Can this equation be solved for x? If yes, how does the solution look? Thank you. Karl.

 Risposta accettata

Things become complicated very easy with high degree polynomials.
First, this is an example for qudratic equation:
syms a b c x
% Define the equation
eqn = a*x^2 + b*x + c == 0;
% Solve the equation symbolically
sol = solve(eqn, x);
pretty(sol)
/ 2 \ | b + sqrt(b - 4 a c) | | - -------------------- | | 2 a | | | | 2 | | b - sqrt(b - 4 a c) | | - -------------------- | \ 2 a /
and this is for 4th order:
syms a b c d x
% Define the equation
eqn = a*x^4-b*x^2-c*x+d == 0;
% Solve the equation symbolically
% The option specifies the max degree
% of polynomials for which the solver tries to find and return
% an explicit solutions
sol = solve(eqn, x,'MaxDegree',4);
pretty(sol(1)) % only first solution
/ / 3 2 \ \ | | 2 b 27 c 72 b d | | | sqrt(6) c sqrt| 3 sqrt(3) #4 - ---- + ----- + ------ | 3 | | 1/3 2 | 3 2 2 | | | b #1 #2 12 d #1 12 b #1 2/3 \ a a a / | sqrt| ------------- - ------- - ----- - 9 #1 #2 - -------------------------------------------------------- | | a a 2 a | #1 \ a / - ------- - --------------------------------------------------------------------------------------------------------------- 1/6 / 1/3 \1/4 6 #2 | 12 d 2/3 6 b #2 | 1/6 6 | ---- + #3 + 9 #2 + --------- | #2 \ a a / where / 1/3 \ | 12 d 2/3 b #2 6 | #1 == sqrt| ---- + #3 + 9 #2 + --------- | \ a a / 3 2 sqrt(3) #4 b c 4 b d #2 == ---------- - ----- + ---- + ----- 18 3 2 2 27 a 2 a 3 a 2 b #3 == -- 2 a / 4 3 4 3 2 2 2 2 \ | 27 c 256 d 16 b d 4 b c 128 b d 144 b c d | #4 == sqrt| ----- - ------ - ------- - ------- + --------- + ---------- | | 4 3 5 5 4 4 | \ a a a a a a /

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