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. $x^2-2 x-16=0$
• B. $x^2-2 x+12=0$
• C. $x^2-2 x-8=0$
• 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^2-2 x+16=0 $
or$x^2-2 x-8=0$
$\therefore$ Required answer is (c).