A) 3
B) 10
C) 8
D) 2
Correct Answer: A
Solution :
[a] \[x=\sqrt{6+\sqrt{6+\sqrt{6+........\infty }}}\] On squaring, \[{{x}^{2}}=6+\sqrt{6+\sqrt{6+..........\infty }}\] \[\Rightarrow \,\,\,{{x}^{2}}=6+x\] \[\Rightarrow \,\,\,{{x}^{2}}-\,x-\,6=0\] \[\Rightarrow \,\,\,{{x}^{2}}-\,3x+2x-\,6=0\] \[\Rightarrow x\,\,(x-3)+2(x-3)=0\] \[\Rightarrow (x-3)\,\,(x+2)=0\] \[\Rightarrow x=3\,\,because\,\,x\ne -2\] By trick \[3\times 2=6\] Let the number be x and y.You need to login to perform this action.
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