Question
Rationalize the denominator : $\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$
✓
Answer
$\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})} \times \frac{(\sqrt{5}-\sqrt{3})}{(\sqrt{5}-\sqrt{3})}$
...[Multiplying the numerator and denominator by $(\sqrt{5}-\sqrt{3})]$
$ =\frac{(\sqrt{5}-\sqrt{3})^2}{(\sqrt{5})^2-(\sqrt{3})^2}$
$\ldots\left[\because(a-b)(a+b)=a^2-b^2\right]$
$=\frac{(\sqrt{5})^2-2 \times \sqrt{5} \times \sqrt{3}+(\sqrt{3})^2}{5-3}$
$=\frac{5-2 \sqrt{15}+3}{2}$
$=\frac{8-2 \sqrt{15}}{2}$
$=\frac{2(4-\sqrt{15})}{2}=4-\sqrt{15}$
$\therefore \quad \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=4-\sqrt{15} $
...[Multiplying the numerator and denominator by $(\sqrt{5}-\sqrt{3})]$
$ =\frac{(\sqrt{5}-\sqrt{3})^2}{(\sqrt{5})^2-(\sqrt{3})^2}$
$\ldots\left[\because(a-b)(a+b)=a^2-b^2\right]$
$=\frac{(\sqrt{5})^2-2 \times \sqrt{5} \times \sqrt{3}+(\sqrt{3})^2}{5-3}$
$=\frac{5-2 \sqrt{15}+3}{2}$
$=\frac{8-2 \sqrt{15}}{2}$
$=\frac{2(4-\sqrt{15})}{2}=4-\sqrt{15}$
$\therefore \quad \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=4-\sqrt{15} $
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