Question

Let $p$ and $q$ be real numbers such that $p \neq 0, p ^{3} \neq q$ and $p ^{3} \neq- q \cdot$ 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)

$\alpha^{3}+\beta^{3}=q$
$\Rightarrow (\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta)=q$
$\Rightarrow -p^{3}+3 p \alpha \beta=q $
$\Rightarrow \alpha \beta=\frac{q+p^{3}}{3 p}$
$x^{2}-\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right) x+\frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha}=0$
$\Rightarrow x^{2}-\frac{p^{2}-2\left(\frac{p^{3}+q}{3 p}\right)}{\frac{p^{3}+q}{3 p}} x+1=0$
$\Rightarrow x^{2}-\left(\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{\alpha \beta}\right) x+1=0$
$\Rightarrow \left(p^{3}+q\right) x^{2}-\left(3 p^{3}-2 p^{3}-2 q\right) x+\left(p^{3}+q\right)=0$
$\Rightarrow \left(p^{3}+q\right) x^{2}-\left(p^{3}-2 q\right) x+\left(p^{3}+q\right)=0$