If \( \Large \frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}-\sqrt{3-x}}=2 \), then x is equal to

A) \( \Large \frac{5}{12} \)

B) \( \Large \frac{12}{5} \)

C) \( \Large \frac{5}{7} \)

D) \( \Large \frac{7}{5} \)

Correct answer:
B) \( \Large \frac{12}{5} \)

Description for Correct answer:

Given, \( \Large \frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}-\sqrt{3-x}}=2 \)

Let \( \Large \sqrt{3+x}=a \  and \ \sqrt{3-x}=b \)

Then, \( \Large \frac{a+b}{a-b}=\frac{2}{1} \)

\( \Large a+b=2a-2b=a=3b \)

On squaring both sides, we get

\( \Large \sqrt{3+x}= \left(3\sqrt{3-x}\right)^{2} \)

= \( \Large 3+x = 9 \left(3-x\right) \)

= \( \Large 3+x=27-9x \)

= 10x = 24

\( \Large x = \frac{12}{5} \)



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