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