Newsgroups: rec.puzzles
From: hoey@aic.nrl.navy.mil (Dan Hoey)
Date: 1995/07/20
Subject: Re: tough math puzzle ! ?

sch...@isi.ee.ethz.ch (Hanspeter Schmid) writes:

> Take the following equation:

>    f1*f2 + f2*f3 + f3*f1 = 0        (1)

> Each of the fi (i=1..3) is of the form

>    fi = ai + x*bi   (i=1..3)

> Then (1) is a quadratic equation in x. Can you show that (1) always
> has real solutions if all coefficients ai and bi are >=0 ?

No, there are no solutions if all bi=0 and at least two ai>0.

Even if we require all ai,bi > 0, equation (1) may have only one
(real) solution.  For instance, if f1=4+x, f2=8+2x, f3=12+3x, then the
equation is 11(4+x)^2=0, whose only zero is x=-4.

But as others have shown, if the zeroes of the fi are distinct, then
there are two distinct real zeroes of (1).

Dan Hoey
Hoey@AIC.NRL.Navy.Mil