Question

If $\alpha +\beta =-2$ and ${\alpha }^3+{\beta }^3=-56$ , then the quadratic equation whose roots are $\alpha$ and $\beta$ is :

• A. $x^2+2x-16=0$
• B. $x^2+2x+15=0$
• C. $x^2+2x-12=0$
• D. $x^2+2x-8=0$

Answer 1

Answer: (D)

$x^2+2x-8=0$

Given that, $\alpha +\beta =-2$ and ${\alpha }^3+{\beta }^3=-56$

$\Rightarrow (\alpha +\beta )({\alpha }^2+{\beta }^2-\alpha \beta )=-56$

$\Rightarrow -2({\alpha }^2+{\beta }^2-\alpha \beta )=-56$

$\Rightarrow {\alpha }^2+{\beta }^2-\alpha \beta =28$

Also, $(\alpha +\beta {)}^2=(-2{)}^2$

$\Rightarrow {\alpha }^2+{\beta }^2+2\alpha \beta =4$

$\Rightarrow 28+\alpha \beta +2\alpha \beta =4$

$\Rightarrow 28+3\alpha \beta =4$

$\Rightarrow 3\alpha \beta =-24$

$\Rightarrow \alpha \beta =-8$

$\therefore$ Required equation is $x^2+2x-8=0$