How can I find roots of cubic function?

I have a 3 x 3 matrix and with position (1,1) "x" as an unknown variable, after I solved for the invariants my cubic functions becomes y^3 - (x+50)y^2 +(500x -5200)y + (3600x) = 0. How can I solve for the roots in terms of x on matlab?

Answer 1

[James Tursa]

Do you mean like this?

syms x y
solve('y^3 - (x+50)*y^2 +(500*x -5200)*y + (3600*x) = 0',y)

[Walter Roberson]

Note that this will fail in R2018a, which no longer permits strings for equations to be solved. But the error message it gives will be misleading: "List of equations must not be empty."

Replacement is

solve(y^3 - (x+50)*y^2 +(500*x -5200)*y + (3600*x),y)

Answer 2

[Star Strider]

One approach:

syms x y 
Eqn = y^3 - (x+50)*y^2 +(500*x -5200)*y + (3600*x) == 0;
X = solve(Eqn, x)
X_fcn = matlabFunction(X)
Y = vpasolve(X,y)

X =

(- y^3 + 50*y^2 + 5200*y)/(- y^2 + 500*y + 3600)

X_fcn =

    function_handle with value:

      @(y)(y.*5.2e3+y.^2.*5.0e1-y.^3)./(y.*5.0e2-y.^2+3.6e3)

Y =

   -51.321687612368740126279743503002
                                    0
      101.321687612368740126279743503

EDIT —

Solving in terms of ‘x’:

S = vpasolve(Eqn,y)

yields:

S =
root(z^3 - z^2*(x + 50) + z*(500*x - 5200) + 3600*x, z, 1)

so the solutions are the roots of the ‘z’ polynomial.

Answer 3

[Lawrence Besong]

Is it possible to find the roots in terms of "x"?

Answer 4

[Walter Roberson]

solve(y^3 - (x+50)*y^2 +(500*x -5200)*y + (3600*x),y)

or

isolate(y^3 - (x+50)*y^2 +(500*x -5200)*y + (3600*x)==0, y)

if you use solve() then you will get

   root(z^3 - z^2*(x + 50) + z*(500*x - 5200) + 3600*x, z, 1)
   root(z^3 - z^2*(x + 50) + z*(500*x - 5200) + 3600*x, z, 2)
   root(z^3 - z^2*(x + 50) + z*(500*x - 5200) + 3600*x, z, 3)

which is the set of three different specific roots of the cubic; isolate() will say

y == root(z^3 - z^2*(x + 50) + z*(500*x - 5200) + 3600*x, z, 1)

which is only in terms of the primary root. This is documented: "For equations with multiple solutions, isolate returns the simplest solution."