Question

Rationalise the denominator of the following:
$\frac{16}{\sqrt{41}-5}$

Answer 1

$\frac{16}{\sqrt{41}-5}$

Dividing numerator and denominator by $\sqrt{41}+5$

$=\frac{16(\sqrt{41}+5)}{(\sqrt{41}-5)(\sqrt{41}+5)}$

$=\frac{16\sqrt{41}+80}{41-25}$

$=\frac{16\sqrt{41}+80}{16}$

$=\sqrt{41}+5$