Question

Rationalize the denominator$\frac{3-2\sqrt{2}}{3+2\sqrt{2}}$.

Answer 1

Rationalization is a process by which radicals in the denominator of an algebraic fraction are eliminated.
Multiply numerator & denominator by $\left(3-2\sqrt{2}\right)$ we have:

$$\begin{gathered} \frac{\left(3-2\sqrt{2}\right)}{\left(3+2\sqrt{2}\right)}=\frac{\left(3-2\sqrt{2}\right)}{\left(3+2\sqrt{2}\right)}\times \frac{\left(3-2\sqrt{2}\right)}{\left(3-2\sqrt{2}\right)} \\ =\frac{{\left(3-2\sqrt{2}\right)}^2}{{\left(3\right)}^2-{\left(2\sqrt{2}\right)}^2}[\left(a+b\right)\left(a-b\right)=a^2-b^2] \\ =\frac{9-12\sqrt{2}+8}{9-8}[{\left(a-b\right)}^2=a^2-2ab+b^2] \\ =17-12\sqrt{2} \end{gathered}$$

So, after rationalization of denominator the fraction $\frac{3-2\sqrt{2}}{3+2\sqrt{2}}=17-12\sqrt{2}$.