Question

When denominator is rationalised, then the number $\frac{5+2\sqrt{3}}{7+4\sqrt{3}}$ becomes $a-6\sqrt{3}$. Find the value of $a$.

Answer 1

We need to rationalise $\frac{5+2\sqrt{3}}{7+4\sqrt{3}}$

$=\frac{5+2\sqrt{3}}{7+4\sqrt{3}}\times \frac{7-4\sqrt{3}}{7-4\sqrt{3}}$

$=\frac{(5+2\sqrt{3})(7-4\sqrt{3})}{49-48}$

$=35-20\sqrt{3}+14\sqrt{3}-24$

$=11-6\sqrt{3}$

Now comparing this with $a-6\sqrt{3}$, we get

$a=11$