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Mathematics Question on Complex Numbers and Quadratic Equations

If α\alpha and β\beta are roots of the quadratic equation x2+4x+3=0,{{x}^{2}}+4x+3=0, then the equation whose roots are 2α + β2\alpha \,\text{+}\,\beta and α +2 β\alpha \,\text{+2}\,\beta is

A

x2−12x+35=0{{x}^{2}}-12x+35=0

B

x2+12x−33=0{{x}^{2}}+12x-33=0

C

x2−12x−33=0{{x}^{2}}-12x-33=0

D

x2+12x+35=0{{x}^{2}}+12x+35=0

Answer

x2+12x+35=0{{x}^{2}}+12x+35=0

Explanation

Solution

Given α,β\alpha ,\beta are the roots of equation
x2+4x+3=0{{x}^{2}}+4x+3=0
∴\therefore α+β=−4\alpha +\beta =-4
and αβ=3\alpha \beta =3
Now, 2α+β+α+2β=3(α+β)=−122\alpha +\beta +\alpha +2\beta =3(\alpha +\beta )=-12
and (2α+β)(α+2β)=2α2+4αβ+αβ+2β2(2\alpha +\beta )(\alpha +2\beta )=2{{\alpha }^{2}}+4\alpha \beta +\alpha \beta +2{{\beta }^{2}}
=2(α+β)2+αβ=2{{(\alpha +\beta )}^{2}}+\alpha \beta
=2(−4)2+3=35=2{{(-4)}^{2}}+3=35
Hence, required equation is
x2−(sum of roots) x + (product of roots) = 0{{x}^{2}}-(\text{sum of roots) x + (product of roots) = 0}
⇒\Rightarrow x2+12x+35=0{{x}^{2}}+12x+35=0