Rationalising the denominator: $\frac{11}{3\sqrt{3}+7}$

Question

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!

Answer 1

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 $\rm\; (a+b\sqrt 3)\:(a-b\sqrt 3)\ =\ a^2 - 3 b^2$

Answer 2

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.

Answer 3

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.

Answer 4

HINT: The general trick is $$ \frac{1}{\sqrt{a}+b}=\frac{\sqrt{a}-b}{(\sqrt{a}+b)(\sqrt{a}-b)}=\frac{\sqrt{a}-b}{a-b^2}. $$