Rationalise the Denominator

Question

Rationalise the denominator and simplify fully:

$$\dfrac{6}{\sqrt{7} + 2}$$

I got the answer $\dfrac{2 \sqrt{7}}{3}$, but didn't get the mark. Is that not fully simplified?

I did $\displaystyle \frac{6}{\sqrt{7} + 2} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{6 \sqrt{7}}{(7 + 2)} = \frac{6 \sqrt{7}}{9} = \frac{2 \sqrt{7}}{ 3}$ .

Answer 1

$$\frac{6}{\sqrt{7}+2}=\frac{6(2-\sqrt{7})}{(2+\sqrt{7})(2-\sqrt{7})} = \frac{12-6\sqrt{7}}{4-7} = \frac{3(4-2\sqrt{7})}{-3} = 2\sqrt{7}-4$$

Note that in general $$\frac{a}{b+c\sqrt{d}} = \frac{a(b-c\sqrt{d})}{(b+c\sqrt{d})(b-c\sqrt{d})} = \frac{ab-ac\sqrt{d}}{b^2-c^2d}$$

Notice that the denominator is now rational, i.e. in particular for this case it has no surds in it.

Answer 2

I'll assume your expression was $\frac{6}{\sqrt{7}+2}$, since it's the only way you obtain that denominator.

$$\frac{6}{\sqrt{7}+2}\times\frac{\sqrt{7}-2}{\sqrt{7}-2}=\frac{6\sqrt{7}-12}{\sqrt{7}^2-2^2}=\frac{6\sqrt{7}-12}{3}={2\sqrt{7}-4}$$