A) \[{{x}^{2}}+x+1=0\]
B) \[{{x}^{2}}-x-1=0\]
C) \[{{x}^{2}}+x-1=0\]
D) None of these
Correct Answer: B
Solution :
Since,\[\alpha \]and\[\beta \]are the roots of\[{{x}^{2}}-3x+1=0\], then \[\alpha +\beta =3\]and\[\alpha \beta =1\] Now, \[S=\frac{1}{\alpha -2}+\frac{1}{\beta -2}\] \[=\frac{\alpha +\beta -4}{\alpha \beta -2(\alpha +\beta )+4}\] \[=\frac{3-4}{1-2\cdot 3+4}=1\] and \[P=\frac{1}{\alpha -2}\times \frac{1}{\beta -2}\] \[=\frac{1}{\alpha \beta -2(\alpha +\beta -2)}\] \[=1\] \[\therefore \]Required equation is \[{{x}^{2}}-Sx+P=0\] \[\Rightarrow \] \[{{x}^{2}}-x-1=0\]
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