From help-request at octave dot org Sun Feb  6 19:57:32 2005
Subject: Re: two quadratic eq, please help
From: Vic Norton <vic at norton dot name>
To: help at octave dot org
Date: Sun, 6 Feb 2005 21:02:34 -0500

So we have proved that a quadratic equation can have two solutions. I 
will really be impressed if octave can find more than two! ;-)

My question is this. Why should one bother using a high powered 
numerical language for an elementary exercise?

This is a philosophical point. Personally I believe it is important 
to recognize trivialities, no matter the power of the machinery you 
have at hand. Two circles can intersect in two points, one point, or 
they might not intersect at all. That is all we are talking about 
here. Anybody with a hand calculator should be able to figure out 
where these two circles intersect (if they intersect at all). To hell 
with high level numerical languages.

That is my two bits worth.

Regards,

Vic

At 5:15 PM -0700 2/6/05, Dmitri A. Sergatskov wrote:
>Geordie McBain wrote:
>
>>
>>Create a function to return the residual:
>>
>>   function z = f (x)
>>     z = [(12-x(1))^2+x(2)^2-15^2, (16+x(2))^2+x(1)^2-25^2];
>>   endfunction
>>
>>and fsolve it:
>>
>>octave> fsolve ("f", [0, 0])
>>ans =
>>
>>    1.8887e-15
>>    9.0000e+00
>>
>
>and:
>
>octave:5> fsolve ("f", [24, -9])
>ans =
>
>    24
>    -9
>
>:)
>
>and so forth :)
>
>>Geordie McBain
>>www.aeromech.usyd.edu.au/~mcbain
>>
>
>Regards,
>
>Dmitri.
>--



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