Question
Rationalize the denominator : $\frac{3}{2 \sqrt{5}-3 \sqrt{2}}$
✓
Answer
$\frac{3}{2 \sqrt{5}-3 \sqrt{2}}=\frac{3}{(2 \sqrt{5}-3 \sqrt{2})} \times \frac{(2 \sqrt{5}+3 \sqrt{2})}{(2 \sqrt{5}+3 \sqrt{2})}$
...[Multiplying the numerator and denominator by $(2 \sqrt{5}+3 \sqrt{2})]$
$\begin{aligned}
& =\frac{3(2 \sqrt{5}+3 \sqrt{2})}{(2 \sqrt{5})^2-(3 \sqrt{2})^2} \\
& \ldots\left[\because(a-b)(a+b)=a^2-b^2\right] \\
& =\frac{3(2 \sqrt{5}+3 \sqrt{2})}{(4 \times 5)-(9 \times 2)} \\
& =\frac{3(2 \sqrt{5}+3 \sqrt{2})}{20-18} \\
\therefore \quad \frac{3}{2 \sqrt{5}-3 \sqrt{2}} & =\frac{3(2 \sqrt{5}+3 \sqrt{2})}{2}
\end{aligned}$
...[Multiplying the numerator and denominator by $(2 \sqrt{5}+3 \sqrt{2})]$
$\begin{aligned}
& =\frac{3(2 \sqrt{5}+3 \sqrt{2})}{(2 \sqrt{5})^2-(3 \sqrt{2})^2} \\
& \ldots\left[\because(a-b)(a+b)=a^2-b^2\right] \\
& =\frac{3(2 \sqrt{5}+3 \sqrt{2})}{(4 \times 5)-(9 \times 2)} \\
& =\frac{3(2 \sqrt{5}+3 \sqrt{2})}{20-18} \\
\therefore \quad \frac{3}{2 \sqrt{5}-3 \sqrt{2}} & =\frac{3(2 \sqrt{5}+3 \sqrt{2})}{2}
\end{aligned}$
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