why I get imaginary part using solve function

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Sarah Alhabbas
Sarah Alhabbas le 14 Juin 2022
Commenté : Walter Roberson le 14 Juin 2022
I am trying to use the solve function but somehow I keep getting more than one answer with imaginary parts and negative numbers
the correct answer should be the second answer = 0.85
  1 commentaire
Torsten
Torsten le 14 Juin 2022
Modifié(e) : Torsten le 14 Juin 2022
If you multiply eq4 by (1+y*m4^2)^2, you get a polynomial equation of degree 4 in m4. This equation has 4 zeros (which are listed in the output of vpasolve). Two of them are purely imaginary, two of them are real. One of the solution is the one you want (the second one).

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Walter Roberson
Walter Roberson le 14 Juin 2022
You have an expression of the form f(x^4)/g(x^2) + b = 0
Multiply through by g(x^2) (assuming nonzero) to get
f(x^4) + b*g(x^2) = 0
collect x terms to get a polynomial in x^4.
Solve the degree 4 polynomial, getting four solutions.
Therefore "the answer" is all four solutions, not just a single solution.
If you have constraints on the outputs, such as being real valued, then filter the results.
  3 commentaires
Torsten
Torsten le 14 Juin 2022
y = 1.4;
to3 = 300;
t_star = 400;
syms m4
eq4 = (((2*(y+1)*m4^2*(1+(y-1)/2)*m4^2))/(1+y*m4^2)^2) - to3/t_star;
m4 = vpasolve(eq4,m4);
m4 = m4(abs(imag(m4)) < eps & real(m4) > 0)
m4 = 
0.85395841885781973634525216634996
Walter Roberson
Walter Roberson le 14 Juin 2022
y = 1.4;
to3 = 300;
t_star = 400;
syms m4 positive
eq4 = (((2*(y+1)*m4^2*(1+(y-1)/2)*m4^2))/(1+y*m4^2)^2) - to3/t_star;
m4 = solve(eq4,m4);
m4
m4 = 
vpa(m4)
ans = 
0.85395841885781973634525216634996

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Plus de réponses (1)

David Hill
David Hill le 14 Juin 2022
y=1.4;
to3=300;
t_star=400;
eq4=@(m4)(((2*(y+1)*m4^2*(1+(y-1)/2)*m4^2))/(1+y*m4^2)^2)-to3/t_star;
m_4=fzero(eq4,.8)

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