The number obtained on rationalising
Question:

The number obtained on rationalising the denominator of $\frac{1}{\sqrt{7}-2}$ is

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

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

(c) $\frac{\sqrt{7}+2}{5}$

(d) $\frac{\sqrt{7}+2}{45}$

Solution:

(a) $\frac{1}{\sqrt{7}-2}=\frac{1}{\sqrt{7}-2} \cdot \frac{\sqrt{7}+2}{\sqrt{7}+2} \quad$ [multiplying numerator and denominator by $\left.\sqrt{7}+2\right]$

$=\frac{\sqrt{7}+2}{(\sqrt{7})^{2}-(2)^{2}}=\frac{\sqrt{7}+2}{7-4}=\frac{\sqrt{7}+2}{3} \quad$ [using identity $(a-b)(a+b)=a^{2}-b^{2}$ ]