Question

Rationalize the denominator of $\frac{4}{2+\surd 3+\surd 7}$

Answer 1

$\frac{4}{2+\sqrt{3}+\sqrt{7}}$

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

The rationalizing factor of the denominator is $\left[\right(2+\sqrt{3})-\sqrt{7}]$

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

$=\frac{4\times (2+\sqrt{3}-\sqrt{7})}{(2+\sqrt{3}{)}^2-7}$

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

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

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

The rationalizing factor of the denominator is $\sqrt{3}$.

$=\frac{(2+\sqrt{3}-\sqrt{7})}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$

$=\frac{\sqrt{3}\times (2+\sqrt{3}-\sqrt{7})}{3}$
$=\frac{2\sqrt{3}+3-\sqrt{2}1)}{3}$
Hence, the fraction with a rational denominator is$=\frac{2\sqrt{3}+3-\sqrt{2}1)}{3}$.