A) 6
B) 5
C) 4
D) 3
Correct Answer: D
Solution :
Here, \[{{x}^{2}}=6+\sqrt{6+\sqrt{6+\sqrt{6+...\infty }}}\] So, \[{{x}^{2}}=6+\sqrt{{{x}^{2}}}\Rightarrow {{x}^{2}}=6+x\] \[\Rightarrow {{x}^{2}}-x-6=0\Rightarrow {{x}^{2}}+2x-3x-6=0\] \[\Rightarrow \,x(x+2)-3(x+2)=0\] \[\Rightarrow \,(x-3)(x+2)=0\therefore x=\]You need to login to perform this action.
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