Using "solve" on symbolic equation returns complex numbers when it shouldn't.

I have created a symbolic equation "sfun". When I try to use the "solve" function on sfun I get a complex (imaginary) number. However, when I run "fplot" on the same function you can clearly see that y = 0 around when x = 550. Why am I getting complex numbers? Thanks, Matt

    global Cp T R;
%constants
    rbar = 8.314;
    M = 28.97;
    A = 3.653;
    B = -1.337*10^-3;
    C = 3.294*10^-6;
    D = -1.913*10^-9;
    E = 0.2763*10^-12;
    Ph = 990;
    Pl = 100;
    T1 = 288.15; 
%derived constants 
    R = rbar/M;
    Rp = Ph/Pl;
%variables
    syms T;
%equations
    Cp = R*(A + B*T + C*T^2 + D*T^3 + E*T^4);
%main    
    h1 = deltah(0, T1);
    h1 = double(h1);

    sfun = deltas(288.15, T, 100, 990);
    T2s = solve(sfun == 0, T)

    fplot(sfun, [400, 600])
%functions
function h = deltah(T1, T2)
    global Cp T; 
    h = int(Cp, T, [T1 T2]);
end 
function s = deltas(T1, T2, P1, P2)
    global Cp T R;
    s = int(Cp/T, T, [T1 T2]) - R*reallog(P2/P1);
end

Answer 1

[Aquatris]

Because there is an imaginary root to that equation as well. If you were to evaluate your sfun using T2s you will see the result is pretty much zero, although not exactly due to numeric issues.

usubs(sfun,'T',T2s) 
ans = 
  1.4e-33+6.4e-35i

If your symbolic variable T is real, you can define T as;

syms T real

which gave

 T2s_Real = 550.71701819094995388330808691848

and after substituting the value to sfun gave

subs(sfun,'T',T2s_real) 
ans = 
  1.45e-33

which is higher than the imaginary number solution.

[Matthew Bartell]

Thank you very much! That makes a lot of sense.