A) \[{{x}^{2}}+\alpha x-\beta =0\]
B) \[{{x}^{2}}-[(\alpha +\beta )+\alpha \beta ]x-\alpha \beta (\alpha +\beta )=0\]
C) \[{{x}^{2}}+[(\alpha +\beta )+\alpha \beta ]x+\alpha \beta (\alpha +\beta )=0\]
D) \[{{x}^{2}}+[\alpha \beta +(\alpha +\beta )]x-\alpha \beta (\alpha +\beta )=0\]
Correct Answer: C
Solution :
\[\alpha +\beta =-\]b and \[\alpha \beta =-c\] Now \[b+c=-[(\alpha +\beta )+\alpha \beta ],bc=(\alpha +\beta )(\alpha \beta )\] Hence required equation is \[{{x}^{2}}+[(\alpha +\beta )+\alpha \beta ]x+\alpha \beta (\alpha +\beta )=0\].
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