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

Answer 1

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

We have,
$\alpha +\beta =-p$
${\alpha }^3+{\beta }^3=q=(\alpha +\beta {)}^3-3\alpha \beta (\alpha +\beta )=-p^3+3p(\alpha \beta )$
$\Rightarrow \alpha \beta =\frac{p^3+q}{3p}$
The quadratic equation with $\frac{\alpha }{\beta }$ and $\frac{\beta }{\alpha }$ as roots is
$x^2-(\frac{\alpha }{\beta }+\frac{\beta }{\alpha })x+\frac{\alpha }{\beta }.\frac{\beta }{\alpha }=0$
$\Rightarrow x^2-\left(\frac{(\alpha +\beta {)}^2-2\alpha \beta }{\alpha \beta }\right)x+1=0$
$\Rightarrow x^2-\frac{p^2-2\frac{p^3+q}{3p}}{\frac{p^3+q}{3p}}x+1=0$
$\Rightarrow (p^3+q)x^2-(p^3-2q)x+(p^3+q)=0$