A) \[{{x}^{2}}-12x+35=0\]
B) \[{{x}^{2}}+12x-33=0\]
C) \[{{x}^{2}}-12x-33=0\]
D) \[{{x}^{2}}+12x+35=0\]
Correct Answer: D
Solution :
Given \[\alpha ,\beta \] are the roots of equation \[{{x}^{2}}+4x+3=0\] \[\therefore \] \[\alpha +\beta =-4\] and \[\alpha \beta =3\] Now, \[2\alpha +\beta +\alpha +2\beta =3(\alpha +\beta )=-12\] and \[(2\alpha +\beta )(\alpha +2\beta )=2{{\alpha }^{2}}+4\alpha \beta +\alpha \beta +2{{\beta }^{2}}\] \[=2{{(\alpha +\beta )}^{2}}+\alpha \beta \] \[=2{{(-4)}^{2}}+3=35\] Hence, required equation is \[{{x}^{2}}-(\text{sum of roots) x + (product of roots) = 0}\] \[\Rightarrow \] \[{{x}^{2}}+12x+35=0\]
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