Method of Sotta

The method of Sotta , imagined and developed by Bernard Sotta, makes it possible to solve all the cubic equations and can spread with certain equations of degree equal to or higher than 4 if the coefficients of these equations check certain conditions.

These equations provide examples of equations which, although having a degree equal to or higher than 5, have a Groupe of resolvable Welshman. We know indeed that the equations of degree equal to or higher than 5 inevitably do not have a group of resolvable Welshman. What makes it possible to affirm that there does not exist general method to solve them. (see Theory of Welshman).

Principle of the method

In all this article N is an integer representing the degree of the equation to be solved.

All the other letters represent complex numbers.

By convention $\backslash \; sqrt\; \{has\}$ indicates any of N n^ème roots of has, it is the same of $\backslash \; sqrt\; \{F\}$

Let us consider an equation of degree N with N > 2:

$\backslash \; qquad\; a\_n\; x^n\; +\; a\_\; \{N\; -\; 1\}\; x^\; \{N\; -\; 1\}\; +\; \backslash \; cdots\; +\; a\_1\; X\; +\; a\_0\; =\; 0$

We will call resolvent equation of Sotta associated with the preceding equation, the quadratic equation following:

$\backslash \; qquad\; (n-1)\; (N2)\; X^2\; +\; 2\; (n-1)\; X\; +\; 6\; (n-1)\; a\_\; \{n-1\}\; a\_\; \{n-3\}\; -\; 4\; (N2)\; a\_\; \{N2\}\; ^2=\; 0$

We have the following theorem then (theorem of Sotta):

If the equation:

$\backslash \; qquad\; a\_n\; x^n\; +\; a\_\; \{N\; -\; 1\}\; x^\; \{N\; -\; 1\}\; +\; \backslash \; cdots\; +\; a\_1\; X\; +\; a\_0\; =\; 0$

admits roots in the form:

$\backslash \; qquad\; \backslash \; frac\; \{B\; \backslash \; sqrt\; \{has\}\; +\; C\; \backslash \; sqrt\; \{F\}\}\; \{D\; \backslash \; sqrt\; \{has\}\; +\; E\; \backslash \; sqrt\; \{F\}\}$

(D and E not no one).

then $\backslash \; frac\; \{B\}\; \{D\}$ and $\backslash \; frac\; \{C\}\; \{E\}$ is the two roots of the resolvent equation.

has and F are then given by the two relations:

• $\backslash \; qquad\; F\; =\; (-\; 1)\; ^n\; \backslash \; frac\; \{da\_\; \{n-1\}\; +nba\_n\}\; \{ne^\; \{n-1\}\; (cd.-Be)\}$

• $\backslash \; qquad\; has\; =\; \backslash \; frac\; \{a\_n+\; (-\; 1)\; ^nfe^n\}\; \{d^n\}$

N roots of the equation suggested will be then:

$\backslash \; qquad\; x\_k\; =\; \backslash \; frac\; \{be^\; \{\backslash \; frac\; \{2ki\; \backslash \; pi\}\; \{N\}\}\; \backslash \; sqrt\; \{has\}\; +\; C\; \backslash \; sqrt\; \{F\}\}\; \{de^\; \{\backslash \; frac\; \{2ki\; \backslash \; pi\}\; \{N\}\}\; \backslash \; sqrt\; \{has\}\; +\; E\; \backslash \; sqrt\; \{F\}\}$ with K successively taking all the whole values of 0 with n-1

Application to the resolution of the equations of degree 3

All the equations of degree 3 admit roots in the form:

$\backslash \; frac\; \{B\; \backslash \; sqrt\; \{has\}\; +\; C\; \backslash \; sqrt\; \{F\}\}\; \{D\; \backslash \; sqrt\; \{has\}\; +\; E\; \backslash \; sqrt\; \{F\}\}\; ~$

consequently, the method of Sotta makes it possible to solve all the equations of degree 3.

That is to say thus the following equation:

$a\_3\; x^3\; +\; a\_2\; x^2\; +\; a\_1\; X\; +\; a\_0\; =\; 0\; ~$

First case: If $(3a\_3a\_1-a\_2^2)\; \backslash \; not\; =\; 0$ and $(3a\_0a\_2-a\_1^2)\; \backslash \; not\; =\; 0$ (condition so that the resolvent one is of the second degree with nonnull roots).

Resolvent of the Sotta associated one will be:

$(3a\_3a\_1-a\_2^2)\; X^2\; +\; (9a\_3a\_0-a\_2a\_1)\; X\; +\; 3a\_2a\_0\; -\; a\_1^2=\; 0\; ~$

It is thus enough to choose B, C, D, E such as $\backslash \; frac\; \{B\}\; \{D\}$ and $\backslash \; frac\; \{C\}\; \{E\}$ is the roots of the resolvent one. One calculates then has and F using the formulas:

• $F\; =\; \backslash \; frac\; \{da\_2+3ba\_3\}\; \{3e^2\; (Be-cd.)\}\; ~$

• $has\; =\; \backslash \; frac\; \{a\_3-fe^3\}\; \{d^3\}\; ~$

The three roots of the equation to be solved will be then:

$\backslash \; qquad\; x\_1\; =\; \backslash \; frac\; \{B\; \backslash \; sqrt\; \{has\}\; +\; C\; \backslash \; sqrt\; \{F\}\}\; \{D\; \backslash \; sqrt\; \{has\}\; +\; E\; \backslash \; sqrt\; \{F\}\}$

$\backslash \; qquad\; x\_2\; =\; \backslash \; frac\; \{bj\; \backslash \; sqrt\; \{has\}\; +\; C\; \backslash \; sqrt\; \{F\}\}\; \{dj\; \backslash \; sqrt\; \{has\}\; +\; E\; \backslash \; sqrt\; \{F\}\}$

$\backslash \; qquad\; x\_3\; =\; \backslash \; frac\; \{bj^2\; \backslash \; sqrt\; \{has\}\; +\; C\; \backslash \; sqrt\; \{F\}\}\; \{dj^2\; \backslash \; sqrt\; \{has\}\; +\; E\; \backslash \; sqrt\; \{F\}\}$

Second case: If $(3a\_3a\_1-a\_2^2)\; =\; 0$ (One is in the case: D or E null)

One multiplies by 3a₁a₂ all the terms of the equation:

$a\_3\; x^3\; +\; a\_2\; x^2\; +\; a\_1\; X\; +\; a\_0\; =\; 0\; ~$

One obtains:

$3a\_1a\_2a\_3x^3\; +\; 3a\_1a\_2^2\; x^2\; +\; 3a\_1^2a\_2\; X\; +\; 3a\_1a\_2a\_0\; =\; 0\; ~$

Like:

$3a\_3a\_1\; =\; a\_2^2\; ~$

The equation becomes:

$a\_2^3\; x^3\; +\; 3a\_1a\_2^2\; x^2\; +\; 3a\_1^2a\_2\; X\; +\; 3a\_0a\_1a\_2\; =\; 0\; ~$

Who puts himself in the form:

$\backslash \; left\; (a\_2x\; +\; a\_1\; \backslash \; right)\; ^3\; =\; a\_1^3\; -\; 3a\_0a\_1a\_2\; ~$

One from of deduced the three roots of the equation to be solved:

$x\_1\; =\; \backslash \; frac\; \{1\}\; \{a\_2\}\; \backslash \; left\; (\backslash \; sqrt\; \{a\_1^3\; -\; 3a\_0a\_1a\_2\}\; -\; a\_1\; \backslash \; right)\; ~$

$x\_2\; =\; \backslash \; frac\; \{1\}\; \{a\_2\}\; \backslash \; left\; (J\; \backslash \; sqrt\; \{a\_1^3\; -\; 3a\_0a\_1a\_2\}\; -\; a\_1\; \backslash \; right)\; ~$

$x\_3\; =\; \backslash \; frac\; \{1\}\; \{a\_2\}\; \backslash \; left\; (j^2\; \backslash \; sqrt\; \{a\_1^3\; -\; 3a\_0a\_1a\_2\}\; -\; a\_1\; \backslash \; right)\; ~$

Third case: If $(3a\_0a\_2-a\_1^2)\; =\; 0$ (One is in the case: B or C null)

One multiplies by 3a₁a₂ all the terms of the equation:

$a\_3\; x^3\; +\; a\_2\; x^2\; +\; a\_1\; X\; +\; a\_0\; =\; 0\; ~$

One obtains:

$3a\_1a\_2a\_3x^3\; +\; 3a\_1a\_2^2\; x^2\; +\; 3a\_1^2a\_2\; X\; +\; 3a\_0a\_1a\_2\; =\; 0\; ~$

Like:

$3a\_0a\_2\; =\; a\_1^2\; ~$

The equation becomes:

$3a\_1a\_2a\_3x^3\; +\; 3a\_1a\_2^2\; x^2\; +\; 3a\_1^2a\_2\; X\; +\; a\_1^3\; =\; 0\; ~$

Now let us divide each term by x³, one obtains:

$3a\_1a\_2a\_3\; +\; 3\; \backslash \; left\; (\backslash \; frac\; \{a\_1\}\; \{X\}\; \backslash \; right)\; a\_2^2\; +\; 3\; \backslash \; left\; (\backslash \; frac\; \{a\_1\}\; \{X\}\; \backslash \; right)\; ^2a\_2\; +\; \backslash \; left\; (\backslash \; frac\; \{a\_1\}\; \{X\}\; \backslash \; right)\; ^3\; =\; 0\; ~$

Who puts himself in the form:

$\backslash \; left\; (\backslash \; frac\; \{a\_1\}\; \{X\}\; +\; a\_2\; \backslash \; right)\; ^3\; =\; a\_2^3\; -\; 3a\_1a\_2a\_3\; ~$

One from of deduced the three roots of the equation to be solved:

$x\_1\; =\; \backslash \; frac\; \{a\_1\}\; \{\backslash \; sqrt\; \{a\_2^3\; -\; 3a\_1a\_2a\_3\}\; -\; a\_2\}\; ~$

$x\_2\; =\; \backslash \; frac\; \{a\_1\}\; \{J\; \backslash \; sqrt\; \{a\_2^3\; -\; 3a\_1a\_2a\_3\}\; -\; a\_2\}\; ~$

$x\_3\; =\; \backslash \; frac\; \{a\_1\}\; \{j^2\; \backslash \; sqrt\; \{a\_2^3\; -\; 3a\_1a\_2a\_3\}\; -\; a\_2\}\; ~$

Application to the resolution of the equations of degree 4

Equations of degree 4:

$\backslash \; qquad\; a\_4\; x^4\; +\; a\_3\; x^3\; +\; a\_2\; x^2\; +\; a\_1\; X\; +\; a\_0\; =\; 0$

roots in the form admit:

$\backslash \; qquad\; \backslash \; frac\; \{B\; \backslash \; sqrt\; \{has\}\; +\; C\; \backslash \; sqrt\; \{F\}\}\; \{D\; \backslash \; sqrt\; \{has\}\; +\; E\; \backslash \; sqrt\; \{F\}\}$

only if:

• $\backslash qquad\; 27a\_4a\_1^2-72a\_4a\_2a\_0+2a\_2^3-9a\_3a\_2a\_1+27a\_3^2a\_0=0$

Consequently, the method of Sotta makes it possible to solve only the equations of degree 4 checking this condition of resolvability.

That is to say thus the following equation:

$\backslash \; qquad\; a\_4\; x^4\; +\; a\_3\; x^3\; +\; a\_2\; x^2\; +\; a\_1\; X\; +\; a\_0\; =\; 0$

First case: If $(8a\_4a\_2-3a\_3^2)\; \backslash \; not\; =\; 0$ (condition so that the resolvent one is of the second degree).

Resolvent of the Sotta associated one will be:

$\backslash \; qquad\; (24a\_4a\_2-9a\_3^2)\; X^2\; +\; (36a\_4a\_1-6a\_3a\_2)\; X\; +\; 9a\_3a\_1\; -\; 4a\_2^2=\; 0$

It is thus enough to choose B, C, D, E such as $\backslash \; frac\; \{B\}\; \{D\}$ and $\backslash \; frac\; \{C\}\; \{E\}$ is the roots of the resolvent one. One calculates then has and F using the formulas:

• $\backslash \; qquad\; F\; =\; \backslash \; frac\; \{da\_3+4ba\_4\}\; \{4e^3\; (cd.-Be)\}$

• $\backslash \; qquad\; has\; =\; \backslash \; frac\; \{a\_4-fe^4\}\; \{d^4\}$

The four roots of the equation to be solved will be:

$\backslash \; qquad\; x\_k\; =\; \backslash \; frac\; \{be^\; \{\backslash \; frac\; \{2ki\; \backslash \; pi\}\; \{4\}\}\; \backslash \; sqrt\; \{has\}\; +\; C\; \backslash \; sqrt\; \{F\}\}\; \{de^\; \{\backslash \; frac\; \{2ki\; \backslash \; pi\}\; \{4\}\}\; \backslash \; sqrt\; \{has\}\; +\; E\; \backslash \; sqrt\; \{F\}\}$ with K taking values 0,1,2,3 successively.

Second case: If $(8a\_4a\_2-3a\_3^2)\; =\; 0$ (One is in the case: D or E null).

See the paragraph Complements at the end of the article.

Application to the resolution of the equations of degree 5

Equations of degree 5:

$\backslash \; qquad\; a\_5\; x^5\; +\; a\_4\; x^4\; +\; a\_3\; x^3\; +\; a\_2\; x^2\; +\; a\_1\; X\; +\; a\_0\; =\; 0$

roots in the form admit:

$\backslash \; qquad\; \backslash \; frac\; \{B\; \backslash \; sqrt\; \{has\}\; +\; C\; \backslash \; sqrt\; \{F\}\}\; \{D\; \backslash \; sqrt\; \{has\}\; +\; E\; \backslash \; sqrt\; \{F\}\}$

only if:

• $\backslash qquad\; 10a\_5a\_2^2-20a\_5a\_3a\_1+a\_3^3-4a\_4a\_3a\_2+8a\_4^2a\_1=0$
• $\backslash qquad\; 8a\_4a\_1^2-20a\_4a\_2a\_0+a\_2^3-4a\_3a\_2a\_1+10a\_3^2a\_0=0$

Consequently, the method of Sotta makes it possible to solve only the equations of degree 5 checking these conditions of resolvability.

That is to say thus the following equation:

$\backslash \; qquad\; a\_5\; x^5\; +\; a\_4\; x^4\; +\; a\_3\; x^3\; +\; a\_2\; x^2\; +\; a\_1\; X\; +\; a\_0\; =\; 0,$

First case: If $(5a\_5a\_3-2a\_4^2)\; \backslash \; not\; =\; 0$ (condition so that the resolvent one is of the second degree).

Resolvent of the Sotta associated one will be:

$\backslash \; qquad\; (10a\_5a\_3-4a\_4^2)\; X^2\; +\; (10a\_5a\_2-2a\_4a\_3)\; X\; +\; 2a\_4a\_2\; -\; a\_3^2=\; 0$

It is thus enough to choose B, C, D, E such as $\backslash \; frac\; \{B\}\; \{D\}$ and $\backslash \; frac\; \{C\}\; \{E\}$ is the roots of the resolvent one. One calculates then has and F using the formulas:

• $\backslash \; qquad\; F\; =\; \backslash \; frac\; \{da\_4+5ba\_5\}\; \{5e^4\; (Be-cd.)\}$

• $\backslash \; qquad\; has\; =\; \backslash \; frac\; \{a\_5-fe^5\}\; \{d^5\}$

The five roots of the equation to be solved will be:

$\backslash \; qquad\; x\_k\; =\; \backslash \; frac\; \{be^\; \{\backslash \; frac\; \{2ki\; \backslash \; pi\}\; \{5\}\}\; \backslash \; sqrt\; \{has\}\; +\; C\; \backslash \; sqrt\; \{F\}\}\; \{de^\; \{\backslash \; frac\; \{2ki\; \backslash \; pi\}\; \{5\}\}\; \backslash \; sqrt\; \{has\}\; +\; E\; \backslash \; sqrt\; \{F\}\}$ with K taking values 0,1,2,3,4 successively.

Second case: If $(5a\_5a\_3-2a\_4^2)\; =\; 0$ (One is in the case: D or E null).

See the paragraph Complements at the end of the article.

Application to the resolution of the equations of degree 6

Equations of degree 6:

$\backslash \; qquad\; a\_6\; x^6\; +\; a\_5\; x^5\; +\; a\_4\; x^4\; +\; a\_3\; x^3\; +\; a\_2\; x^2\; +\; a\_1\; X\; +\; a\_0\; =\; 0$

roots in the form admit:

$\backslash \; qquad\; \backslash \; frac\; \{B\; \backslash \; sqrt\; \{has\}\; +\; C\; \backslash \; sqrt\; \{F\}\}\; \{D\; \backslash \; sqrt\; \{has\}\; +\; E\; \backslash \; sqrt\; \{F\}\}$

only if:

• $\backslash qquad\; 135a\_6a\_3^2-240a\_6a\_4a\_2+16a\_4^3-60a\_5a\_4a\_3+100a\_5^2a\_2=0$
• $\backslash qquad\; 160a\_5a\_2^2-300a\_5a\_3a\_1+27a\_3^3-96a\_4a\_3a\_2+160a\_4^2a\_1=0$
• $\backslash qquad\; 100a\_4a\_1^2-240a\_4a\_2a\_0+16a\_2^3-60a\_3a\_2a\_1+135a\_3^2a\_0=0$

Consequently, the method of Sotta makes it possible to solve only the equations of degree 6 checking these conditions of resolvability.

That is to say thus the following equation:

$\backslash \; qquad\; a\_6\; x^6\; +\; a\_5\; x^5\; +\; a\_4\; x^4\; +\; a\_3\; x^3\; +\; a\_2\; x^2\; +\; a\_1\; X\; +\; a\_0\; =\; 0,$

First case: If $(12a\_6a\_4-5a\_5^2)\; \backslash \; not\; =\; 0$ (condition so that the resolvent one is of the second degree).

Resolvent of the Sotta associated one will be:

$\backslash \; qquad\; (120a\_6a\_4-50a\_5^2)\; X^2\; +\; (90a\_6a\_3-20a\_5a\_4)\; X\; +\; 15a\_5a\_3\; -\; 8a\_4^2=\; 0$

It is thus enough to choose B, C, D, E such as $\backslash \; frac\; \{B\}\; \{D\}$ and $\backslash \; frac\; \{C\}\; \{E\}$ is the roots of the resolvent one. One calculates then has and F using the formulas:

• $\backslash \; qquad\; F\; =\; \backslash \; frac\; \{da\_5+6ba\_6\}\; \{6e^5\; (cd.-Be)\}$

• $\backslash \; qquad\; has\; =\; \backslash \; frac\; \{a\_6+fe^6\}\; \{d^6\}$

The six roots of the equation to be solved will be:

$\backslash \; qquad\; x\_k\; =\; \backslash \; frac\; \{be^\; \{\backslash \; frac\; \{2ki\; \backslash \; pi\}\; \{6\}\}\; \backslash \; sqrt\; \{has\}\; +\; C\; \backslash \; sqrt\; \{F\}\}\; \{de^\; \{\backslash \; frac\; \{2ki\; \backslash \; pi\}\; \{6\}\}\; \backslash \; sqrt\; \{has\}\; +\; E\; \backslash \; sqrt\; \{F\}\}$ with K taking values 0,1,2,3,4,5 successively.

Second case: If $(12a\_6a\_4-5a\_5^2)\; =\; 0$ (One is in the case: D or E null).

See the paragraph Complements at the end of the article.

Application to the resolution of the equations of degree 7

Equations of degree 7:

$\backslash \; qquad\; a\_7\; x^7\; +\; a\_6\; x^6\; +\; a\_5\; x^5\; +\; a\_4\; x^4\; +\; a\_3\; x^3\; +\; a\_2\; x^2\; +\; a\_1\; X\; +\; a\_0\; =\; 0$

roots in the form admit:

$\backslash \; qquad\; \backslash \; frac\; \{B\; \backslash \; sqrt\; \{has\}\; +\; C\; \backslash \; sqrt\; \{F\}\}\; \{D\; \backslash \; sqrt\; \{has\}\; +\; E\; \backslash \; sqrt\; \{F\}\}$

only if:

• $\backslash qquad\; 189a\_7a\_4^2-315a\_7a\_5a\_3+25a\_5^3-90a\_6a\_5a\_4+135a\_6^2a\_3=0$
• $\backslash qquad\; 135a\_6a\_3^2-225a\_6a\_4a\_2+27a\_4^3-90a\_5a\_4a\_3+125a\_5^2a\_2=0$
• $\backslash qquad\; 125a\_5a\_2^2-225a\_5a\_3a\_1+27a\_3^3-90a\_4a\_3a\_2+135a\_4^2a\_1=0$
• $\backslash qquad\; 135a\_4a\_1^2-315a\_4a\_2a\_0+25a\_2^3-90a\_3a\_2a\_1+189a\_3^2a\_0=0$

Consequently, the method of Sotta makes it possible to solve only the equations of degree 7 checking these conditions of resolvability.

That is to say thus the following equation:

$\backslash \; qquad\; a\_7\; x^7\; +\; a\_6\; x^6\; +\; a\_5\; x^5\; +\; a\_4\; x^4\; +\; a\_3\; x^3\; +\; a\_2\; x^2\; +\; a\_1\; X\; +\; a\_0\; =\; 0,$

First case: If $(7a\_7a\_5-3a\_6^2)\; \backslash \; not\; =\; 0$ (condition so that the resolvent one is of the second degree).

Resolvent of the Sotta associated one will be:

$\backslash \; qquad\; (105a\_7a\_5-45a\_6^2)\; X^2\; +\; (63a\_7a\_4-15a\_6a\_5)\; X\; +\; 9a\_6a\_4\; -\; 5a\_5^2=\; 0$

It is thus enough to choose B, C, D, E such as $\backslash \; frac\; \{B\}\; \{D\}$ and $\backslash \; frac\; \{C\}\; \{E\}$ is the roots of the resolvent one. One calculates then has and F using the formulas:

• $\backslash \; qquad\; F\; =\; \backslash \; frac\; \{da\_6+7ba\_7\}\; \{7e^5\; (Be-cd.)\}$

• $\backslash \; qquad\; has\; =\; \backslash \; frac\; \{a\_7+fe^7\}\; \{d^7\}$

The septs roots of the equation to be solved will be:

$\backslash \; qquad\; x\_k\; =\; \backslash \; frac\; \{be^\; \{\backslash \; frac\; \{2ki\; \backslash \; pi\}\; \{7\}\}\; \backslash \; sqrt\; \{has\}\; +\; C\; \backslash \; sqrt\; \{F\}\}\; \{de^\; \{\backslash \; frac\; \{2ki\; \backslash \; pi\}\; \{7\}\}\; \backslash \; sqrt\; \{has\}\; +\; E\; \backslash \; sqrt\; \{F\}\}$ with K taking values 0,1,2,3,4,5,6 successively.

Second case: If $(7a\_7a\_5-3a\_6^2)\; =\; 0$ (One is in the case: D or E null).

See the paragraph Complements hereafter.

Complements

This paragraph examines more in detail the case or the resolvent one is not of the second degree. I.e. if:

$2na\_na\_\; \{N2\}\; -\; (n-1)\; a\_\; \{n-1\}\; ^2\; =\; 0\; ~$

We have two possibilities then.

First case: the constant term of resolvent is not null.

I.e.:

$6\; (n-1)\; a\_\; \{n-1\}\; a\_\; \{n-3\}\; -\; 4\; (N2)\; a\_\; \{N2\}\; ^2\; \backslash \; not\; =\; 0\; ~$

We notice whereas if $\backslash \; frac\; \{B\}\; \{D\}$ and $\backslash \; frac\; \{C\}\; \{E\}$ are root of an equation of the form:

$AX^2\; +\; BX\; +\; C\; =\; 0\; ~$

then $\backslash \; frac\; \{D\}\; \{B\}$ and $\backslash \; frac\; \{E\}\; \{C\}$ is root of the equation:

$CX^2\; +\; BX\; +\; HAS\; =\; 0\; ~$

obtained by reversing the coefficient dominating and the constant term.

Consequently, we will be able always to then find B, C, D, E by considering that $\backslash \; frac\; \{D\}\; \{B\}$ and $\backslash \; frac\; \{E\}\; \{C\}$ is root of a new resolvent equation obtained by reversing the coefficient dominating and the constant term.

Second case: the constant term of resolvent is also null.

I.e.:

$6\; (n-1)\; a\_\; \{n-1\}\; a\_\; \{n-3\}\; -\; 4\; (N2)\; a\_\; \{N2\}\; ^2\; =\; 0\; ~$

In this case two conditions:

$2na\_na\_\; \{N2\}\; -\; (n-1)\; a\_\; \{n-1\}\; ^2\; =\; 0\; ~$

$6\; (n-1)\; a\_\; \{n-1\}\; a\_\; \{n-3\}\; -\; 4\; (N2)\; a\_\; \{N2\}\; ^2\; =\; 0\; ~$

Which one adds the conditions of resolvability for each degree are the requirements and sufficient so that a polynomial of degree N is put in the form:

$(ax\; +\; b)^n\; ~$

In this case present the equation to be solved will thus be put in the form:

$(ax\; +\; b)^n\; =\; 0\; ~$

and only one multiple root of order N will thus admit which it will be easy to determine.

Examples

The two examples which follow were selected so that the resolvent equation has a discriminant in the form of perfect square in order to simplify calculations. But the method applies as well when the discriminant is not a perfect square, is negative, or is an unspecified complex number.

Example 1

That is to say to solve the equation:

$\backslash \; qquad\; 6x^3\; -\; 6x^2\; +\; 12x\; +\; 7\; =\; 0$

The resolvent one of Sotta is:

$\backslash \; qquad\; 2X^2\; +\; 5X\; -\; 3\; =\; 0$

who has as a root:

$\backslash \; qquad\; \backslash \; frac\; \{B\}\; \{D\}\; =\; \backslash \; frac\; \{1\}\; \{2\}\; and\; \backslash \; frac\; \{C\}\; \{E\}\; =\; -3$

One can choose:

$\backslash \; qquad\; B\; =\; 1,\; C\; =\; -3,\; D\; =\; 2,\; E\; =\; 1$

from where:

$\backslash \; qquad\; F\; =\; \backslash \; frac\; \{da\_2+3ba\_3\}\; \{3e^2\; (Be-cd.)\}\; =\; \backslash \; frac\; \{2\}\; \{7\}$

$\backslash \; qquad\; has\; =\; \backslash \; frac\; \{a\_3-fe^3\}\; \{d^3\}\; =\; \backslash \; frac\; \{5\}\; \{7\}$

While posing:

$\backslash \; qquad\; J\; =\; e^\; \{\backslash \; frac\; \{2i\; \backslash \; pi\}\; \{3\}\}$

The three following roots are obtained:

$\backslash \; qquad\; x\_1\; =\; \backslash \; frac\; \{\backslash \; sqrt\; \{\backslash \; frac\; \{5\}\; \{7\}\}\; -\; 3\; \backslash \; sqrt\; \{\backslash \; frac\; \{2\}\; \{7\}\}\}\; \{2\; \backslash \; sqrt\; \{\backslash \; frac\; \{5\}\; \{7\}\}\; +\; \backslash \; sqrt\; \{\backslash \; frac\; \{2\}\; \{7\}\}\}\; =\; \backslash \; frac\; \{\backslash \; sqrt\; \{5\}\; -\; 3\; \backslash \; sqrt\; \{2\}\}\; \{2\; \backslash \; sqrt\; \{5\}\; +\; \backslash \; sqrt\; \{2\}\}$

$\backslash \; qquad\; x\_2\; =\; \backslash \; frac\; \{J\; \backslash \; sqrt\; \{\backslash \; frac\; \{5\}\; \{7\}\}\; -\; 3\; \backslash \; sqrt\; \{\backslash \; frac\; \{2\}\; \{7\}\}\}\; \{2j\; \backslash \; sqrt\; \{\backslash \; frac\; \{5\}\; \{7\}\}\; +\; \backslash \; sqrt\; \{\backslash \; frac\; \{2\}\; \{7\}\}\}\; =\; \backslash \; frac\; \{J\; \backslash \; sqrt\; \{5\}\; -\; 3\; \backslash \; sqrt\; \{2\}\}\; \{2j\; \backslash \; sqrt\; \{5\}\; +\; \backslash \; sqrt\; \{2\}\}$

$\backslash \; qquad\; x\_3\; =\; \backslash \; frac\; \{j^2\; \backslash \; sqrt\; \{\backslash \; frac\; \{5\}\; \{7\}\}\; -\; 3\; \backslash \; sqrt\; \{\backslash \; frac\; \{2\}\; \{7\}\}\}\; \{2j^2\; \backslash \; sqrt\; \{\backslash \; frac\; \{5\}\; \{7\}\}\; +\; \backslash \; sqrt\; \{\backslash \; frac\; \{2\}\; \{7\}\}\}\; =\; \backslash \; frac\; \{j^2\; \backslash \; sqrt\; \{5\}\; -\; 3\; \backslash \; sqrt\; \{2\}\}\; \{2j^2\; \backslash \; sqrt\; \{5\}\; +\; \backslash \; sqrt\; \{2\}\}$

Example 2

That is to say to solve the equation:

$\backslash \; qquad\; 14x^5\; -\; 36x^4\; +\; 32x^3\; -\; 24x^2\; -\; 2x\; -\; 3\; =\; 0$

One has then:

• $\backslash \; qquad\; a\_5\; =\; 14$

• $\backslash \; qquad\; a\_4\; =\; -36$
• $\backslash \; qquad\; a\_3\; =\; 32$
• $\backslash \; qquad\; a\_2\; =\; -24$
• $\backslash \; qquad\; a\_1\; =\; -2$
• $\backslash \; qquad\; a\_0\; =\; -3$

To know if the equation is resolvable by the method of Sotta, we must check the conditions of resolvability.

$\backslash qquad\; 10a\_5a\_2^2-20a\_5a\_3a\_1+a\_3^3-4a\_4a\_3a\_2+8a\_4^2a\_1=10*14*24^2+20*14*32*2+32^3-4*36*32*24-8*36^2*2=0$ $\backslash qquad\; 8a\_4a\_1^2-20a\_4a\_2a\_0+a\_2^3-4a\_3a\_2a\_1+10a\_3^2a\_0=-8*36*2^2+20*36*24*3-24^3-4*32*24*2-10*32^2*3=0$

The resolvent one of Sotta is:

$\backslash \; qquad\; 2X^2\; +\; 3X\; -\; 2=0$

who has as a root:

$\backslash \; qquad\; \backslash \; frac\; \{B\}\; \{D\}\; =\; \backslash \; frac\; \{1\}\; \{2\}\; and\; \backslash \; frac\; \{C\}\; \{E\}\; =\; -2$

One can choose:

$\backslash \; qquad\; B\; =\; 1,\; C\; =\; 2,\; D\; =\; 2,\; E\; =\; -1$

from where:

$\backslash \; qquad\; F\; =\; \backslash \; frac\; \{da\_2+3ba\_3\}\; \{3e^2\; (Be-cd.)\}\; =\; \backslash \; frac\; \{2\}\; \{25\}$

$\backslash \; qquad\; has\; =\; \backslash \; frac\; \{a\_3-fe^3\}\; \{d^3\}\; =\; \backslash \; frac\; \{11\}\; \{25\}$

One of the roots of the equation will be:

$\backslash \; qquad\; x\_1\; =\; \backslash \; frac\; \{\backslash \; sqrt\; \{\backslash \; frac\; \{11\}\; \{25\}\}\; +\; 2\; \backslash \; sqrt\; \{\backslash \; frac\; \{2\}\; \{25\}\}\}\; \{2\; \backslash \; sqrt\; \{\backslash \; frac\; \{11\}\; \{25\}\}\; -\; \backslash \; sqrt\; \{\backslash \; frac\; \{2\}\; \{25\}\}\}\; =\; \backslash \; frac\; \{\backslash \; sqrt\; \{11\}\; +\; 2\; \backslash \; sqrt\; \{2\}\}\; \{2\; \backslash \; sqrt\; \{11\}\; -\; \backslash \; sqrt\; \{2\}\}$

One obtains the cinqs following roots then:

$\backslash \; qquad\; x\_1\; =\; \backslash \; frac\; \{\backslash \; sqrt\; \{11\}\; +\; 2\; \backslash \; sqrt\; \{2\}\}\; \{2\; \backslash \; sqrt\; \{11\}\; -\; \backslash \; sqrt\; \{2\}\}$

$\backslash \; qquad\; x\_2\; =\; \backslash \; frac\; \{e^\; \{\backslash \; frac\; \{2i\; \backslash \; pi\}\; \{5\}\}\; \backslash \; sqrt\; \{11\}\; +\; 2\; \backslash \; sqrt\; \{2\}\}\; \{2e^\; \{\backslash \; frac\; \{2i\; \backslash \; pi\}\; \{5\}\}\; \backslash \; sqrt\; \{11\}\; -\; \backslash \; sqrt\; \{2\}\}$

$\backslash \; qquad\; x\_3\; =\; \backslash \; frac\; \{e^\; \{\backslash \; frac\; \{4i\; \backslash \; pi\}\; \{5\}\}\; \backslash \; sqrt\; \{11\}\; +\; 2\; \backslash \; sqrt\; \{2\}\}\; \{2e^\; \{\backslash \; frac\; \{4i\; \backslash \; pi\}\; \{5\}\}\; \backslash \; sqrt\; \{11\}\; -\; \backslash \; sqrt\; \{2\}\}$

$\backslash \; qquad\; x\_4\; =\; \backslash \; frac\; \{e^\; \{\backslash \; frac\; \{6i\; \backslash \; pi\}\; \{5\}\}\; \backslash \; sqrt\; \{11\}\; +\; 2\; \backslash \; sqrt\; \{2\}\}\; \{2e^\; \{\backslash \; frac\; \{6i\; \backslash \; pi\}\; \{5\}\}\; \backslash \; sqrt\; \{11\}\; -\; \backslash \; sqrt\; \{2\}\}$

$\backslash \; qquad\; x\_5\; =\; \backslash \; frac\; \{e^\; \{\backslash \; frac\; \{8i\; \backslash \; pi\}\; \{5\}\}\; \backslash \; sqrt\; \{11\}\; +\; 2\; \backslash \; sqrt\; \{2\}\}\; \{2e^\; \{\backslash \; frac\; \{8i\; \backslash \; pi\}\; \{5\}\}\; \backslash \; sqrt\; \{11\}\; -\; \backslash \; sqrt\; \{2\}\}$

Other methods of solution of equations

• Method of Tschirnhaus

• Method of Bézout
• Method of Cardan joint
• Method of Ferrari
• Method of Descartes

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