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```Date: 06/24/2003 at 05:09:38
From: Jim

Show that quadratics with odd coefficients have no rational roots.
```

```
Date: 06/24/2003 at 05:35:39
From: Doctor Jacques

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.

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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