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 $(a-b)(a+b)=a^{2}-b^{2}$ ]
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