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MathGroup Archive 2005

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Re: problem solving polynomial equations


On 15 Oct 2005, at 11:22, wtplasar at ehu.es wrote:

>
> Hi,
>
>
> I have  two equations. The first one is
>
> eqy=(y*(3*b^2 + 12*b*y^2 + 12*y^4 + 3*b^2*z^4 + 4*b*y^2*z^4 +
> 3*r*y*Sqrt[4*y^4 + 4*b*y^2*(1 + z^4) + b^2*(1 + 3*z^4) + z^2*(8*b*y^2
> + 4*y^4 + b^2*(3 + z^4))*Sign[k]]*
>     Sign[s] + z^2*Sign[k]*(6*b^2 + 16*b*y^2 + 8*y^4 + r*y*Sqrt[4*y^4 +
> 4*b*y^2*(1 + z^4) + b^2*(1 + 3*z^4) + z^2*(8*b*y^2 + 4*y^4 + b^2*(3 +
> z^4))*Sign[k]]*
>       Sign[s])))/(3*(3*b^2 + 8*b*y^2 + 4*y^4 + 3*b^2*z^4 + 2*b*(3*b +
> 4*y^2)*z^2*Sign[k] +
>    2*r*y*Sqrt[4*y^4 + 4*b*y^2*(1 + z^4) + b^2*(1 + 3*z^4) + z^2*
> (8*b*y^2 + 4*y^4 + b^2*(3 + z^4))*Sign[k]]*Sign[s]));
>
>
> and the second one is
>
> eqz=(y*z*(4*b*y + 8*y^3 + 4*b*y*z^4 + r*Sqrt[4*y^4 + 4*b*y^2*(1 + z^4)
> + b^2*(1 + 3*z^4) + z^2*(8*b*y^2 + 4*y^4 + b^2*(3 + z^4))*Sign[k]] 
> *Sign
> [s] +
>    z^2*Sign[k]*(8*b*y + 8*y^3 + r*Sqrt[4*y^4 + 4*b*y^2*(1 + z^4) +  
> b^2*
> (1 + 3*z^4) + z^2*(8*b*y^2 + 4*y^4 + b^2*(3 + z^4))*Sign[k]]*Sign
> [s])))/
>  (3*(3*b^2 + 8*b*y^2 + 4*y^4 + 3*b^2*z^4 + 2*b*(3*b + 4*y^2)*z^2*Sign
> [k] +
>    2*r*y*Sqrt[4*y^4 + 4*b*y^2*(1 + z^4) + b^2*(1 + 3*z^4) + z^2*
> (8*b*y^2 + 4*y^4 + b^2*(3 + z^4))*Sign[k]]*Sign[s]));
>
> Now when I do
>
> Solve[{eqy == 0, eqzsubs == 0}, {y, z}] /. Sign[s]^2 -> 1
>
> I get the result fairly quickly, but if I do
>
> Solve[{Numerator[eqy]== 0, Numerator[eqz] == 0}, {y, z}]/. Sign[s]^2 -
>
>> 1
>>
>
> it seems to get stuck. I may get an answer eventually, but it seemed
> to be taking too long and aborted it.
>
> Any clues? Thanks.
>
> Ruth
>
>


What is eqzsubs?


Andrzej Kozlowski
Tokyo, Japan




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