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Unformatted text preview: A possible algebraic solution to one of our in-class examples Math 8 2009, Chris Lyons On October 16, we gave a proof of the following, as one of our in-class examples: Theorem 1. Suppose we have a polynomial f ( x ) = a + a 1 x + ... + a d x d with d > . If a a d < , then f has at least one positive root. We proved this by using calculus... in fact, we used Bolzano’s theorem. But you might ask: Question. Can we prove using only algebra? This seems like a valid question... after all, polynomials are a subject we all learn in algebra class, so perhaps resorting to the fancy calculus theorem isn’t really necessary. Let’s look at the first two cases of d : • d = 1: Let’s change some letters: let a = C and a 1 = B . Then f ( x ) looks like f ( x ) = Bx + C. The only root is x =- C/B , which is positive since B and C have opposite signs by assumption. So the case d = 1 is easy! • d = 2: This case is a little more tricky, but not difficult. Let’s write f ( x ) = Ax 2 + Bx + C, so that our assumption on f becomes AC < 0. We know the roots of f are x =- B ± √ B 2- 4 AC 2 A . Is one of these roots positive? Yes! The algebraic proof of this is straightforward, but a little bit cumbersome (i.e., boring), so I’ve placed it at the end of this handout, if you want to read it.cumbersome (i....
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