A) \[-2+\sqrt{3}\]
B) \[-\sqrt{3}-2\]
C) \[2-\sqrt{3}\]
D) \[\sqrt{3}+1\]
Correct Answer: C
Solution :
| Let \[\alpha ,\,\,\beta\] are the roots of the given equation \[2-\sqrt{3}\] |
| Comparing the given polynomial \[{{x}^{2}}-4x+1\] with the standard equation \[a{{x}^{2}}+bx+c\] |
| we get a = 1, b = - 4 and c = 1. Sum of zeroes |
| \[\alpha +\beta =\frac{-b}{a}=4\] |
| \[2+\sqrt{3}+\beta =4\,\Rightarrow \,\beta =2-\sqrt{3}\] |
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