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Re: Finding all Solutions

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From	Axel Vogt <&noreply@axelvogt.de>
Newsgroups	comp.soft-sys.math.maple
Subject	Re: Finding all Solutions
Date	Tue, 06 Jan 2015 21:59:23 +0100
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On 06.01.2015 11:26, Thomas D. Dean wrote:
> I have two equations,
>
> eq1:=y=-2*x^2 + 10*x - 6;
> eq2:=y=6*x^3 + 5*x^2 - 8;

Besides what was said/written: this is for same y,
so -2*x^2 + 10*x - 6 =  6*x^3 + 5*x^2 - 8. Bring it
to 'one side' to have a cubic = p.

Plot it in your expected range. In x=0 and x=-1 one
can evaluate and as p goes to +-oo for x --> +-oo
it is easy to see that there must be 2 real zeros.

It follows (from algebra) that the 3rd zero is real
as well. Which justifies the usage of Real methods.

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Finding all Solutions "Thomas D. Dean" <tomdean@speakeasy.org> - 2015-01-06 02:26 -0800
  Re: Finding all Solutions "Nasser M. Abbasi" <nma@12000.org> - 2015-01-06 06:22 -0600
  Re: Finding all Solutions acer <maple@rogers.com> - 2015-01-06 08:09 -0800
  Re: Finding all Solutions Axel Vogt <&noreply@axelvogt.de> - 2015-01-06 21:59 +0100

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