solve function depend on z

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Cahit Semih
Cahit Semih 2025-11-2，9:08
回答： Walter Roberson 2025-11-2，9:32
Ran in:
clear all;
clc;
syms s m b T B K d;
%d=e^(-s(T/2))
A=m*T*s^3 + b*T*s^2+K+d*(B*s)+d^2*(-K)+d^3*(-B*s);
A_tilde=-m*T*s^3*d^3 + b*T*s^2*d^3+K*d^3-d^2*(B*s)+d*(-K)-(-B*s);
res=resultant(A,A_tilde,d);
res_s=solve(res==0,s)
res_s = 
I want the roots of the polynomial s obtained using the solve function to be in the form only m b B T K, but they are coming out as z-dependent.

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采纳的回答

Walter Roberson
Walter Roberson 2025-11-2，9:32
You can divide res by s^4, and you can then substitute s = sqrt(S), and then collect on S. You will obtain a degree 7 polynomial in S.
But then you are stuck. There are no general solutions to polynomials of degree 5 or higher, and you did not happen to get lucky to be able to factorize into polynomials of degree 4 or less. You will not be able to find solutions in terms of m b B T K

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