Question

rationalise denominator of
$\frac{4}{2 + \sqrt{3} + \sqrt{7} }$
​

Answer 1

Solution :

Now, $\displaystyle\frac{4}{2+\sqrt{2}+\sqrt{3}}$

To rationalize the denominator, we multiply both the numerator and the denominator by $\displaystyle(2+\sqrt{3}-\sqrt{7})$

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

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

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

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

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

Again, to rationalize the denominator, we multiply both the numerator and the denominator by $\displaystyle(\sqrt{3})$

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

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

$\displaystyle\to \boxed{\frac{4}{2+\sqrt{2}+\sqrt{3}}=\frac{\sqrt{3}(2+\sqrt{3}-\sqrt{7})}{3}}$

which is the required rationalisation.