How to solve a fourth order algebraic equation?

How to solve the value of x for the equation 2x^4+x=34.

Can we solve the fourth order algebraic equations using solve command?

Thanks

[Jaan Raa]

How to solve the value of x for the equations: a_0*x^n + a_1*x^(n-1) +...+ a_(n-1)*x + a_n = 0

Can we solve the nth degree using solve commands?

[Walter Roberson]

For degree 5 or higher, solve() will typically return a data structure the "stands in" for the roots. There are some polynomials of degree 5 or higher that solve() is able to provide exact solutions for, but most of them it cannot handle. This is not due to a limitation of solve() : it has been mathematically proven that degree 5 and higher is not certain to have solutions that are "algebraic numbers"

For degree 3 and degree 4, solve() will sometimes return exact solutions, and other times it will returns the kind of data structure mentioned above that "stands in" for the roots. In the cases where it does not return exact roots, you can pass 'MaxDegree', 3 or 'MaxDegree', 4 to solve() to get exact solutions. However exact solutions for degree 4 are long and there are very few humans that can make sense of the resulting expressions just by reading them.

For degree 2, solve() nearly always returns exact solutions, but every once in a while returns the kind of data structure mentioned above.

In most cases, the data structure "standing in" for the roots is more useful to carry around, at least until you need final answers.

Answer 1

[Star Strider]

It’s easier than that. Create a vector of its coefficients and use the roots function:

% 2x^4+x=18 -> 2x^4 + x -18 = 0
coefvct = [2  0  0  1 -18];     % Coefficient Vector
x = roots(coefvct)              % Solution

produces:

x =

      -1.7732e+000 +             i
      41.6666e-003 +  1.7326e+000i
      41.6666e-003 -  1.7326e+000i
       1.6899e+000 +             i

Two real and two complex roots.

[John D'Errico]

Actually that should be 0i on those first and last roots. As you have shown it, all roots have a non-zero imaginary part.

roots([2 0 0 1 -18])
ans =
      -1.7732 +          0i
     0.041667 +     1.7326i
     0.041667 -     1.7326i
       1.6899 +          0i

[Star Strider]

That’s what R2014b returned to me in format short eng.

[R7 DR]

Thanks for the answer. I have another question. How to solve the equation if I am having varying constant.

For example,

2x^4+x=C

C=20,22,24,26,28,30........50

Now, how to find x value at different C values.

Thanks

[Star Strider]

My pleasure.

A loop is the easiest way:

C = [20:2:50];                                  % Define ‘C’
X = nan(4, length(C));                          % Preallocate ‘X’
for k1 = 1:length(C)
    X(:,k1) = roots([2  0  0  1 -C(k1)]);        % Coefficient Matrix
end

The ‘X’ matrix is a (4x16) array with each column the coefficients corresponding the the same element in ‘C’.

[Star Strider]

R7 DR’s ‘Answer’ moved here...

Hi Thanks for the reply.

My equation got changed, now I have to solve two varying constants.

For example

ax^4+x=C

C=20,22,24,26,28,30........50

a=1,2,3...16

Could you please tell me, how to find the 'x' value at different 'C' and 'a' values.

I wrote the code like this...

C = [20:2:50]; % Define ‘C’
X = nan(4, length(C)); 
a=[1:1:length(C)]                               % Define ‘a’
for i = 1:length(C)
  X(:,i) = roots([a(i)  0  0  1 -C(i)]);        % Coefficient Matrix
end

Thanks

[Star Strider]

My pleasure.

Don’t use ‘i’ and ‘j’ as variables. MATLAB uses them for its imaginary operators, and using them as variables will cause confusion.

This is easy. You need two nested loops and a redefined preallocation:

C = [20:2:50];                                          % Define ‘C’
a = 1:16;                                               % Define ‘a’
X = nan(4, length(C), length(a));                       % Preallocate ‘X’
for k1 = 1:length(C)
    for k2 = 1:length(a)
        X(:,k1,k2) = roots([a(k2)  0  0  1 -C(k1)]);    % Coefficient Matrix
    end
end