Question

Rationalise the denominator $\frac{1}{\sqrt{7}+\sqrt{6}-\sqrt{13}}$

Answer 1

Given:

$\frac{1}{\sqrt{7}+\sqrt{6}-\sqrt{13}}$

Since, the conjugate surd of $(\sqrt{7}+\sqrt{6})-\sqrt{13}$ is

$(\sqrt{7}+\sqrt{6})+\sqrt{13}$

So, lets multiply both numerator and denominator by $(\sqrt{7}+\sqrt{6})+\sqrt{13}$

$\frac{1}{\sqrt{7}+\sqrt{6}-\sqrt{13}}\times \frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{\sqrt{7}+\sqrt{6}+\sqrt{13}}$

$=\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{(\sqrt{7}+\sqrt{6}{)}^2-13}$ [$\because (a+b)(a-b)=a^2-b^2$]

$=\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{7+6+2\sqrt{7\times 6}-13}$

$=\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{2\sqrt{42}}$

$=\frac{\sqrt{7}}{2\sqrt{42}}+\frac{\sqrt{6}}{2\sqrt{42}}+\frac{\sqrt{13}}{2\sqrt{42}}$

$=\frac{\sqrt{6}}{12}+\frac{\sqrt{7}}{14}+\frac{\sqrt{13}}{2\sqrt{42}}\times \frac{\sqrt{42}}{\sqrt{42}}$

$=\frac{\sqrt{6}}{12}+\frac{\sqrt{7}}{14}+\frac{\sqrt{546}}{84}$

$=\frac{7\sqrt{6}+6\sqrt{7}+\sqrt{546}}{84}$

Here, the denominator is a rational number.

Hence, rationalized the denominator.