Question

Rationalise the denominator of $\frac{4}{2+\sqrt{3}+\sqrt{7}}$

Answer 1

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

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

Rationalize the denominator

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

Denominator is in the form of $(a+b)(a-b)=a^2-b^2$

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

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

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

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

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

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

Rationalize the denominator

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

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

$\Rightarrow \frac{2\sqrt{3}+3-\sqrt{21}}{3}$