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Topic: easiest way to solve rational polynomial inequality p(x)/q(x) > 0 is
by solving p(x)q(x) > 0

Replies: 12   Last Post: Aug 23, 2013 4:22 PM

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vrut25@gmail.com

Posts: 5
Registered: 12/1/12
Re: Solving rational polynomials
Posted: Aug 23, 2013 12:39 AM
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On Wednesday, August 21, 2013 3:43:24 AM UTC-7, Kobu wrote:
> On Saturday, August 17, 2013 4:26:18 AM UTC-4, Virgil wrote:
>

> > In article <0388f571-326f-437f-990f-b3000253e952@googlegroups.com>,
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> >
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> > Kobu <kobu.selva@gmail.com> wrote:
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> > > Thm: Let p and q be functions. Then,
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> > > p(x)/q(x) > 0 iff p(x)*q(x) > 0
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> > Not quite, since when q(x) = 0,
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> > p(x)/q(x) is not defined whereas
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> > whereas p(x)*q(x) is defined and zero, so
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> > "p(x)/q(x) > 0" is indeterminate but
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> > "p(x)*q(x) > 0" is false


Look up the definition of the "iff' statement.
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> Ok, but if someone has an equation like p(x)/q(x) > 0, then all the solutions of p(x)*q(x) > 0 will be solutions of the original one.
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> So this is a good way to solve things where numerator and denominator are polynomials.


Or any other arbitrary functions



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