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# 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) $$\Large \frac{7}{5}$$

 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

$$\Large x = \frac{12}{5}$$

Part of solved Linear Equations questions and answers : >> Aptitude >> Linear Equations

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