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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/ 
Associated Topics:
High School Polynomials

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