Question

Rationalise the denominator of the following:
$\frac{2+\sqrt{3}}{2-\sqrt{3}}$

Answer 1

$\frac{2+\sqrt{3}}{2-\sqrt{3}}$

Dividing numerator and denominator by $2+\sqrt{3}$

$=\frac{(2+\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}$

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

$=\frac{4+4\sqrt{3}+3}{4-3}$

$=\frac{7+4\sqrt{3}}{1}$

$=7+4\sqrt{3}$