Problem: (1992 Brazil National Olympiad - #1) The equation $ x^3+px+q = 0 $ has three distinct real roots. Show that $ p < 0 $.
Solution: Let $ a,b,c $ represent the roots of the equation. By Vieta's Formulas, we have
$ a+b+c = 0 $ and $ ab+bc+ca = p $.
But since $ (a-b)^2+(b-c)^2+(c-a)^2 > 0 \Rightarrow ab+bc+ca < \frac{(a+b+c)^2}{3} = 0 $, we have $ p < 0 $ as desired. QED.
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Comment: Not a difficult problem, but an interesting result (rather general). Manipulating Vieta's Formulas is often useful when dealing with polynomials (particularly with real roots).
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Practice Problem: Find a calculus proof of the above problem.
Three distinct real roots means the first derivative must have two distinct real roots; 3x^2 + p = 0 implies p < 0
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