## Saturday, March 18, 2006

### Degree Three. Topic: Algebra/Polynomials. Level: AIME/Olympiad.

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.

#### 1 comment:

1. Three distinct real roots means the first derivative must have two distinct real roots; 3x^2 + p = 0 implies p < 0