Question : If $x=\frac{4\sqrt{15}}{\sqrt{5}+\sqrt{3}}$, the value of $\frac{x+\sqrt{20}}{x–\sqrt{20}}+\frac{x+\sqrt{12}}{x–\sqrt{12}}$ is:
Option 1: $1$
Option 2: $2$
Option 3: $\sqrt{3}$
Option 4: $\sqrt{5}$
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Correct Answer: $2$
Solution :
Given: $x=\frac{4\sqrt{15}}{\sqrt{5}+\sqrt{3}}=\frac{4\sqrt{5}\sqrt{3}}{\sqrt{5}+\sqrt{3}}$
Now, $\frac{x}{\sqrt{20}}=\frac{4\sqrt{5}\sqrt{3}}{\sqrt{20}(\sqrt{5}+\sqrt{3})} = \frac{2\sqrt{3}}{(\sqrt{5}+\sqrt{3})}$
Using componendo and dividendo
$\frac{x+\sqrt{20}}{x-\sqrt{20}}=\frac{\sqrt{5}+3\sqrt{3}}{(\sqrt{3}-\sqrt{5})}$
Similarly, we can find $\frac{x+\sqrt{12}}{x-\sqrt{12}}=\frac{\sqrt{3}+3\sqrt{5}}{(\sqrt{5}-\sqrt{3})}$
Now, $\frac{x+\sqrt{20}}{x–\sqrt{20}}+\frac{x+\sqrt{12}}{x–\sqrt{12}}=\frac{\sqrt{5}+3\sqrt{3}}{(\sqrt{3}-\sqrt{5})}+\frac{\sqrt{3}+3\sqrt{5}}{(\sqrt{5}-\sqrt{3})} =\frac{2\sqrt{3}-2\sqrt{5}}{(\sqrt{3}-\sqrt{5})}$ =2
Hence, the correct answer is $2$.
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