Question

$\frac{1}{\left(\sqrt{4}-\sqrt{3}\right)}=?$
(a) $\left(2+\sqrt{3}\right)$
(b) $\left(2-\sqrt{3}\right)$
(c) 1
(d) none of these

Answer 1

(a) (2 + $\sqrt{3}$)
On rationalising the denominator, we get:
$$\begin{gathered} \frac{1}{\sqrt{4}-\sqrt{3}} \\ =\frac{1}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}} \\ =\frac{2+\sqrt{3}}{{\left(2\right)}^2-{\left(\sqrt{3}\right)}^2} \\ =\frac{2+\sqrt{3}}{4-3} \\ =2+\sqrt{3} \end{gathered}$$