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

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

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

(C) $(p^{3}-q)x^{2}-(5p^{3}-2q)x+(p^{3}-q)=0$

(D) $(p^{3}-q)x^{2}-(5p^{3}+2q)x+(p^{3}-q)=0$

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Quadratic Equations and Inequalities

Let

p

and

q

be real numbers such that

p \neq = 0, p^{3} \neq = q

and

p^{3} \neq = - q

. If

α

and

β

are nonzero complex numbers satisfying

α + β = - p

and

α^{3} + β^{3} = q

, then a quadratic equation having

\frac{α}{β}

and

\frac{β}{α}

as its roots is

Answer: (B)

Given:

α + β = - p

and

α^{3} + β^{3} = q

.

Using the identity:

α^{3} + β^{3} = (α + β)^{3} - 3 α β (α + β)

q = (- p)^{3} - 3 α β (- p) = - p^{3} + 3 p α β

α β = \frac{q + p ^{3}}{3 p}

For the quadratic with roots

\frac{α}{β}

and

\frac{β}{α}

Sum of roots: $\frac{α}{β} + \frac{β}{α} = \frac{α ^{2} + β ^{2}}{α β} = \frac{( α + β ) ^{2} - 2 α β}{α β} = \frac{p ^{2} - 2 \cdot \frac{q + p ^{3}}{3 p}}{\frac{q + p ^{3}}{3 p}} = \frac{3 p ^{3} - 2 q - 2 p ^{3}}{q + p ^{3}} = \frac{p ^{3} - 2 q}{p ^{3} + q}$

The equation is:

x^{2} - \frac{p ^{3} - 2 q}{p ^{3} + q} x + 1 = 0

Multiplying by

(p^{3} + q)

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

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Single correct option