• 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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