Question

Let p and q be real numbers such that $p\ne 0$, $p^3\ne 2q$ and $p^3\ne -q$. If $\alpha$ and $\beta$ are non zero 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. $(p^3-q)x^2-(p^3+2q)x+(p^3-q)=0$
• 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. $(p^3+q)x^2-(p^3-2q)x+(p^3+q)=0$

Answer 1

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

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