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 $,x^2-\frac{(p^3-2q)x}{p^3+q}+1=0$
$\Rightarrow {(p^3+q)x^2-}{(p^3-2q)x+(p^3+q)}=0$