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

Answer 1

Answer: (B)

Sum of roots=$\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }$ and product $=1$
Given, $\alpha +\beta =-p$ and ${\alpha }^3+{\beta }^3=q$
$\Rightarrow ,(\alpha +\beta )({\alpha }^2-\alpha \beta +{\beta }^2)=q$
$\therefore ({\alpha }^2+{\beta }^2-\alpha \beta )=\frac{-q}{p}...(i)$
and ${\alpha }^2+{\beta }^2=p^2$
$\Rightarrow {\alpha }^2+{\beta }^2+2\alpha \beta =p^2...(ii)$
From Eqs. (i) and (ii), we get
${\alpha }^2+{\beta }^2=\frac{p^3-2q}{3p}$ and $\alpha \beta =\frac{p^3+q}{3p}$
$\therefore$ Required equation is $,x^2-\frac{(p^3-2q)x}{p^3+q}+1=0$
$\Rightarrow (p^3+q)x^2-(p^3-2q)x+(p^3+q)=0$