Can Graphing Reveal All Solutions to Complex Polynomial Systems?

[lape99]

I don't know if I'm posting in the wright section, but i need to solve this system of equations:

{5x^4 - 4x + 5y = 0
5y^4 + 5x - 4y = 0}

I know one answer is (0;0), but i don't know how to show it, also there is another point.
Maybe someone can help me?

 

[arildno]

If you subtract the equations from each other, you get the following:

5x^4-5y^4-9(x-y)=0

Now, you have:
x^4-y^4=(x-y)(x^3+x^2y+xy^2+y^3), and thus, you may factorize the above equation as:
(x-y)*(5(x^3+x^2y+xy^2+y^3)-9)=0

 

[lape99]

Thank you very much :)

 

[Ray Vickson]

I don't know if I'm posting in the wright section, but i need to solve this system of equations:

{5x^4 - 4x + 5y = 0
5y^4 + 5x - 4y = 0}

I know one answer is (0;0), but i don't know how to show it, also there is another point.
Maybe someone can help me?

One can use Groebner basis methods to get at all the solutions. Using the Groebner package in Maple 11, I get:
sys:=[5*x^4-4*x+5*y,5*y^4+5*x-4*y]:lprint(sys); <== input
[5*x^4-4*x+5*y, 5*y^4+5*x-4*y]
with(Groebner): <=== load the Groebner package
B:=Basis(sys,plex(x,y)):lprint(B);
[225*y+756*y^4-1280*y^7+2400*y^10-2000*y^13+625*y^16, 5*y^4+5*x-4*y]

This B is the output. The two original equations are equivalent to setting the two components of B to zero (although, of course, there may be some extraneous roots, so those must be checked in the original system). Note that the first component of B is a polynomial in y alone, so we can find roots of the two-equation system by first solving the polynomial in y, then solving for x by setting the second expression in B to zero.

f1y:=B[1]:lprint(f1y);
225*y+756*y^4-1280*y^7+2400*y^10-2000*y^13+625*y^16 <--- set to zero

Let's try to factor the polynomial.

F:=factor(f1y):lprint(F);
y*(5*y^3+1)*(125*y^12-425*y^9+565*y^6-369*y^3+225) <=== the factors

So, y = 0 or 5*y^3 + 1 = 0, or
125*y^12 -425*y^9 + 565*y^6 -360*y^3 + 225 = 0, which is a 4th degree polynomial in z = y^3.

After solving for y, we get x from 5*x = 4*y - 5*y^4.

Note: doing anything like this by hand would take months or years of work and require hundreds of pages of algebraic work. Use of a computer algebra system is essential.

RGV

 

[CellCoree]

I would approach this problem by using mathematical methods to solve the system of equations. One approach could be to use substitution, where one variable is isolated in one equation and then substituted into the other equation. This would involve manipulating the equations algebraically to eliminate one variable and solve for the other. Another approach could be to use elimination, where the two equations are combined in a way that eliminates one variable when they are subtracted or added together. This would also involve algebraic manipulation to simplify the equations and solve for the remaining variable.

In order to show that the point (0,0) is a solution to this system of equations, we can substitute x=0 and y=0 into both equations and show that they are equal to 0. This would demonstrate that (0,0) satisfies both equations and is therefore a solution.

Additionally, there may be other methods such as graphing or using matrices to solve this system of equations. It is important to choose the method that is most appropriate and efficient for the specific problem at hand. I would recommend consulting with a mathematics expert or using a reliable software program to solve this system of equations accurately and efficiently.