Question

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

The correct option is D x${}^2$+2x-8=0

$\Rightarrow$ $\alpha +\beta =-2$ ----- ( 1 )

$\Rightarrow$ ${\alpha }^3+{\beta }^3=-56$ ----- ( 2 )

$\Rightarrow$ $(\alpha +\beta {)}^3={\alpha }^3+{\beta }^3+3{\alpha }^2\beta +3\alpha {\beta }^2$

$\Rightarrow$ $(\alpha +\beta {)}^3={\alpha }^3+{\beta }^3+3\alpha \beta (\alpha +\beta )$

$\Rightarrow$ $(-2{)}^3=(-56)+3\alpha \beta (-2)$ [ Using ( 1 ) and ( 2 ) ]

$\Rightarrow$ $-8+56=-6\alpha \beta$

$\Rightarrow$ $48=-6\alpha \beta$

$\therefore$ $\alpha \beta =-8$ ------ ( 3 )

The required equation,

$x^2-(\alpha +\beta )x+\left(\alpha \beta \right)=0$

Now, From ( 1 ) and ( 3 ) we get,

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