Question : If $\frac{\sqrt{5+x}+\sqrt{5-x}}{\sqrt{5+x}-\sqrt{5-x}}=3$, what is the value of $x$?
Option 1: $\frac{5}{2}$
Option 2: $\frac{25}{3}$
Option 3: $4$
Option 4: $3$
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Correct Answer: $3$
Solution :
Given:
$\frac{\sqrt{5+x}+\sqrt{5-x}}{\sqrt{5+x}-\sqrt{5-x}}=3$
Take $a=(\sqrt{5+x}+\sqrt{5-x})$, $b=(\sqrt{5+x}-\sqrt{5-x}), m=3$ and $n=1$.
Applying the rule of componendo and dividendo,
$\frac{a}{b}=\frac{m}{n}$ ⇒ $\frac{a+b}{a-b}=\frac{m+n}{m-n}$
Now, $\frac{\sqrt{5+x}+\sqrt{5-x}}{\sqrt{5+x}-\sqrt{5-x}}=3$
⇒ $\frac{\sqrt{5+x}+\sqrt{5-x}+\sqrt{5+x}-\sqrt{5-x}}{\sqrt{5+x}+\sqrt{5-x}- \sqrt{5+x}+\sqrt{5-x}}=\frac{3+1}{3-1}$
⇒ $\frac{\sqrt{5+x}+\sqrt{5+x}}{\sqrt{5-x}+\sqrt{5-x}}=\frac{4}{2}$
⇒ $\frac{2\sqrt{5+x}}{2\sqrt{5-x}}=\frac{4}{2}$
⇒ $\frac{\sqrt{5+x}}{\sqrt{5-x}}=2$
Squaring both sides, we get,
⇒ $\left(\frac{\sqrt{5+x}}{\sqrt{5-x}}\right)^{2}=2^{2}$
⇒ $\frac{5+x}{5-x}=4$
⇒ $5+x=20-4x$
⇒ $5x=15$
$\therefore x=3$
Hence, the correct answer is $3$.
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