Question

Let $p$ and $q$ be real numbers such that $p\ne 0,p^3\ne q$ and $p^3\ne -q\cdot$ 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. $\left(p^3+q\right)x^2-\left(p^3+2q\right)x+\left(p^3+q\right)=0$
• 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)

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