JEE Advance - Mathematics (2010 - Paper 1 Offline - No. 6)
Let $$p$$ and $$q$$ be real numbers such that $$p \ne 0,\,{p^3} \ne q$$ and $${p^3} \ne - q.$$ If $${p^3} \ne - q.$$ and $$\,\beta $$ are nonzero complex numbers satisfying $$\alpha \, + \beta = - p\,$$ and $${\alpha ^3} + {\beta ^3} = q,$$ then a quadratic equation having $${\alpha \over \beta }$$ and $${\beta \over \alpha }$$ as its roots is
$$\left( {{p^3} + q} \right){x^2} - \left( {{p^3} + 2q} \right)x + \left( {{p^3} + q} \right) = 0$$
$$\left( {{p^3} + q} \right){x^2} - \left( {{p^3} - 2q} \right)x + \left( {{p^3} + q} \right) = 0$$
$$\left( {{p^3} - q} \right){x^2} - \left( {5{p^3} - 2q} \right)x + \left( {{p^3} - q} \right) = 0$$
$$\left( {{p^3} - q} \right){x^2} - \left( {5{p^3} + 2q} \right)x + \left( {{p^3} - q} \right) = 0$$
Explanation
Sum of roots = $${{{\alpha ^2} + {\beta ^2}} \over {\alpha \beta }}$$ and product = 1
Given, $$\alpha$$ + $$\beta$$ = $$-$$ p and $$\alpha$$3 + $$\beta$$3 = q
$$ \Rightarrow (\alpha + \beta )({\alpha ^2} - \alpha \beta + {\beta ^2}) = q$$
$$\therefore$$ $${\alpha ^2} + {\beta ^2} - \alpha \beta = {{ - q} \over p}$$ ..... (i)
and $${(\alpha + \beta )^2} = {p^2}$$
$$ \Rightarrow {\alpha ^2} + {\beta ^2} + 2\alpha \beta = {p^2}$$ ..... (ii)
From Eq. (i) and (ii), we get
$${\alpha ^2} + {\beta ^2} = {{{p^3} - 2q} \over {3p}}$$
and $$\alpha \beta = {{{p^3} + q} \over {3p}}$$
$$\therefore$$ Required equation is
$${x^2} - {{({p^2} - 2q)x} \over {({p^3} + q)}} + 1 = 0$$
$$ \Rightarrow ({p^3} + q){x^2} - ({p^3} - 2q)x + ({p^3} + q) = 0$$
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