}{27 - 49} = \frac{33\sqrt{3} - 77}{-22} = -\frac{33\sqrt{3} - 77}{22}$$
</s>