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

Answer 1

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

Let the new equation be $x^2+Bx+C=0$

$B=-\left(\frac{\alpha }{\beta }+\frac{\beta }{\alpha }\right)$

$C=\frac{\alpha }{\beta }.\frac{\beta }{\alpha }$

$x^2-\left(\frac{\alpha }{\beta }+\frac{\beta }{\alpha }\right)x+\frac{\alpha }{\beta }\cdot \frac{\beta }{\alpha }=0$

$x^2-\frac{({\alpha }^2+{\beta }^2)}{\alpha \beta }x+1=0$

$x^2-\left(\frac{{(\alpha +\beta )}^2-2\alpha \beta }{\alpha \beta }\right)x+1=0$

Now, we have ${\alpha }^3+{\beta }^3=-q$

${(\alpha +\beta )}^3-3\alpha \beta (\alpha +\beta )=-q$

$-p^3+3p\alpha \beta =-q$

$\alpha \beta =\frac{-q+p^3}{3p}$

Substituting the value of $\alpha \beta$ in the equation,

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

$(p^3-q)x^2-(3p^3-2p^3+2q)x+(p^3-q)=0$

$(p^3-q)x^2-(p^3+2q)x+(p^3-q)=0$