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=$\frac{8}{5}$