Question

Let $p$ and $q$ be real numbers such that $p\ne 0,p^3\ne q$ and $p^3\ne -q$ . If $\alpha$ and $\beta$ are nonzero complex numbers satisfying $\alpha +\beta =-p$ and ${\alpha }^3+{\beta }^3=q$ , then a quadratic equation having $\frac{\alpha }{\beta }$ and $\frac{\beta }{\alpha }$ as its roots is -

• A. $(p^3+q)x^2-(5p^3-2q)x+(p^3-q)=0$
• B. $(p^3+q)x^2-(p^3+2q)x+(p^3+q)=0$
• C. $(p^3+q)x^2-(p^3-2q)x+(p^3+q)=0$
• D. $(p^3-q)x^2-(5p^3+2q)x+(p^3-q)=0$

Answer 1

The correct option is C $(p^3+q)x^2-(p^3-2q)x+(p^3+q)=0$

Given that $\alpha +\beta =-p$ and ${\alpha }^3+{\beta }^3=q$
$\Rightarrow (\alpha +\beta {)}^3-3\alpha \beta (\alpha +\beta )=q$
$\Rightarrow -p^3-3\alpha \beta (-p)=q\Rightarrow \alpha \beta =\frac{p^3+q}{3p}$
Now for required quadratic equation,
Sum of the roots $=\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{(\alpha +\beta {)}^2-2\alpha \beta }{\alpha \beta }=\frac{p^2-2(\frac{p^3+q}{3p})}{\frac{p^3+q}{3p}}=\frac{3p^3-2p^3-2q}{p^3+q}=\frac{p^3-2q}{p^3+q}$
And product of the roots $=\frac{\alpha }{\beta }.\frac{\beta }{\alpha }=1$
$\therefore$ The required equation is $x^2-(\frac{p^3-2q}{p^3+q})x+1=0$
or $(p^3+q)x^2-(p^3-2q)x+(p^3+q)=0$