Question

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

Answer 1

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

As we know that the rationalization factor for $\frac{1}{\sqrt{a}-b}$ is $(\sqrt{a}+b)$.

$\therefore \frac{16}{\sqrt{41}-5}=\frac{16}{\sqrt{41}-5}\times \frac{\sqrt{41}+5}{\sqrt{41}+5}$

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

$\Rightarrow =\frac{16(\sqrt{41}+5)}{{\left(\sqrt{41}\right)}^2-{\left(5\right)}^2}(\text{Using identity}:a^2-b^2=(a+b)(a-b))$

$\Rightarrow =\frac{16(\sqrt{41}+5)}{41-25}$

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

$\Rightarrow =\sqrt{41}+5$