Question:
Find the value of $\frac{6}{\sqrt{5}-\sqrt{3}}$, it being given that $\sqrt{3}=1.732$ and $\sqrt{5}=2.236$.
Solution:
Given,
$\frac{6}{\sqrt{5}-\sqrt{3}}$
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor
$\sqrt{5}+\sqrt{3}$ for $\frac{1}{\sqrt{5}-\sqrt{3}}$
$=\frac{6(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}$
Since, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{6 \sqrt{5}+6 \sqrt{3}}{5-3}$
$=\frac{6 \sqrt{5}+6 \sqrt{3}}{2}$
$=3(\sqrt{5}+\sqrt{3})$
$=3(2.236+1.732)=3(3.968)=11.904$
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