@shakuntala 

filename contains the name of the temporary file. It must be in any directory with write permission.
Unfortunately I am not sure whether exportplot and plot/background is available in Maple 15. If not, you will have to see the links to older posts Acer has provided above. 

You use strangely the "dimension".
The 1st solve gives a 1-dimensional (linear) subspace in R^3  (or C^3 if a,b,c are considered in C) i.e. a straight line thru 0.

The 2nd solve gives the union of two such 1-dimensional (linear) subspaces; the topological dimension of the union is still 1.

@Preben Alsholm 

So, evalf prevents the evaluation of the first argument, just like
evalf := proc(x::uneval) ... end proc;

to generate polynomials with real roots would be to construct orthogonal polynomials wrt a (random) measure.
This way you can prescribe an interval containing the roots.

@Preben Alsholm 

It is very nice that Maple was able to compute the limit, but I am not at all sure that limits such as

limit( RootOf( f(x,y,_Z) ), y=1 );

are to be trusted mathematically, without a serious analysis.

@Carl Love 

But it's OK. I truly like your style.

@Carl Love 

You are (as always) a perfectionist :-)
 

@Preben Alsholm 

The typsetting mechanism is supposed to be used for display purposes. Now it seems to be very intrusive and alters also the results!

@vv 

I must add that  evalf[10](frac(Pi^20)) = 23  is not technically a bug. It's, let's say, a "psychological bug" because nobody expects frac(...) be >= 1. This happens because `evalf/frac` is not implemented and the number of Digits is only 10; note that frac(Pi^20) returns (correctly) Pi^20-8769956796.
Maple's evalf is reliable if used correctly!

@Christopher2222 

Just set the rankings  to -infinity (or better, some very large <0)  rather than 0.
This should probably solve the problem; however, when two such countries with negative ratings play each other, one of them must win.
 

@Markiyan Hirnyk 

It's easy to elucidate your doubts: open the implicitplot help page in a worksheet window, insert trace(fsolve)  and execute all.

Near the curve the sqrt's are almost 0 and may take imaginary values. So, the sign changes are absent and the implicitplot algorithm does not work well in this case. That's why Acer's direct solution is probably the best!

(Of course, the algorithm could implement a sign check for Re and Im).

@acer 

Strangely,

plots:-implicitplot(Re(ee)=0, b=0..5, p=0..5,gridrefine=1, crossingrefine=7);
also works, but
plots:-implicitplot(abs(ee)=0, b=0..5, p=0..5,gridrefine=1, crossingrefine=7);
fails.

(Actually for abs it's not that strange because the needed sign changes are absent).

@Markiyan Hirnyk 

The parametrization covers the whole curve. This follows from the solve result wrt {b,p} .

@Markiyan Hirnyk 

It's simply a substitution; other substitutions are also possible.
What for? We want a parametrization, isn't it?