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i am new to mathcad and want to know if mathcad can solf a quadratic system like this And know how many answers it gives.
2x1^2-x1+x2^2-3x2+4x3^2+5x3=53
x1^2+3x1-2x2^2+x2-2x3^2+x3=-17
-x1^2-3x1+2x2^2-4x2+x3^2-x3=2
THanks
Try this but I think you must use a numerical not symbolic solution use (Find function)
You can get two sets of real number solutions as follows (I've used M15):
Alan
On contrary to the example Valery posted in your system you have summands linear in x1, x2, x3, too.
This would mean that Mathcad has to solve a generic equation of eight order for which no general solution exists.
Therefore I guess the symbolic solver will fail and you have to resort to a numeric solve block which gives you just one solution triple depending on the guess values you have to provide. Different guesses may give you different solutions.
WE
EDIT: Looks like I was a bit late as Alan already solved the problem in an elegant way.
For completeness sake here are all eight solutions (just found by trying different guesses 😉
Hi Rob. If you like programming, you can solve your system in this way:
The system has eight solutions (see above) not four.
Given that all coefficients are real, with any complex solution the conjugate is solution, too.
So you just missed the first set of complex solutions given in my answer above.
But probably Rob is interested in the two real solutions only anyway.
WE
The method I adopted can be extended to find all 8 roots by making use of the function polyroots to find the roots of f(_z1).
ALan
The method I adopted can be extended to find all 8 roots by making use of the function polyroots to find the roots of f(_z1).
Combining symbolic solver and numeric methods that way sure is a clean, clever , sound way of solving that task. On of those "Why didn't I thought of that" things. I guess 😉
Werner
I leave to the reader the task to search other solutions.
I would be pleased if you were also interested in another similar example in which the search of the roots is more difficult:
thank you