Question

Let $p$ and $q$ be real numbers such that $p$ $\neq 0, p^3 \neq q$ and $p^3 \neq-q$. 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. $\left(p^3+q\right) x^2-\left(p^3+2 q\right) x+\left(p^3+q\right)=0$
• B. $\left(p^3+q\right) x^2-\left(p^3-2 q\right) x+\left(p^3+q\right)=0$
• C. $\left(p^3-q\right) x^2-\left(5 p^3-2 q\right) x+\left(p^3-q\right)=0$
• D. $\left(p^3-q\right) x^2-\left(5 p^3+2 q\right) x+\left(p^3-q\right)=0$

Answer 1

Answer: (B)

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