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 nonzero 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. $\left(p^3+q\right)x^2-\left(p^3+2q\right)x+\left(p^3+q\right)=0$
• B. $\left(p^3+q\right)x^2-\left(p^3-2q\right)x+\left(p^3+q\right)=0$
• C. $\left(p^3-q\right)x^2-\left(5p^3-2q\right)x+\left(p^3-q\right)=0$
• D. $\left(p^3-q\right)x^2-\left(5p^3+2q\right)x+\left(p^3-q\right)=0$

Answer 1

Answer: (B)

$q={\alpha }^3+{\beta }^3=(\alpha +\beta {)}^3-3\alpha \beta (\alpha +\beta )$ $=-p^3+3\alpha \beta p$
$\Rightarrow \alpha \beta =\frac{p^3+q}{3p}$
We have
$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }$
$=\frac{(\alpha +\beta {)}^2-2\alpha \beta }{\alpha \beta }=\frac{(\alpha +\beta {)}^2}{\alpha \beta }-2$
$=\frac{p^2}{\left(p^3+q\right)/3p}-2$
$=\frac{3p^3-2p^3-2q}{p^3+q}=\frac{p^3-2q}{p^3+q}$
$\text{ and }\frac{\alpha }{\beta }\cdot \frac{\beta }{\alpha }=1$
Thus, required quadratic equation is
$x^2-\frac{\left(p^3-2q\right)}{p^3+q}x+1=0$
or $\left(p^3+q\right)x^2-\left(p^3-2q\right)x+\left(p^3+q\right)=0$