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Brazilian Mathematical Olympiad

Algebra Difficulty 4.5 AIME Prove it Brazil

The polynomial x3+px+qx^3 + px + q has three distinct real roots. Show that p<0p < 0.

Solution

The sum of the squares of the roots α,β,γ\alpha, \beta, \gamma is s2=α2+β2+γ2=(α+β+γ)2−2(αβ+βγ+γα)=02−2ps_2 = \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = 0^2 - 2p. Since the roots are distinct, s2>0  ⟺  p<0s_2 > 0 \iff p < 0.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.