A) 3
B) 4
C) \[4.5\]
D) \[3.5\]
Correct Answer: B
Solution :
Let \[x=\sqrt{12+\sqrt{12}+\sqrt{12}}+.......\infty \] \[x=\sqrt{12+x}\] Squaring both sides \[\Rightarrow \,\,\,\,{{x}^{2}}=12+x\] \[{{x}^{2}}-x-12=0\Rightarrow {{x}^{2}}-4x+3x-12=0\] \[x(x-4)+3(x-4)=0\Rightarrow (x+3)\,(x-4)=0\] So, \[x=-3,4\,\,\,\Rightarrow \,x=\underline{\mathbf{4}}\]You need to login to perform this action.
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