@acer 

This is no doubt the most natural solution, and appears clearly in the help file.
Unfortunately it seems to me that the search mechanism in the help system and its organization are now worse than they used to be. I even found parameters not described in the help page but they appear in a tutorial!

Thank you for the answer. I was looking for a quick method to test/modify/debug one procedure without having do execute all of them. Not a big deal, the execution group can be split of course.

@tomleslie

@Markiyan Hirnyk 

In worksheet mode, which I almost always use, it is OK.

@H-R 

It's not enough to assume that u is real; you must work in a domain where all radicals have nonnegative arguments and unambiguous signums for their subexpressions (factors).

Otherwise, you could be right, but:

1. This kind of things happen in any CAS.
2. There are workarounds.
3. Using complex numbers has many advantages and should not scare.
4. I would not switch to Mathematica just for that. It's your choice though.

@asa12 

What composition? Do you mean multiplication? What terminology do you use? Indicate a reference. Your math seems very strange to me.

If it is multiplication, see the PolynomialIdeals package.

@asa12 

Why don't you use standard math terms?

If a and b are system of polynomials (so, probably, lists of univariate polynomials here)  what do you mean by a o b ? What kind of composition do you have in mind?

@asa12 

"the only way i think that can do" what? You did not present mathematically the problem, so I can't help you, sorry.

@asa12 

_Z is the unknown in RootOf; see ?RootOf in the help.
You can obtain radicals instead of RootOf if you use

result := solve({subs(c1 = c1sol, sys1), subs(c1 = c1sol, sys2)},{c3,c4}, explicit);

But the problem (yours, not mine) is the significance of the result and the role of diff here.

Your intentions are not clear, at least for me. You shoud explain mathematically. After that, the Maple code will be straightforward.

If it's a part of your own quartic solver, then follow the well known methods in
https://en.wikipedia.org/wiki/Quartic_function

@Doug Meade 

You have assumed in the worksheet that x,y do not depend on a.
It seems to me that this is impossible.
So, actually x = x(a), y=y(a).

A numerical investigation will be very difficult.
Note that x(a)=y(a)=1/2, F(a)=1 almost satisfy the system, because
[eq1,eq2,eq3] reduces to

[ 10^(-53)*(1+a)^3 = 0, 0=0, 0=0].

@Kitonum

An irritating fact is that

a:=0;
plots[inequal](
sqrt((x-1)^2+(y-1)^2 ) < sqrt ((x+1)^2+(y+1)^2 +a),
x=-5..5, y=-5..5, color="LightGreen");

does not work properly, but it is ok for any a<>0. Maple can be so unpredictable sometimes!

Of course kitonum's solution does not contain proofs. My point was that each step can be easily formulated with full mathematical rigor. I am sorry if you don't see this, it's not my fault.

@Markiyan Hirnyk 

kitonum's solution is rigorous. The regions are actually the connected components of the plane \ {five lines}.
In a connected set, the signum of a continuos <>0 function is constant. Please don't try to convince me about an elementary fact in mathematical analysis because I work and teach in this domain for a long time.

@Markiyan Hirnyk 

Yes, I know, I am never right :-).

But please note that kitonum's solution (which is btw very nice) is practically a "by hand" one.
The regions can be drawn by hand at once, and the inequalities in the chosen points can also be verified easily becasuse the initial inequality is equivalent to

in those points.

This type of problems are much easier to solve by hand than using Maple.
It would be possible to give a Maple solution, but only imitating the "by hand" one (which must be done by the user).