First slide

Theory of equations

Question

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 $α / β and β / α$ as its roots is

Moderate

A

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

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

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

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

Solution

$\begin{array}{l} α^{3} + β^{3} = q \\ \Rightarrow (α + β)^{3} - 3 αβ (α + β) = q \\ \Rightarrow - p^{3} + 3 pαβ = q \Rightarrow αβ = \frac{q + p^{3}}{3 p} \end{array}$
Required equation is
$\begin{array}{l} x^{2} - (\frac{α}{β} + \frac{β}{α}) x + \frac{α}{β} \cdot \frac{β}{α} = 0 \\ x^{2} - \frac{(α^{2} + β^{2})}{αβ} x + 1 = 0 \\ \Rightarrow x^{2} - (\frac{(α + β)^{2} - 2 αβ}{αβ}) x + 1 = 0 \\ \Rightarrow x^{2} - \frac{p^{2} - 2 (\frac{p^{3} + q}{3 p})}{\frac{p^{3} + q}{3 p}} x + 1 = 0 \\ \Rightarrow (p^{3} + q) x^{2} - (3 p^{3} - 2 p^{3} - 2 q) x + (p^{3} + q) = 0 \\ \Rightarrow (p^{3} + q) x^{2} - (p^{3} - 2 q) x + (p^{3} + q) = 0 \end{array}$

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