## Friday, May 12, 2006

### Really Rooty Real Roots. Topic: Algebra/Calculus. Level: AIME/Olympiad.

Problem: (Problem-Solving Through Problems - 6.5.5(b)) If $a_0, a_1, \ldots, a_n$ are real numbers satisfying

$\frac{a_0}{1}+\frac{a_1}{2}+\cdots+\frac{a_n}{n+1} = 0$,

show that the equation $a_0+a_1x+\cdots+a_nx^n = 0$ as at least one real root.

Solution: Consider integrating this function over the interval $[0,1]$. Let $f(x) = a_0+a_1x+\cdots+a_nx^n = 0$. We have

$\displaystyle \int_0^1 (a_0+a_1x+\cdots+a_nx^n)dx = \left[\frac{a_0x}{1}+\frac{a_1x^2}{2}+\cdots+\frac{a_nx^{n+1}}{n+1}\right]^1_0 = \frac{a_0}{1}+\frac{a_1}{2}+\cdots+\frac{a_n}{n+1} = 0$.

If $f$ had no real roots on the interval $[0,1]$ it would be strictly positive or negative. Then the integral of $f$ over $[0,1]$ would also be strictly positive or negative, respectively. Hence $f$ must take on both positive and negative values. But since $f$ is continuous, we know that there must be a $c \in [0,1]$ such that $f(c) = 0$ and therefore $f$ has a real root. QED.

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Comment: The condition involving the fractions should give away integration; from there, it's not hard to see that we want the interval $[0,1]$ and this gives us the result immediately.

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Practice Problem: (Problem-Solving Through Problems - 6.5.5(a)) Show that $5x^4-4x+1$ has a root between $0$ and $1$.