Question
Let $$p$$ and $$q$$ be real numbers such that $$p \ne 0,{p^3} \ne q\,\,{\text{and }}{p^3} \ne - q.$$ If $$\alpha \,\,{\text{and }}\beta $$ are non-zero complex numbers satisfying $$\alpha \,{\text{ + }}\,\beta = - p\,\,{\text{and}}\,\,{\alpha ^3} + {\beta ^3} = q,$$ then a quadratic equation having $$\frac{\alpha }{\beta }\,\,{\text{and }}\frac{\beta }{\alpha }$$ 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.
$$\left( {{p^3} - q} \right){x^2} - \left( {5{p^3} - 2q} \right)x + \left( {{p^3} - q} \right) = 0$$
D.
$$\left( {{p^3} - q} \right){x^2} - \left( {5{p^3} + 2q} \right)x + \left( {{p^3} - q} \right) = 0$$
Answer :
$$\left( {{p^3} + q} \right){x^2} - \left( {{p^3} - 2q} \right)x + \left( {{p^3} + q} \right) = 0$$
Solution :
Given that
$$\eqalign{
& \alpha {\text{ + }}\beta = - p\,\,{\text{and }}{\alpha ^3} + {\beta ^3} = q \cr
& \Rightarrow \,\,{\left( {\alpha + \beta } \right)^3} - 3\alpha \beta \left( {\alpha + \beta } \right) = q \cr
& \Rightarrow \,\, - {p^3} - 3\alpha \beta \left( { - p} \right) = q \cr
& \Rightarrow \,\,\alpha \beta = \frac{{{p^3} + q}}{{3p}} \cr} $$
Now for required quadratic equation,
sum of roots $$ = \frac{\alpha }{\beta } + \frac{\beta }{\alpha } = \frac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }}$$
$$\eqalign{
& = \frac{{{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta }}{{\alpha \beta }} \cr
& = \frac{{{p^2} - 2\left( {\frac{{{p^3} + q}}{{3p}}} \right)}}{{\frac{{{p^3} + q}}{{3p}}}} \cr
& = \,\frac{{3{p^3} - 2{p^3} - 2q}}{{{p^3} + q}} \cr
& = \frac{{{p^3} - 2q}}{{{p^3} + q}} \cr} $$
and product of roots $$ = \frac{\alpha }{\beta }.\frac{\beta }{\alpha } = 1$$
∴ Required equation is $${x^2} - \left( {\frac{{{p^3} - 2q}}{{{p^3} + q}}} \right)x + 1 = 0$$
or $$\left( {{p^3} + q} \right){x^2} - \left( {{p^3} - 2q} \right)x + \left( {{p^3} + q} \right) = 0$$