A) \[-1\]
B) \[\sqrt{2}\]
C) 2
D) \[1/2\]
Correct Answer: C
Solution :
\[y=\sqrt{2+\sqrt{2+\sqrt{2}}}+...\] \[y=\sqrt{2+y}\] Squaring of both sides \[{{y}^{2}}=2+y\Rightarrow \] \[{{y}^{2}}-y-2=0\] Since \[y>0\,{{y}^{2}}-2y+y-2=0\] \[y(y-2)+1(y-2)=0\] \[y=2,-1\]You need to login to perform this action.
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