Quadratics with Odd Coefficients
Date: 06/24/2003 at 05:09:38 From: Jim Subject: Quadratic roots Show that quadratics with odd coefficients have no rational roots.
Date: 06/24/2003 at 05:35:39 From: Doctor Jacques Subject: Re: Quadratic roots Hi Jim, Let our quadratic be given by f(x) = 0. "Odd" only applies to integers, so f(x) has integer coefficients, and we assume that these coefficients are odd. Let us assume by contradiction that f(x) has two (not necessarily distinct) rational roots a/b and c/d, where a,b,c,d are integers. This means that f(x) is divisible by (x - a/b) and (x - c/d); if we multiply by bd, we find that the equation can be written as: f(x) = m*(bx - a)*(dx - c) where m is an integer. If you multiply this, you should be able to prove that at least one of the coefficients of f(x) must be even, contradicting our initial assumption. Does this help? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/
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