Question

If $\alpha$ and $\beta$ are roots of the quadratic equation $ {{x}^{2}}+4x+3=0, $ then the equation whose roots are $ 2\alpha \,\text{+}\,\beta $ and $ \alpha \,\text{+2}\,\beta $ is

• A. $ {{x}^{2}}-12x+35=0 $
• B. $ {{x}^{2}}+12x-33=0 $
• C. $ {{x}^{2}}-12x-33=0 $
• D. $ {{x}^{2}}+12x+35=0 $

Answer 1

Answer: (D)

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 $