Question

Two complex numbers $\alpha$ and $\beta$ are such that $\alpha +\beta =2$ and ${\alpha }^4+{\beta }^4=272$, then the quadratic equation whose roots are $\alpha$ and $\beta$ can be

• A. $x^2-2x-16=0$
• B. $x^2-2x+12=0$
• C. $x^2-2x-8=0$
• D. none of these

Answer 1

Answer: (C)

$\alpha +\beta =2\Rightarrow {\alpha }^2+{\beta }^2+2\alpha \beta =4$
$\Rightarrow {\alpha }^2+{\beta }^2=4-2\alpha \beta$ $\Rightarrow {\left({\alpha }^2+{\beta }^2\right)}^2=16-16\alpha \beta +4{\alpha }^2{\beta }^2$
$\Rightarrow {\alpha }^4+{\beta }^4+2{\alpha }^2{\beta }^2=16-16\alpha \beta +4{\alpha }^2{\beta }^2$
$\Rightarrow 272+2(\alpha \beta {)}^2=16-16\alpha \beta +4(\alpha \beta {)}^2$
$\Rightarrow 2(\alpha \beta {)}^2-16(\alpha \beta )-256=0$
$\Rightarrow (\alpha \beta {)}^2-8(\alpha \beta )-128=0$
$\Rightarrow (\alpha \beta -16)(\alpha \beta +8)=0$
$\Rightarrow \alpha \beta =16\text{ or }\alpha \beta =-8$
Thus, required equation is either
$x^2-2x+16=0$
or$x^2-2x-8=0$
$\therefore$ Required answer is (c).