A) \[-1\]
B) \[1\]
C) \[2\]
D) \[3\]
Correct Answer: C
Solution :
Given,\[x=\sqrt{2+\sqrt{2+\sqrt{2...}}}\] \[\Rightarrow \] \[x=\sqrt{2+x}\] On squaring both sides, we get \[{{x}^{2}}=2+x\] \[\Rightarrow \] \[{{x}^{2}}-x-2=0\] \[\Rightarrow \] \[(x-2)(x+1)=0\Rightarrow x=2,\,\,-1\] But\[\sqrt{2+\sqrt{2...}}\ne -1\] So, it is equal to\[2\].
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