A) \[2{{x}^{2}}-7x+6=0\]
B) \[3{{x}^{2}}+9x+11=0\]
C) \[9{{x}^{2}}-27x+20=0\]
D) \[4{{x}^{2}}+22x+15=0\]
E) \[3{{x}^{2}}-12x-5=0\]
Correct Answer: E
Solution :
\[\therefore \]\[\alpha +\beta =4\]and\[{{\alpha }^{3}}+{{\beta }^{3}}=44\] \[\Rightarrow \] \[(\alpha +{{\beta }^{3}})-3\,\alpha \beta (\alpha +\beta )=44\] \[\Rightarrow \] \[64-44=12\alpha \beta \] \[\Rightarrow \] \[\alpha \beta =\frac{20}{12}=\frac{5}{3}\] \[\therefore \]Required equation is \[{{x}^{2}}-(\alpha +\beta )x+\alpha \beta =0\] \[\Rightarrow \] \[{{x}^{2}}-4x+\frac{5}{3}=0\] \[\Rightarrow \] \[3{{x}^{2}}-12x+5=0\]
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