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I have a set of 3 quadratic equtions:

$A_1µ_1^2 + B_1µ_1µ_2 + C_1µ_2^2 + D_1µ_1 + E_1µ_2 + F_1 = 0$ $A_2µ_2^2 + B_2µ_2µ_3 + C_2µ_3^2 + D_2µ_2 + E_2µ_3 + F_2 = 0$ $A_3µ_3^2 + B_3µ_3µ_1 + C_3µ_1^2 + D_3µ_3 + E_3µ_1 + F_3 = 0$

I want to write a C-code, that find solutions for $µ_1$, $µ_2$ and $µ_3$ with input $A_1$, $A_2$, ...

Any ideas on where to start? Is there a good method for solving this problem, or exist already a good free code?

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Are all the parameters and solutions (of interest) real, complex, integers? Floating point? Sounds like homework to me. –  Kevin O'Bryant Aug 6 '12 at 17:07
    
They are all real (parameters and solution). And it isn't homework, it's a little problem at work. –  Jakube Aug 7 '12 at 7:34

1 Answer 1

up vote 6 down vote accepted

The magic word is "elimination". The obvious thing to do is to solve the first equation for $\mu_2$ in terms of $\mu_1.$ You will have two roots. Plug in each one into the second equation, expressing $\mu_3$ in terms of $\mu_1,$ then you will have eight possible equations satisfied by $\mu_1.$ The equation will be somewhat scary beasts with nested square roots. This is why mankind (actually, J.J. Sylvester) had invented the "resultant" (magic word number two), which, when used twice, will give you an equation of degree eight satisfied by $\mu_1.$ You have to be able to solve this equation (probably GSL has code for this), and then work backwards. But first, google for "elimination" and "resultant".

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Thanks, that "resultant" aproach is very useful. I never heard about it before. –  Jakube Aug 7 '12 at 9:04
    
Igor is right. Also, this is the kind of thing that a Gröbner basis is supposed to be good for, but in practice computing the damn thing is tough. (Mathematica can't do it for this problem, but maybe I should buy a more expensive computer.) –  TerronaBell Aug 7 '12 at 13:59

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