A) \[6,-4,1\]
B) \[4,6,-1\]
C) \[3,-2,1\]
D) \[6,4,1\]
Correct Answer: A
Solution :
Given quadratic equation is \[{{\operatorname{ax}}^{2}} + bx + c = 0\] whose one root is \[\frac{1}{2-\sqrt{-\,2}}\] Consider \[\frac{1}{2-\sqrt{-\,2}}=\frac{1}{2-\sqrt{2i}}\times \frac{2+\sqrt{2i}}{2+\sqrt{2}i}\] \[=\,\,\,\frac{2+\sqrt{2}i}{4+2}=\frac{2+\sqrt{2}\,i}{6}\] \[\therefore \] Another root will be \[\frac{2-\sqrt{2}i}{6}\] (\[\because \] complex roots always occurs in pairs) Thus, sum of roots \[=\,\,\left( \frac{2+\sqrt{2}i}{6} \right)\left( \frac{2-\sqrt{2}i}{6} \right)\] \[=\,\,\frac{4+2}{36}=\frac{1}{6}\] \[\therefore \] Required equation is \[{{\operatorname{x}}^{2}}- \left( sum of roots \right)x + \left( product of roots \right) = 0\] \[{{x}^{2}}-\frac{4}{6}x+\frac{1}{6}=0\] \[\Rightarrow \,\,\,6{{x}^{2}}-4x+1=0\] Thus, the values of a, b, c are 6, - 4, 1 respectively
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