Method of Cardan joint

The method of Cardan joint , imagined and developed by Jerome Cardan joint in his work Ars Magna published in 1545, is a method making it possible to solve all the cubic equations.

This method makes it possible to set up formulas called formulas of Cardan joint giving according to p and Q the solutions of the equation:

x^3+px + q= 0 ~

It makes it possible to prove that the equations of degree 3 are resolvable by radicals. It is pointed out that only the equations of degree 1,2,3,4 are génériquement resolvable by radicals, i.e. only these equations have general methods of resolutions giving the solutions according to the coefficients of the polynomial by using only the four usual operations on the rational numbers, and the extraction of square, cubic roots and quartics.

Formulas of Cardan joint

Let us consider the equation

x^3+px + q= 0 \,

One calculates the discriminant

\ Delta = q^2 + \ frac {4} {27} p^3 \,

and one studies his sign. Note: There exists also the concept of " discriminant writing réduite" by posing p = 3p' and Q = 2q', that is written

\ Delta = q'^2 + p'^3 \,

If one leaves the general equation ^{$a z^3 + B z^2 + C Z + D = 0 \,$}, one is reduced to the form reduced while posing $\ textstyle {X = Z + \ frac {B} {3a}}$ , $p = - \ frac {b^2} {3a^2} + \ frac {C} {has}$ and $Q = \ frac {B} {27a} \ left (\ frac {2b^2} {a^2} - \ frac {9c} {has} \ right) + \ frac {D} {has}$ .

If Δ is positive

The equation has then a real solution and two complexes . One poses

U = \ sqrt {\ frac {- Q + \ sqrt {\ Delta}} {2}} \ mbox {and} v = \ sqrt {\ frac {- Q - \ sqrt {\ Delta}} {2}}

The only real solution is then $x_1 = U + v \,$ . There exist also two combined complex solutions one of the other $\ begin {boxes} x_2= J U + \ bar {J} v \ \ x_3= j^2u + \ overline {j^2} v \ end {boxes}$ where ^{$j = - \ frac {1} {2} + I \ frac {\ sqrt {3}} {2} = e^ {I \ frac {2 \ pi} {3}}$}.

If Δ is null

The equation then has two solutions real , simple and a double:

\ begin {boxes} x_1= 2 \ sqrt {\ frac {- Q} {2}} = -2 \ sqrt {\ frac {- p} {3}} = \ frac {3q} {p} \ \ x_2=x_3= - \ sqrt {\ frac {- Q} {2}} = \ sqrt {\ frac {- p} {3}} = \ frac {- 3q} {2p} \ end {boxes}

If Δ is negative

The equation then has three solutions real . However, it is necessary to make an incursion into the complexes for all to find them (see notices historical). The solutions are the sums of two combined complexes

j^ku \,

and

\ overline {j^ku}

where ^{$u= \ sqrt {\ frac {- Q + I \ sqrt {- \ Delta}} {2}}$ ^{and $k= \ {0,1,2 \} \,$ , is the following unit: $\ begin {boxes} x_1 = U + \ bar {U} \ \ x_2 = J U + \ overline {ju} \ \ x_3= j^2u + \ overline {j^2u} \ end {boxes}$

The real form of the solutions is obtained by writing $j^ku$ under the trigonometrical form, which gives:

$x_k = 2 \ sqrt {\ frac {- p} {3}} \ cos {\ frac13 \ left (\ arccos {\ left (\ frac {- Q} {2} \ sqrt {\ frac {27} {- p^3}} \ right)}+ 2k \ pi \ right)} \ mbox {with} k= \ {1,2,3 \} \,$

Principle of the method

Let us consider the general equation of the third degree following: ^{$a x^3 + B x^2 + C X + D = 0 \,$}.

By posing $\ textstyle {X = Z - \ frac {B} {3a}}$ , one is brought back to an equation of the form: $z^3 + p Z + Q = 0 \,$

where $p = - \ frac {b^2} {3a^2} + \ frac {C} {has}$
and $Q = \ frac {B} {27a} \ left (\ frac {2b^2} {a^2} - \ frac {9c} {has} \ right) + \ frac {D} {has}$ .

One now will pose $Z = U + v \,$ with U and v complexes, in order to have two unknown factors instead of one and to thus give the possibility of laying down later on a condition on U and v allowing to simplify the problem. The equation $z^3 + p Z + Q = 0 \,$ becomes thus
$(u+v) ^3 + p (u+v) + Q = 0 \,$ .
This equation changes easily in the following form: $u^3+v^3+ (3uv+p) (u+v) +q=0 \,$
The condition of announced simplification will be then $3uv+p=0 \,$ . What gives us on the one hand $u^3+v^3+q=0 \,$ and on the other hand $uv=- \ frac {p} {3} \,$ , which, by raising the two members with power 3 gives ^{$u^3v^3=- \ frac {p^3} {27} \,$}.

We obtain finally the system nap-product of the two unknown factors $u^3$ and $v^3$ according to:
$\ begin {boxes} u^3+v^3&=-q \ \ u^3v^3&=- \ frac {p^3} {27} \ end {boxes}$
The unknown factors $u^3$ and $v^3$ being two complexes which one knows the sum and the product, they are thus the roots of the quadratic equation: $X^2+qX-\frac{p^3}{27}=0$

The discriminant of this equation is $\ Delta = q^2 + \ frac {4} {27} p^3 \,$ and the roots are
$\ begin {boxes} u^3 = \ frac {- Q + \ sqrt {\ Delta}} {2} \ quad \! \ mbox {and} v^3 = \ frac {- Q - \ sqrt {\ Delta}} {2}, & \ mbox {if} \ Delta \ mbox {is positive} \ \ u^3 = \ frac {- Q + I \ sqrt {- \ Delta}} {2} \ mbox {and} v^3 = \ frac {- Q - I \ sqrt {- \ Delta}} {2}, & \ mbox {if} \ Delta \ mbox {is N} \ acute {E} \ mbox {gatif} \ \ u^3 = v^3 = \ frac {-Q} {2}, & \ mbox {if} \ Delta \ mbox {is null} \ end {boxes}$
It is then enough to associate the three cubic roots of $u^3$ and $v^3$ two by two in order to obtain three couples (U, v) such as $uv=- \ frac {p} {3}$ , then to defer the three couples of values found for $u$ and $v$ in the expression $Z = U + v \,$ .

Lastly, one returns to the first change of variable $x = Z - \ frac {B} {3a}$ to have the three roots of the cubic equation posed at the beginning.

Examples

Example 1

Let us consider for example the equation $x^3 = 18x + 35 \,$ or $x^3 - 18x - 35 = 0 \,$ .
There is $p = - 18 \,$ and $q = - 35 \,$ , therefore:
$u^3v^3 = {18^3 \ over 27} = 216 \,$ and $u^3 + v^3 = 35 \,$ thus $u^3 \,$ and $v^3 \,$ is roots of the equation $X^2 - 35X + 216 = 0 \,$ , of which the roots are 27 and 8. Thus U and v is worth 3 and 2 and the sought solution is $x = U + v = 5 \,$ .

If one places oneself in $\ mathbb C$ , then the other roots are $u = 3 \, J \,$ and $v = 2 \, j^2 \,$ , where $j = \ exp \ left (\ frac {2i \ pi} {3} \ right)$ , or $u = 3j^2 \,$ and $v = 2j \,$ . One thus obtains like other roots:
$z = 3j + 2j^2 = - {5 \ over 2} + I {\ sqrt {3} \ over 2}$
$z = 3j^2 + 2j = - {5 \ over 2} - I {\ sqrt {3} \ over 2}$

Let us notice that before launching out in such calculations, it is " better; tester" a little the result using the rule Wheeler. A graph gives already the rule of Descartes: there will be only one real root; it lies between 4 and 6. And $5^3 = 125$ and $18 \ times 5= 90$ thus 5 is root.

The remainder results from this: $P (X) = (x-5) (x^2+5x+7)$ , which is studied more easily.

Example 2

That is to say to solve the equation:

$6x^3-6x^2+12x+7=0$

Let us pose:

$x = Z + \ frac {1} {3}$

One obtains while replacing and while developing:

$54z^3+90z+95 = 0$

Let us pose then:

$z = U + v$

One obtains:

$54 (u+v) ^3+90 (u+v) + 95 = 0$

Who is written:

$54 (u^3+v^3) + (162uv+90) (u+v) +95=0$

The condition of simplification will be thus:

$162uv+90 = 0$

I.e.:

$uv = - \ frac {5} {9}$

One thus has:

$u^3+v^3 = - \ frac {95} {54}$

$u^3v^3 = - \ frac {125} {729}$

u³ and v³ are thus the roots of the equation:

$X^2 + \ frac {95} {54} X - \ frac {125} {729} = 0$

The two roots of this equation are:

$u^3 = \ frac {5} {2 \ cdot 27}$

$v^3 = - \ frac {50} {27}$

The three couples (U, v) checking:

$uv = - \ frac {5} {9}$

are thus:

$u_1 = \ frac {1} {3} \ sqrt {\ frac {5} {2}}$ and $v_1 = - \ frac {1} {3} \ sqrt {50}$

$u_2 = \ frac {J} {3} \ sqrt {\ frac {5} {2}}$ and $v_2 = - \ frac {j^2} {3} \ sqrt {50}$

$u_3 = \ frac {j^2} {3} \ sqrt {\ frac {5} {2}}$ and $v_3 = - \ frac {J} {3} \ sqrt {50}$

While deferring in:

$z = U + v$

One obtains:

$z_1 = \ frac {1} {3} \ sqrt {\ frac {5} {2}} - \ frac {1} {3} \ sqrt {50}$

$z_2 = \ frac {J} {3} \ sqrt {\ frac {5} {2}} - \ frac {j^2} {3} \ sqrt {50}$

$z_3 = \ frac {j^2} {3} \ sqrt {\ frac {5} {2}} - \ frac {J} {3} \ sqrt {50}$

And while deferring in:

$x = Z + \ frac {1} {3}$

One obtains finally the three solutions of the equation which one had been given to solve:

$x_1 = \ frac13 \ left (\ sqrt {\ frac52} - \ sqrt {50} + 1 \ right)$

$x_2 = \ frac13 \ left (J \ sqrt {\ frac52} - j^2 \ sqrt {50} + 1 \ right)$

$x_3 = \ frac13 \ left (j^2 \ sqrt {\ frac52} - J \ sqrt {50} + 1 \ right)$

Example 3

Let us consider the equation:

$x^3-6x^2+9x-1 = 0$

By translation $P (X) = P (z+2) = z^3-3z+1 = Q (Z)$ .

Let us pose then:

$z = U + v$

One obtains:

$(u+v) ^3-3 (u+v) + 1 = 0$

Who is written:

$u^3+v^3+ (3uv-3) (u+v) +1=0$

The condition of simplification will be thus:

$3uv-3 = 0$

I.e.:

$uv = 1$

One thus has:

$u^3+v^3 = -1$

$u^3v^3 = 1$

$u^3$ and $v^3$ are thus the roots of the equation:

$X^2 + X + 1 = 0$

The two roots of this equation are:

$u^3 = \ frac {- 1+i \ sqrt {3}} {2} = e^ {\ frac {2i \ pi} {3}} = J$

$v^3 = \ frac {- 1-i \ sqrt {3}} {2} = e^ {\ frac {- 2i \ pi} {3}} = j^2$

The three couples (U, v) checking:

$uv = 1$

are thus:

$u_1 = e^ {\ frac {2i \ pi} {9}}$ and $\ qquad v_1 = e^ {\ frac {- 2i \ pi} {9}}$

$u_2 = je^ {\ frac {2i \ pi} {9}}$ and $\ qquad v_2 = j^2e^ {\ frac {- 2i \ pi} {9}}$

$u_3 = j^2e^ {\ frac {2i \ pi} {9}}$ and $\ qquad v_3 = je^ {\ frac {- 2i \ pi} {9}}$

While deferring in:

$z = U + v$

One obtains:

$z_1 = e^ {\ frac {2i \ pi} {9}} + e^ {\ frac {- 2i \ pi} {9}} = 2 \ cos \ left (\ frac {2 \ pi} {9} \ right)$

$z_2 = je^ {\ frac {2i \ pi} {9}} + j^2e^ {\ frac {- 2i \ pi} {9}} = 2 \ cos \ left (\ frac {8 \ pi} {9} \ right)$

$z_3 = j^2e^ {\ frac {2i \ pi} {9}} + je^ {\ frac {- 2i \ pi} {9}} = 2 \ cos \ left (\ frac {4 \ pi} {9} \ right)$

From where solutions in X.

Let us notice that in practice, on this level of mathematics, one (E) student (E) raises the question reasonably: can't one better do, at least deferred action? The solution makes think of $x^9-1 = (x^3-1) (x^3-j) (x^3-j^2) isn't$ , and possible of réagencer this polynomial by finding $x^ {- 3} x+1$ ? This concept of réagencement, of Re group lies of the roots should guide. One can also think of $Q (Z) = Q (iy) = I R (there)$ .

Notice historical

A polemic concerning the paternity of this method exists.

It is told that the method was previously discovered by the Italian mathematician Tartaglia. At that time, the mathematicians launched out challenges to solve cubic equations and Tartaglia solved them all. Intrigued, Cardan asked to him whether it would not have found methods. After being itself made request and have received the insurance that Cardan would not reveal them with anybody, Tartaglia entrusted to him. Which was not its surprise to see Cardan publishing them in 1545.

One often calls from now on these formulas the formulas of Tartaglia-Cardan joint.

The use of the formulas of Cardan joint requires sometimes the use of complex numbers, even to find solutions real. In fact, the imaginary numbers precisely were born with this occasion.

In the example X ³ = 15 X + 4 or X ³ - 15 X - 4 = 0, one has p = - 15 and Q = -4, therefore:
$u^3v^3 = {15^3 \ over 27} = 125$ and U ³ + v ³ = 4 thus U ³ and v ³ is roots of the X² equation - 4X + 125 = 0, of which the roots do not exist. However, there is well a solution X with the initial equation; it is X = 4.
It is Bombelli which will overcome this difficulty by proposing for the first time a calculation on the imaginary numbers. The formal resolution of the X² equation - 4X + 125 = 0 gives for $u^3 roots = 2 + \ sqrt {- 121} = 2 + 11 \ sqrt {- 1}$ and $v^3 = 2 - \ sqrt {- 121} = 2 - 11 \ sqrt {- 1}$ , but Bombelli realizes that the cube of $2 + \ sqrt {- 1}$ is worth
$2 + \ sqrt {- 121}$ and that the cube of $2 - \ sqrt {- 1}$ is worth $2 - \ sqrt {- 121}$ . It from of deduced that $u = 2 + \ sqrt {- 1}$ and that $v = 2 - \ sqrt {- 1}$ and he finds well like final solution X = U + v = 4.

The imaginary numbers were born.

Other methods of solution of equations of the 3rd degree

This new method was preceded at the time of the price in Fermat junior of mathematics 1995: http://spoirier.lautre.net/equation.pdf

Other methods of solution of equations

Method of Tschirnhaus
Method of Bézout
Method of Sotta
Method of Ferrari
Method of Descartes

Applet Java using the method to solve cubic equations

http://www.java-cardan.fr

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