A) \[{{x}^{2}}+3x-1=0\]
B) \[{{x}^{2}}+3x-2=0\]
C) \[{{x}^{2}}+3x+2=0\]
D) \[{{x}^{2}}-3x+2=0\]
Correct Answer: D
Solution :
Given equation is \[3{{x}^{2}}-6x+5=0\] \[\therefore \] \[\alpha +\beta =2\] and \[\alpha \,\,\beta =\frac{5}{3}\] Now, \[\alpha +\beta +\frac{2}{\alpha +\beta }=2+\frac{2}{2}=3\] and \[(\alpha +\beta )\times \frac{2}{\alpha +\beta }=2\] \[\therefore \] Required equation is \[{{x}^{2}}-\left( (\alpha +\beta )+\frac{2}{\alpha +\beta } \right)x+\left( (\alpha +\beta )\times \frac{2}{\alpha +\beta } \right)=0\] \[\Rightarrow \] \[{{x}^{2}}-3x+2=0\]
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