Method of Ferrari

The method of Ferrari imagined and developed by Ludovico Ferrari makes it possible to solve the equations of the fourth degree, i.e. to write the solutions like a combination of additions, subtractions, multiplications, divisions, and roots square, cubic and quartics of rational numbers.

Principle of the method

That is to say the general equation of the fourth degree following: $\backslash \; qquad\; has\; x^4\; +\; B\; x^3\; +\; C\; x^2\; +\; D\; X\; +\; E\; =\; 0,$

The change of variable refines $\backslash \; qquad\; X\; =\; Z\; -\; \backslash \; frac\; \{B\}\; \{4a\}$ watch which the equation is equivalent to an equation of the form $\backslash \; qquad\; z^4\; +\; p\; z^2\; +\; Q\; z+\; R\; =\; 0$, which is still written: $\backslash \; qquad\; z^4\; =\; -\; p\; z^2\; -\; Q\; Z\; -\; R$.

$\backslash \; qquad\; p=\; \backslash \; frac\; \{-\; 3b^2\}\; \{8a^2\}\; +\; \backslash \; frac\; \{C\}\; \{has\}$

$\backslash \; qquad\; q=\; \backslash \; frac\; \{(B/2)\; ^3\}\; \{a^3\}\; -\; \backslash \; frac\; \{(1/2)\; bc\}\; \{a^2\}\; +\; \backslash \; frac\; \{D\}\; \{has\}$

$\backslash \; qquad\; r=\; -3\; (\backslash \; frac\; \{B/4\}\; \{has\})\; ^4\; +\; C\; (\backslash \; frac\; \{(B/4)\; ^2\}\; \{a^3\})\; -\; \backslash \; frac\; \{(1/4)\; data\; base\}\; \{a^2\}\; +\; \backslash \; frac\; \{E\}\; \{has\}$

The central point of the method consists in replacing the students' rag procession z⁴ by the polynomial (z²+y) ² , parameterized by there , and to find a value of there suitable, which allows, via a Identité remarkable to factorize the expression:

(z^2+y) ^2 + p z^2 + Q Z + R - 2yz^2 - y^2 \ \

&= (z^2+y) ^2 - (2y - p) z^2 + qz - y^2 + R. \ \ \ end {array}

More precisely, it is a question of determining there so that the term $(2y-p)\; z^2-qz+y^2-r$, seen like polynomial of the second degree in Z, is written in the form of square, which is equivalent to the cancellation of sound Discriminant: $q^2\; -\; 4\; (2y\; -\; p)\; (y^2\; -\; R)$, i.e. with:

$\backslash \; qquad\; 8y^3\; -\; 4py^2\; -\; 8ry\; +\; 4rp\; -\; q^2\; =\; 0$

This equation is solved by radicals while using, for example, the Méthode of Cardan joint, which gives at least an actual value y₀ suitable. By deferring the value y₀ obtained in the preceding equation, one obtains:

$\backslash \; qquad\; (z^2\; +\; y\_0)\; ^2\; =\; (a\_0z\; +\; b\_0)\; ^2$ with has (real or complex) such as $a\_0^2\; =\; -\; p+2y\_0$ and B such as $b\_0\; =\; -\; \backslash \; frac\; \{Q\}\; \{2a\_0\}$; and, by a remarkable Identity, this equality is equivalent to:

$\backslash \; qquad\; (z^2\; +\; y\_0\; -\; a\_0z\; -\; b\_0)\; (z^2\; +\; y\_0\; +\; a\_0z\; +\; b\_0)\; =\; 0$

what is equivalent to the cancellation of one of both factors of the second degree in Z :

$\backslash \; qquad\; z^2\; +\; y\_0\; -\; a\_0z\; -\; b\_0\; =\; 0$

$\backslash \; qquad\; z^2\; +\; y\_0\; +\; a\_0z\; +\; b\_0\; =\; 0$

Each one of these two equations provides two values for Z , that is to say four values in very for Z , and the values of X solutions of the initial equation result some.

Example

We propose to solve the equation:

$x^4\; +\; 4x^3\; +\; 3x^2\; -\; 8x\; -\; 10\; =\; 0\; ~$

Let us pose:

$X\; =\; Z\; -\; 1\; \backslash \; qquad\; (*)\; ~$

While replacing in the equation, one obtains:

$z^4\; -\; 3z^2\; -\; 6z\; -\; 2\; =\; 0\; \backslash \; qquad\; (**)\; ~$

there being an unspecified complex number, let us develop the expression now:

$(z^2\; +\; there)\; ^2\; =\; z^4\; +\; 2yz^2\; +\; y^2~$

Who, taking into account (**), can be written:

$(z^2\; +\; there)\; ^2\; =\; 3z^2\; +\; 6z\; +\; 2\; +\; 2yz^2\; +\; y^2~$

Let us put the second member in the form of polynomial in Z:

$(z^2\; +\; there)\; ^2\; =\; (3\; +\; 2y)\; z^2\; +\; 6z\; +\; y^2\; +\; 2\; \backslash \; qquad\; (***)\; ~$

If it is wanted that the second member also puts himself in the form of square, there is necessary to find a value of such as the polynomial in Z is a double root. Either a value of there such as the discriminant of the polynomial in Z or no one, i.e. such as:

$6^2\; -4\; (3\; +\; 2y)\; (y^2\; +\; 2)\; =\; 0\; ~$

Who is simplified in the form:

$2y^3\; +\; 3y^2\; +\; 4y\; -3\; =\; 0\; ~$

We notice that this cubic equation in admits the obvious root there:

$there\; =\; 0.5\; ~$

(What avoids us in this example using a more elaborate method like the Méthode of Cardan joint to solve it).

One fixes there at this value which one defers in (***), we obtain:

$(z^2\; +\; 0.5)\; ^2\; =\; 4z^2\; +\; 6z\; +\; 2.25\; ~$

And as envisaged, the second member puts himself well in the form of square:

$(z^2\; +\; 0.5)\; ^2\; =\; 2z\; +\; 1.5\; ^2\; ~$

Who is written:

$(z^2\; +\; 0.5)\; ^2\; -\; 2z\; +\; 1.5\; ^2\; =\; 0\; ~$

We have a difference in square which one can thus write:

$+\; 2z\; +\; 2\; -\; 2z\; -\; 1\; =\; 0\; ~$

We are thus brought back to solve the two equations:

$z^2\; +\; 2z\; +\; 2\; =\; 0\; ~$

$z^2\; -\; 2z\; -\; 1\; =\; 0\; ~$

The discriminants of these two equations are:

For the first equation:

$\backslash \; triangle\_1\; =\; 4\; -\; 4*2\; =\; -4\; ~$

For the second equation:

$\backslash \; triangle\_2\; =\; 4\; -\; 4*\; (-\; 1)\; =\; 8\; ~$

We deduce the four possible values from them from Z.

For the first equation:

$z\_1\; =\; \backslash \; frac\; \{-\; 2\; -\; 2i\}\; \{2\}\; =\; -\; 1\; -\; I\; ~$

$z\_2\; =\; \backslash \; frac\; \{-\; 2\; +\; 2i\}\; \{2\}\; =\; -\; 1\; +\; I\; ~$

For the second equation:

$z\_3\; =\; \backslash \; frac\; \{2\; -\; 2\; \backslash \; sqrt\; \{2\}\}\; \{2\}\; =\; 1\; -\; \backslash \; sqrt\; \{2\}\; ~$

$z\_4\; =\; \backslash \; frac\; \{2\; +\; 2\; \backslash \; sqrt\; \{2\}\}\; \{2\}\; =\; 1\; +\; \backslash \; sqrt\; \{2\}\; ~$

By deferring these four values of Z in (*), one obtains:

$x\_1\; =\; -2\; -\; I\; ~$

$x\_2\; =\; -2\; +\; I\; ~$

$x\_3\; =\; -\; \backslash \; sqrt\; \{2\}\; ~$

$x\_4\; =\; \backslash \; sqrt\; \{2\}\; ~$

Who are well the four roots of the equation to be solved.

Other methods of solution of equations

• Method of Tschirnhaus
• Method of Bézout
• Method of Cardan joint
• Method of Sotta
• Method of Descartes

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