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

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Karl on 30 May 2023
Edited: Askic V on 30 May 2023
a*x^4-b*x^2-c*x+d=0

Askic V on 30 May 2023
Edited: Askic V on 30 May 2023
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 /