Problem: $\frac{\sqrt{2+x+\sqrt{2-x}}}{\sqrt{2+x-\sqrt{2-x}}}$ = 2. Find the value of x.

View all: BANGLADESH BANK OFFICER | WRITTEN QUESTION (MATH) SOLVE | 2015

Correct Answer: ans , $\frac{8}{5}$

Explanation:

$\frac{\sqrt{2+x+\sqrt{2-x}}}{\sqrt{2+x-\sqrt{2-x}}}$ = 2

=> $\frac{\sqrt{2+x+\sqrt{2-x+\sqrt{2+x-\sqrt{2-x}}}}}{\sqrt{2+x+\sqrt{2-x-\sqrt{2+x+\sqrt{2-x}}}}}$ = $\frac{2+1}{2-1}$

=> $\frac{2\sqrt{2+x}}{2\sqrt{2-x}} =\frac{3}{1}$

=> $(\frac{\sqrt{2+x}}{\sqrt{2-x}})^{2}=3^{2}$

=> $\frac{2+x}{2-x}=9$

=> 2+x=18-9x

=> 10x=l6

=> x=$\frac{16}{10}$

x=$\frac{8}{5}$

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