Roots of a polynomial with variables

For some problems, we have to to study some notions of stablility and zero polynomials in two variables, my que'stion how we can find the roots or zero polynomials in two variables. for example:

P(x,y)=3*xy -5y^2+7*x^2y

or a nother polynom

Answer 1

[Sulaymon Eshkabilov]

One of the viable ways to solve such polynomial type equations is to setp up the solution space within which you are seeking the roots to compute and solve them using fzero(). E.g.:

x=linspace(-2,2):  % Choose the necessary solution space
for t=1:100
    EQ= @(y)(3*x(t)*y-5*y.^2+7*(x(t)^2)*y); 
    y_roots = fzero(EQ,0);
end

[Lewis Fer]

but this method in reality doesn't making a difference to search and find th pole of some matrices like transfer function in two dimensional case

Answer 2

[Paul]

Don't know the scope of the actual problems of interest, but for the two examples in the question:

syms x y
sol = solve(3*x*y - 5*y^2 + 7*x^2*y == 0,[x y],'ReturnConditions',true);
[sol.x sol.y sol.conditions]

ans = 

syms z1 z2
sol = solve(1 - z1*z2 - 1/2*z1^2 - 1/2*z2^2 + z1^2*z2^2 == 0,[z1 z2],'ReturnConditions',true);
[sol.z1 sol.z2 sol.conditions]

ans = 

[Lewis Fer]

my real problem is the Generalized Eigenvalues problem in two dimensional system; for example

for a given 2*2 matrix A, we compute some matrix : K=[p 0; 0 s]-A and the eigenvalue of the matrice K is the pair (p,s), and to compute this eigenvalues we have to compute the determinant of the matrix K and when we do we obtain a polynome with two variable.

[Paul]

I'm not sure what the bold part of the comment is intended to convey. The question gave two examples for the equations of interest, and the answer showed how to solve them. So I'm not sure what else you're looking for. If it's the first part about taking the determinant, then maybe something like this?

A = [1 2;3 4]; % example
syms p s
eq = det(diag([p s]) - A) == 0

eq = 

sol = solve(eq,[p s],'ReturnConditions',true)

sol = struct with fields:
             p: [1×1 sym]
             s: [1×1 sym]
    parameters: [1×1 sym]
    conditions: [1×1 sym]

[sol.p sol.s sol.conditions]

ans = 

[Lewis Fer]

Yes this is what I want to do exactly, but tje solution which is represented by the pair ( p,s) ???

[Paul]

Apparently there are many solutions to this problem, i.e., many pairs (p,s) that make the determinant equal to zero. The pair (p,s) can be expressed as ( (z+2)/(z-4) , z) for z any number not equal to four. Check

A = [1 2;3 4];
p = @(z)((z+2)./(z-4));
s = @(z)(z);
z = 1;
det(diag([p(z) s(z)]) - A)

ans = -3.3307e-16

z = 8;
det(diag([p(z) s(z)]) - A)

ans = 0

z = 1 + 1i;
det(diag([p(z) s(z)]) - A)

ans = -1.3323e-15 + 3.3307e-16i

[Lewis Fer]

thank', Paul for your answers your idea help me for my work.