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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