| \[\sqrt{12+\sqrt{12+\sqrt{12+...}}}\]is equal to |
A) 3
B) 4
C) 6
D) 2
Correct Answer: A
Solution :
| Let \[x=\sqrt{12+\sqrt{12+\sqrt{12+...}}}\] |
| On squaring both sides, we get |
| \[{{x}^{2}}=12+\sqrt{12+\sqrt{12+...}}\] |
| \[\Rightarrow \] \[{{x}^{2}}-12=x\] |
| \[\Rightarrow \] \[{{x}^{2}}-x-12=0\] |
| \[\Rightarrow \]\[{{x}^{2}}-4x+3x-12=0\] |
| \[\Rightarrow \]\[x\,(x-4)+3\,(x-4)=0\] |
| \[\Rightarrow \] \[(x+3)(x-4)=0\] |
| \[\Rightarrow \] \[x=4,\]\[-\,\,3\] |
| Since, x will be positive number. |
| \[\therefore \] \[x=4,\]\[x\ne 3\] |
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