Question

$\frac{1}{\left(3+2\sqrt{2}\right)}=?$
(a) $\frac{3-2\sqrt{2}}{17}$
(b) $\frac{\left(3-2\sqrt{2}\right)}{13}$
(c) $\left(3-2\sqrt{2}\right)$
(d) none of these

Answer 1

(c) $\left(3-2\sqrt{2}\right)$
On rationalising, we get:
$$\begin{gathered} \frac{1}{3+2\sqrt{2}} \\ =\frac{1}{3+2\sqrt{2}}\times \frac{3-2\sqrt{2}}{3-2\sqrt{2}} \\ =\frac{3-2\sqrt{2}}{(3{)}^2-{\left(2\sqrt{2}\right)}^2} \\ =\frac{3-2\sqrt{2}}{9-8} \\ =3-2\sqrt{2} \end{gathered}$$