Method of Descartes

The method of Descartes said by unspecified coefficients makes it possible to solve the equations of the second, but especially of the fourth degree.

Rene Descartes uses for this the factorization of the polynomials of degree NR in the form $(X - x_1) (x-x_2) \ cdots (x-x_n)$ with $x_1, \ cdots, x_n$ N real roots or complexes (see Théorème of Alembert-Gauss) which it is then one of the first mathematicians to be controlled.

To solve the second degree, one leaves the relation sometimes known as then of François Viète which one obtains while developing:

x_1 + x_2 = - \ frac Ba ~

x_1x_2 = \ frac ca

One poses

x_1 then = - \ frac {B} {2a} + p ~

x_2 = - \ frac {B} {2a} - p ~

with p a real quantity which one will seek to determine in the other relation.

This easy way is very important: when one has a sum of two numbers has and B being worth real C, one can always write has as the sum of half of C and a certain quantity; B, to maintain the equality has + B = C, will inevitably be worth half of C minus this quantity.

One arrives then at

(- \ frac {B} {2a} + p) (- \ frac {B} {2a} - p) = \ frac Ca ~

what brings by simple development to p, then with the two roots, whose formula is famous!

This method is more particularly used to solve the equations of the fourth degree; it is initially necessary to see the polynomial

X^4 + AX^2 + BX + C ~

(to which one was brought back by division by the coefficient of X then an algebraic translation which removes X with power 3, to see the articles on the Méthode of Ferrari and the Méthode of Cardan joint for more information) like the product of two polynomials of the second degree, and not like the product of four polynomials of the first degree; the coefficients must be found so that while developing, one finds the coefficients well has , B , and C One thus poses

(X^2 + aX +b) (X^2 - aX +c) = X^4 + AX^2 + BX + C ~

This brings to a manageable system, the reader will be able to check while developing:

\ begin {boxes} b+c = a^2+ has & (1) \ \ (C has - b)= B & (2) \ \ bc = C & (3) \ end {boxes} The goal being then not to have more but to solve two quadratic equations to leave the four roots!

Here, one poses

B = \ frac 12 (a^2 + has + p) ~

one calls by convention the unspecified quantity

\ frac 12 p

to simplify calculations. One determines it in the expression (2): the reader will be able to check that

p = - \ frac Ba ~

in the last expression, one expresses B and C entirely according to has, which leads to an equation known as resolvent, while developing a little:

a^6 + 2Aa^4 + (A^2 - 4C) a^2 - B^2 = 0 ~

who, while posing

z = a^2

, brings back himself to a cubic equation, the formula of Cardan joint thus gives to all the blows real a solution of which the square root will be worth has :

z^3 + 2Az^2 + (A^2-4C) z-B^2 = 0 ~

B and C are easily with the relations above and with final, one will find formulas equivalent to those of Ferrari.

Example

We propose to solve the equation:

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

Let us pose:

$X = X - 1 \ qquad (*) ~$

While replacing in the equation, one obtains:

$X^4 - 3X^2 - 6X - 2 = 0 \ qquad (**) ~$

By developing the equation:

(X^2 + aX +b) (X^2 - aX +c) = 0 \ qquad (***) ~

And while identifying with the equation (**), one obtains:

$\ begin {boxes}
b+c = a^2 - 3 & (1) \ \
(C has - b)= -6 & (2) \ \
bc = -2 & (3)
\ end {boxes}$

Let us pose:

$b = \ frac {1} {2} (a^2 - 3 + p) ~$

$c = \ frac {1} {2} (a^2 - 3 - p) ~$

What, for any p, checks (1).

One can now determine p while carrying B and C in (2), one obtains:

$p = \ frac {6} {has} ~$

One thus has:

$b = \ frac {1} {2} (a^2 - 3 + \ frac {6} {has}) ~$

$c = \ frac {1} {2} (a^2 - 3 - \ frac {6} {has}) ~$

Now let us carry these values of B and C in (3), one obtains:

$\ frac {1} {4} (a^2 - 3 + \ frac {6} {has}) (a^2 - 3 - \ frac {6} {has}) = -2 ~$

Who puts himself in the form:

$(a^2-3) ^2 - \ frac {36} {a^2} = -8 ~$

While multiplying by a² and while developing, one obtains:

$a^2 (a^4 - 6a^2 + 9) - 36 = -8a^2 ~$

And finally:

$a^6 - 6a^4 + 17a^2 - 36 = 0 ~$

Let us pose:

$z = a^2 ~$

One obtains:

$z^3 - 6z^2 + 17z - 36 = 0 ~$

This equation admits for obvious root:

$z = 4 ~$

(If there are no obvious roots, one can solve the equation while using, for example, the Méthode of Cardan joint)

One from of deduced:

$a = 2 ~$

And consequently:

$b = \ frac {1} {2} (a^2 - 3 + \ frac {6} {has}) = \ frac {1} {2} (2^2 - 3 + \ frac {6} {2}) = 2 ~$

$c = \ frac {1} {2} (a^2 - 3 - \ frac {6} {has}) = \ frac {1} {2} (2^2 - 3 - \ frac {6} {2}) = -1 ~$

By deferring the values of has, B, C in (***), one obtains:

$(X^2 + 2X +2) (X^2 -2X - 1) = 0 ~$

One is thus brought back to solve the two equations:

$X^2 + 2X +2 = 0 ~$

$X^2 -2X - 1 = 0 ~$

The discriminants of these two equations are:

For the first equation:

$\ triangle_1 = -4 = (2i) ^2 ~$

For the second equation:

$\ triangle_2 = 8 = (2 \ sqrt {2}) ^2 ~$

We deduce the four possible values from them from X.

For the first equation:

$X_1 = \ frac {- 2 + 2i} {2} = -1 + I ~$

$X_2 = \ frac {- 2 - 2i} {2} = -1 - I ~$

For the second equation:

$X_3 = \ frac {2 + 2 \ sqrt {2}} {2} = 1 + \ sqrt {2} ~$

$X_4 = \ frac {2 - 2 \ sqrt {2}} {2} = 1 - \ sqrt {2} ~$

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

$x_1 = -2 + I ~$

$x_2 = -2 - I ~$

$x_3 = \ sqrt {2} ~$

$x_4 = - \ sqrt {2} ~$

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

Method of Descartes

Example

Other methods of solution of equations