Browse · MathNet Print → XIV OBM Brazil algebra Problem The polynomial x3+px+q has three distinct real roots. Show that p<0. Solution — click to reveal The sum of the squares of the roots α,β,γ is s2=α2+β2+γ2=(α+β+γ)2−2(αβ+βγ+γα)=02−2p. Since the roots are distinct, s2>0⟺p<0. Techniques Vieta's formulas ← Previous problem Next problem →