A) \[{{x}^{2}}+2x-16=0\]
B) \[{{x}^{2}}+2x+15=0\]
C) \[~{{x}^{2}}+2x-12=0\]
D) \[{{x}^{2}}+2x-\text{ }8=0\]
Correct Answer: D
Solution :
Given that, \[\alpha +\beta =-2\]and \[{{\alpha }^{3}}+{{\beta }^{3}}=-56\] \[\Rightarrow \]\[(\alpha +\beta )({{\alpha }^{2}}+{{\beta }^{2}}-\alpha \beta )=-56\] \[\Rightarrow \] \[{{\alpha }^{2}}+{{\beta }^{2}}-\alpha \beta =28\] Also, \[{{(\alpha +\beta )}^{2}}={{(-2)}^{2}}\] \[\Rightarrow \] \[{{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta =4\] \[\Rightarrow \] \[28+3\alpha \beta =4\] \[\Rightarrow \] \[\alpha \beta =-8\] \[\therefore \] Required equation is \[{{x}^{2}}+2x-8=0\]
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