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

Answer 1

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

Sum of roots $=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }$ and product $=1$

Given $\alpha +\beta =-p$ and ${\alpha }^3+{\beta }^3=q$

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

$\therefore {\alpha }^2+{\beta }^2-\alpha \beta =\frac{-q}{p}$ and ${(\alpha +\beta )}^2=p^2$ ...(1)

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

From (1) and (2)

${\alpha }^2+{\beta }^2=\frac{p^3-2q}{3p}$ and $\alpha \beta =\frac{p^3+q}{3p}$

Therefore required equation is

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