The equation of the fifth degree. A private solution.

In this material, one of the solutions of the equation of the fifth degree of a particular form is considered. To date, the Abel-Ruffinni theorem states that there is no general solution to the fifth degree equation that can be expressed by a finite number of calculations using arithmetic operations, exponentiation and root extraction. 

At least two assumptions can be made from this theorem:

- there are particular types of an equation of degree 5, the roots of which can be found by substituting into a certain (inherently finite) formula

- if it is impossible to find a general solution where there is a "finite number of operations", then theoretically it is possible to search for a general solution using functions where there is an "infinite number of operations"

I will offer you an online solution of the fifth degree equation of this kind

\(x^5+ax^3+a^2/5x+b\)

To do this, we need to calculate two auxiliary parameters \(F\) and \(T\)

\(\cfrac{-i*c}{2b}\sqrt{\cfrac{125}{a}}=F\)

rather like this

\(F=\cfrac{ib}{2}\sqrt{(\cfrac{5}{a})^5}\)

\(\sqrt{\cfrac{4a}{5}}=T\)

After that, we will be able to find all the roots of such an equation.

It also works when the source data are complex numbers.

Recall that the discriminant of such a polynomial is

\(\cfrac{16 a^{10} + 25000 a^5 b^2 + 9765625 b^4}{3125}\)

or if we simplify

\(\cfrac{(4a^5-3125b^2)^2}{3125}\)

Some examples

\(x^5+ix^3-\cfrac{x}{5}+i=0\)

But it was not without a fly in the ointment. There are errors in the calculations, or rather in the choice of the root sign. Here is one example. 

\(x^5+(tan(2+i))x^3+\cfrac{(tan(2+i))^2}{5}x+(i)=0\)

On this equation, despite the fact that all values coincide, the sign must be changed to the opposite. Why so and what criterion, I have not yet understood.