A) \( \Large \frac{5}{12} \) |
B) \( \Large \frac{12}{5} \) |
C) \( \Large \frac{5}{7} \) |
D) \( \Large \frac{7}{5} \) |
B) \( \Large \frac{12}{5} \) |
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
\( \Large x = \frac{12}{5} \)