Cardan formulas

The Cardan formulas are formulas for solving reduced cubic equations (equations of the 3rd degree). This calculates all zeros of a given cubic polynomial . The formulas were first published in 1545 by the mathematician Gerolamo Cardano in his book Ars magna , together with solution formulas for quartic equations (equations of the fourth degree) . The formula for solving the reduced cubic equations was discovered by Nicolo Tartaglia ; according to Cardano even earlier through Scipione del Ferro . Cardano himself developed the method for reducing the general equation of the third degree to this special case.

The Cardan formulas were an important motivation for the introduction of the complex numbers , since in the case of the casus irreducibilis (Latin for "non-traceable case") one can get to real solutions by taking a square root from a negative number. Franciscus Vieta only managed to solve this case around 1600 using trigonometry.

Today, the Cardan formulas have hardly any practical significance for a purely numerical , ie approximate solution of cubic equations, since the solutions can be approximately more conveniently determined using the Newton method using an electronic computer. On the other hand, they are of considerable importance for an exact calculation of the solutions in radicals . The proof that there can be no corresponding formulas for equations of the fifth or higher degree has, however, had a decisive influence on the development of algebra (see Galois theory ).

Reduction of the general equation of the third degree

The general equation of the third degree

${\displaystyle Ax^{3}+Bx^{2}+Cx+D=0}$

with real numbers , , , and , by division by first in the normal form ${\displaystyle A}$ ${\displaystyle B}$ ${\displaystyle C}$ ${\displaystyle D}$ ${\displaystyle A\neq 0}$ ${\displaystyle A}$

${\displaystyle x^{3}+ax^{2}+bx+c=0}$

be brought with , and . ${\displaystyle a={\tfrac {B}{A}}}$ ${\displaystyle b={\tfrac {C}{A}}}$ ${\displaystyle c={\tfrac {D}{A}}}$

With the help of substitution , the square term in the normal form is eliminated, and the reduced form is obtained : ${\displaystyle x=z-{\tfrac {a}{3}}}$

${\displaystyle z^{3}+pz+q=0,}$

in which

${\displaystyle p=b-{\frac {a^{2}}{3}}={\frac {9AC-3B^{2}}{9A^{2}}}}$     and     ${\displaystyle q={\frac {2a^{3}}{27}}-{\frac {ab}{3}}+c={\frac {2B^{3}-9ABC+27A^{2}D}{27A^{3}}}}$

The reduced form is now solved with the help of the Cardan formula and then the solutions of the original equation are determined by back substitution . ${\displaystyle x=z-{\tfrac {a}{3}}}$

The cardan formula for dissolving the reduced form ${\displaystyle z^{3}+pz+q=0}$

In contrast to the quadratic equation , the cubic equation requires you to consider complex numbers , even if all three solutions are real.

The three solutions result from the substitution : Then and coefficient comparison yields: and . The result is the system of equations and . According to Vieta's theorem , and are solutions of the so-called quadratic resolvents . So you get ${\displaystyle z=u+v}$
${\displaystyle z^{3}=\left(u+v\right)^{3}=u^{3}+3uv\left(u+v\right)+v^{3}=3uvz+u^{3}+v^{3}}$ ${\displaystyle -p=3uv}$ ${\displaystyle -q=u^{3}+v^{3}}$ ${\displaystyle u^{3}+v^{3}=-q}$ ${\displaystyle u^{3}\cdot v^{3}=-\left({\tfrac {p}{3}}\right)^{3}}$ ${\displaystyle u^{3}}$ ${\displaystyle v^{3}}$ ${\displaystyle t^{2}+qt-{\tfrac {p^{3}}{27}}=0}$

${\displaystyle u={\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {\Delta }}}},\quad v={\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {\Delta }}}},}$       in which

${\displaystyle \Delta :=\left({\frac {q}{2}}\right)^{2}+\left({\frac {p}{3}}\right)^{3}={\frac {27A^{2}D^{2}+4B^{3}D-18ABCD+4AC^{3}-B^{2}C^{2}}{108A^{4}}}}$

is the discriminant (of the cubic equation). For their derivation see below . The two complex third roots and must be chosen in such a way that the secondary condition is fulfilled (so there are only three pairs instead of nine ). ${\displaystyle u}$ ${\displaystyle v}$ ${\displaystyle \textstyle u\cdot v=-{\frac {p}{3}}}$ ${\displaystyle (u,v)}$

The other two third roots are then each obtained by multiplying by the two primitive third roots of unity

${\displaystyle \varepsilon _{1}=-{\tfrac {1}{2}}+{\tfrac {1}{2}}\mathrm {i} {\sqrt {3}}}$   and    . ${\displaystyle \varepsilon _{2}=\varepsilon _{1}^{2}=-{\tfrac {1}{2}}-{\tfrac {1}{2}}\mathrm {i} {\sqrt {3}}}$

Because of the secondary condition, the three solutions of the reduced form result in

${\displaystyle {\begin{aligned}z_{1}&=u+v\\z_{2}&=u\varepsilon _{1}+v\varepsilon _{2}\\z_{3}&=u\varepsilon _{2}+v\varepsilon _{1}\end{aligned}}}$

Relationship between the sign of the discriminant and the number of zeros

The solution behavior depends crucially on the sign of the discriminant:

• ${\displaystyle \Delta >0}$: There is exactly one real solution and two really complex solutions (graphic: case B).
• ${\displaystyle \Delta =0}$: There is either a double real solution and a simple real solution (case C) or a triple real solution (case A).
• ${\displaystyle \Delta <0}$: There are three different real solutions (case D).

In the case there are two options for the course of the associated graph: either (case B) or strictly monotonically increasing (not shown in the figure). ${\displaystyle \Delta >0}$

Δ> 0

One chooses the real roots for and in each case. There is exactly one real and two complex conjugate solutions that result from the above formulas ${\displaystyle u}$ ${\displaystyle v}$

${\displaystyle {\begin{aligned}z_{1}&=u+v\\z_{2,3}&=-{\frac {u+v}{2}}\pm {\frac {u-v}{2}}\,\mathrm {i} {\sqrt {3}}\end{aligned}}}$

given are.

So you get the solutions

${\displaystyle {\begin{aligned}x_{1}&=u+v-{\frac {B}{3A}}\\x_{2,3}&=-{\frac {u+v}{2}}-{\frac {B}{3A}}\pm {\frac {u-v}{2}}\,\mathrm {i} {\sqrt {3}}\end{aligned}}}$

However, extracting the cube roots is not always that easy. Cardano cites as an example: . Here we choose and real. Hence, and . Refer to the literature for techniques for pulling out nested roots . ${\displaystyle z^{3}+6z-20=0}$ ${\displaystyle \textstyle u={\sqrt[{3}]{10+{\sqrt {108}}}}=1+{\sqrt {3}}}$ ${\displaystyle \textstyle v={\sqrt[{3}]{10-{\sqrt {108}}}}=1-{\sqrt {3}}}$ ${\displaystyle z_{1}=2}$ ${\displaystyle z_{2,3}=-1\pm 3\mathrm {i} }$

Δ = p = 0

(Then is too .) In this case the only (three-fold) solution is, and we have: ${\displaystyle q=0}$
${\displaystyle z=0}$

${\displaystyle x_{1,2,3}=-{\frac {B}{3A}}}$

Δ = 0 and p ≠ 0

(Then is too .) In this case one chooses real. According to the above formulas there is then a simple real solution ${\displaystyle q\neq 0}$
${\displaystyle u=v}$

${\displaystyle z_{1}=2u={\sqrt[{3}]{-4q}}={\frac {3q}{p}}}$,

and a double real solution

${\displaystyle z_{2,3}=-u={\sqrt[{3}]{\frac {q}{2}}}=-{\frac {3q}{2p}}}$.

So you get the solutions

${\displaystyle {\begin{aligned}x_{1}&={\frac {3q}{p}}-{\frac {B}{3A}}&=&{\frac {B^{3}-4ABC+9A^{2}D}{3A^{2}C-AB^{2}}}\\x_{2,3}&=-{\frac {3q}{2p}}-{\frac {B}{3A}}&=&{\frac {BC-9AD}{6AC-2B^{2}}}\end{aligned}}}$

Δ <0 ( casus irreducibilis )

If you choose and conjugate complexes to each other, then three different real solutions result. ${\displaystyle u}$ ${\displaystyle v}$ ${\displaystyle z=u+v=u+{\overline {u}}=2\operatorname {Re} (u)}$

When determining , however, third roots must be calculated from genuinely complex numbers (e.g. using de Moivre's theorem ). That is why this case is called casus irreducibilis . With the help of the trigonometric functions, however, the solutions can also be calculated in real terms: According to the addition theorems , the relationship holds for all α ${\displaystyle u}$

${\displaystyle \cos ^{3}\alpha ={\frac {\cos 3\alpha +3\cos \alpha }{4}}\qquad \qquad (1)}$

You write

${\displaystyle 0=z^{3}+p\cdot z+q}$

with the help of the approach , results ${\displaystyle z=r\cdot \cos \alpha }$

${\displaystyle 0=r^{3}\cdot \cos ^{3}\alpha +p\cdot r\cdot \cos \alpha +q}$

If you start here , then it arises ${\displaystyle (1)\,}$

${\displaystyle {\begin{aligned}0&=r^{3}\cdot {\frac {\cos 3\alpha +3\cos \alpha }{4}}+p\cdot r\cdot \cos \alpha +q\\&={\frac {r^{3}}{4}}\cos 3\alpha +\left({\frac {3}{4}}r^{2}+p\right)\cdot r\cdot \cos \alpha +q\qquad \quad (2)\\&\,={\sqrt {-{\frac {4}{27}}\,p^{3}}}\cdot \cos 3\alpha +q\end{aligned}}}$

It was   chosen so that the expression in brackets in (2) disappears. It turns out ${\displaystyle \textstyle r={\sqrt {-{\frac {4}{3}}\,p}}}$

${\displaystyle {\begin{aligned}\cos 3\alpha &=-{\frac {q}{2}}\cdot {\sqrt {-{\frac {27}{p^{3}}}}}\\[1em]\iff \quad \alpha &={\frac {1}{3}}\arccos \left(-{\frac {q}{2}}\cdot {\sqrt {-{\frac {27}{p^{3}}}}}\right)+{\frac {2}{3}}k\pi \end{aligned}}}$

with whole numbers . ${\displaystyle k}$

Inserting in provides with and the following three solutions: ${\displaystyle z=r\cdot \cos \alpha }$ ${\displaystyle k=-1,0,1}$ ${\displaystyle \cos(\alpha \pm {\tfrac {2\pi }{3}})=-\cos(\alpha \mp {\tfrac {\pi }{3}})}$

${\displaystyle {\begin{aligned}z_{2}&=-\,{\sqrt {-{\frac {4}{3}}p}}\cdot \cos \left({\frac {1}{3}}\arccos \left(-{\frac {q}{2}}\cdot {\sqrt {-{\frac {27}{p^{3}}}}}\right)+{\frac {\pi }{3}}\right)\\[.7em]z_{1}&=\quad {\sqrt {-{\frac {4}{3}}p}}\cdot \cos \left({\frac {1}{3}}\arccos \left(-{\frac {q}{2}}\cdot {\sqrt {-{\frac {27}{p^{3}}}}}\right)\right)\\[.7em]z_{3}&=-\,{\sqrt {-{\frac {4}{3}}p}}\cdot \cos \left({\frac {1}{3}}\arccos \left(-{\frac {q}{2}}\cdot {\sqrt {-{\frac {27}{p^{3}}}}}\right)-{\frac {\pi }{3}}\right)\end{aligned}}}$

So the equation has the following three solutions: ${\displaystyle Ax^{3}+Bx^{2}+Cx+D=0}$

${\displaystyle {\begin{aligned}x_{2}&=-\,{\sqrt {-{\frac {4}{3}}p}}\cdot \cos \left({\frac {1}{3}}\arccos \left(-{\frac {q}{2}}\cdot {\sqrt {-{\frac {27}{p^{3}}}}}\right)+{\frac {\pi }{3}}\right)-{\frac {B}{3A}}\\[.7em]x_{1}&=\quad {\sqrt {-{\frac {4}{3}}p}}\cdot \cos \left({\frac {1}{3}}\arccos \left(-{\frac {q}{2}}\cdot {\sqrt {-{\frac {27}{p^{3}}}}}\right)\right)-{\frac {B}{3A}}\\[.7em]x_{3}&=-\,{\sqrt {-{\frac {4}{3}}p}}\cdot \cos \left({\frac {1}{3}}\arccos \left(-{\frac {q}{2}}\cdot {\sqrt {-{\frac {27}{p^{3}}}}}\right)-{\frac {\pi }{3}}\right)-{\frac {B}{3A}}\end{aligned}}}$

Derivation of the discriminant via the differential calculus

Relationship between the sign of the discriminant and the number of zeros

To do this, one has to transition to the differential calculation . As can be seen in the graphic, the equation can only have exactly one real solution and two really complex solutions if both extreme points are above or below the -axis or if there are no extreme points, in the case of three different real solutions the high point is ( Extreme point: maximum) above and the lowest point (extreme point: minimum) below the -axis and in the case of multiple real zeros there are extreme points on the -axis. In the case of a double zero, these are high or low points and in the case of a triple zero, saddle points. ${\displaystyle x}$ ${\displaystyle x}$ ${\displaystyle x}$

Extreme points have the property that there the function neither rises nor falls, but its slope is exactly zero. The slope of a function at the point results from the equation: ${\displaystyle f}$ ${\displaystyle P(x_{1}\mid f(x_{1}))}$

${\displaystyle f'(x_{1})=\lim _{x_{1}\to x_{0}}{\frac {f(x_{1})-f(x_{0})}{x_{1}-x_{0}}}=\lim _{h\to 0}{\frac {f(x_{1}+h)-f(x_{1})}{h}}}$   (with ) ${\displaystyle h=x_{1}-x_{0}}$

${\displaystyle f'}$means the first derivative function . describes the second derivative function. applies if and only if there is a turning point. In the case there is a saddle point.
${\displaystyle f''}$ ${\displaystyle f''=0}$
${\displaystyle f''=f'=0}$

If we write as a function , it looks like this: ${\displaystyle z^{3}+pz+q=0}$ ${\displaystyle f(z)}$

${\displaystyle f(z)=z^{3}+pz+q}$

Their first and second derivatives are:

${\displaystyle f'(z)=3z^{2}+p}$ and

${\displaystyle f''(z)=6z}$.

Solve the two differential equations:

Extreme points: and ${\displaystyle f'(z_{E1,E2})=3z^{2}+p=0}$

Turning points : , ${\displaystyle f''(z_{W})=6z}$

so you get:

${\displaystyle z_{E1,E2}=\pm {\sqrt {\left({\dfrac {-p}{3}}\right)}}}$ and

${\displaystyle z_{W}=0}$.

Their functional values ​​are:

${\displaystyle {\begin{aligned}f(z_{E1,E2})&=z_{E1,E2}^{3}+p\cdot z_{E1,E2}+q\\&=(\pm {\sqrt {\left({\dfrac {-p}{3}}\right)}})^{3}+p\cdot (\pm {\sqrt {\left({\dfrac {-p}{3}}\right)}})+q\\&=(\pm {\sqrt {\left({\dfrac {-p}{3}}\right)}})\cdot ((\pm {\sqrt {\left({\dfrac {-p}{3}}\right)}})^{2}+p)+q\\&=(\pm {\sqrt {\left({\dfrac {-p}{3}}\right)}})\cdot ((\left({\dfrac {-p}{3}}\right))+p)+q\\&=(\pm {\sqrt {\left({\dfrac {-p}{3}}\right)}})\cdot ((\left({\dfrac {2p}{3}}\right)))+q\end{aligned}}}$

and . ${\displaystyle f(z_{W})=z_{W}^{3}+p\cdot z_{W}+q=0^{3}+p\cdot 0+q=q}$

The first solution can be reformulated as follows:

${\displaystyle {\begin{array}{rcl}f(z_{E1,E2})=z_{E1,E2}^{3}+p\cdot z_{E1,E2}+q=(\pm {\sqrt {\left({\dfrac {-p}{3}}\right)}})\cdot \left({\dfrac {2p}{3}}\right)+q\end{array}}}$

${\displaystyle (\pm {\sqrt {\left({\dfrac {-p}{3}}\right)}})\cdot \left({\dfrac {2p}{3}}\right)+q=0\Leftrightarrow \ (\pm {\sqrt {\left({\dfrac {-p}{3}}\right)}})\cdot \left({\dfrac {2p}{3}}\right)=-q\Leftrightarrow \ (\mp {\sqrt {\left({\dfrac {-p}{3}}\right)}})\cdot \left({\dfrac {2p}{3}}\right)=q\Leftrightarrow \ \left({\dfrac {-p}{3}}\right)\cdot \left({\dfrac {2p}{3}}\right)^{2}=-2^{2}\cdot \left({\dfrac {p}{3}}\right)^{3}=q^{2}\qquad \qquad (1)}$

${\displaystyle (\pm {\sqrt {\left({\dfrac {-p}{3}}\right)}})\cdot \left({\dfrac {2p}{3}}\right)+q>0\Leftrightarrow \ (\pm {\sqrt {\left({\dfrac {-p}{3}}\right)}})\cdot \left({\dfrac {2p}{3}}\right)>-q\Leftrightarrow \ (\mp {\sqrt {\left({\dfrac {-p}{3}}\right)}})\cdot \left({\dfrac {2p}{3}}\right)<q\qquad \qquad (2)}$

${\displaystyle (\pm {\sqrt {\left({\dfrac {-p}{3}}\right)}})\cdot \left({\dfrac {2p}{3}}\right)+q<0\Leftrightarrow \ (\pm {\sqrt {\left({\dfrac {-p}{3}}\right)}})\cdot \left({\dfrac {2p}{3}}\right)<-q\Leftrightarrow \ (\mp {\sqrt {\left({\dfrac {-p}{3}}\right)}})\cdot \left({\dfrac {2p}{3}}\right)>q\qquad \qquad (3)}$

In cases (2) and (3) one cannot square without problems, since after squaring the relation sign can be reversed according to the inversion rule. again it can be positive or negative, so that one should proceed with the help of (“amount of q ”). A total of four sub-cases can be distinguished. In sub-cases (a) and (b) the left side is positive, in sub-cases (c) and (d) the left side is negative. ${\displaystyle q}$ ${\displaystyle |q|}$

First the case (2):

Left side> 0, q > 0

${\displaystyle (\mp {\sqrt {\left({\dfrac {-p}{3}}\right)}})\cdot \left({\dfrac {2p}{3}}\right)<q\Leftrightarrow \ |({\sqrt {\left({\dfrac {-p}{3}}\right)}})\cdot \left({\dfrac {2p}{3}}\right)|<|q|\Rightarrow \ \left({\dfrac {-p}{3}}\right)\cdot \left({\dfrac {2p}{3}}\right)^{2}=-2^{2}\cdot \left({\dfrac {p}{3}}\right)^{3}<q^{2}\qquad \qquad (2a)}$

Left side> 0, q ≤ 0

${\displaystyle (\mp {\sqrt {\left({\dfrac {-p}{3}}\right)}})\cdot \left({\dfrac {2p}{3}}\right)<q\Leftrightarrow \ 0<|({\sqrt {\left({\dfrac {-p}{3}}\right)}})\cdot \left({\dfrac {2p}{3}}\right)|<-|q|\leq 0}$ is a false statement ${\displaystyle \qquad \qquad (2b)}$

Left side ≤ 0, q> 0

${\displaystyle (\mp {\sqrt {\left({\dfrac {-p}{3}}\right)}})\cdot \left({\dfrac {2p}{3}}\right)<q\Leftrightarrow \ -|({\sqrt {\left({\dfrac {-p}{3}}\right)}})\cdot \left({\dfrac {2p}{3}}\right)|<|q|\Leftrightarrow \ |({\sqrt {\left({\dfrac {-p}{3}}\right)}})\cdot \left({\dfrac {2p}{3}}\right)|>-|q|}$ is always true ${\displaystyle \qquad \qquad (2c)}$

Left side ≤ 0, q ≤ 0

${\displaystyle (\mp {\sqrt {\left({\dfrac {-p}{3}}\right)}})\cdot \left({\dfrac {2p}{3}}\right)<q\Leftrightarrow \ -|({\sqrt {\left({\dfrac {-p}{3}}\right)}})\cdot \left({\dfrac {2p}{3}}\right)|<-|q|\leq 0\Leftrightarrow \ |({\sqrt {\left({\dfrac {-p}{3}}\right)}})\cdot \left({\dfrac {2p}{3}}\right)|>|q|\geq 0\Rightarrow \ \left({\dfrac {-p}{3}}\right)\cdot \left({\dfrac {2p}{3}}\right)^{2}=-2^{2}\cdot \left({\dfrac {p}{3}}\right)^{3}>q^{2}\qquad \qquad (2d)}$

Case (3) leads to similar results, only in a different order.

The reformulation of the equations (first division by 4, then the expression on the left is transferred to the right side) results in: ${\displaystyle p}$

${\displaystyle -2^{2}\cdot \left({\dfrac {p}{3}}\right)^{3}=q^{2}\Leftrightarrow \left({\dfrac {q}{2}}\right)^{2}+\left({\dfrac {p}{3}}\right)^{3}=0\qquad \qquad (1)}$

${\displaystyle -2^{2}\cdot \left({\dfrac {p}{3}}\right)^{3}>q^{2}\Leftrightarrow \left({\dfrac {q}{2}}\right)^{2}+\left({\dfrac {p}{3}}\right)^{3}<0\qquad \qquad (2)}$

${\displaystyle -2^{2}\cdot \left({\dfrac {p}{3}}\right)^{3}<q^{2}\Leftrightarrow \left({\dfrac {q}{2}}\right)^{2}+\left({\dfrac {p}{3}}\right)^{3}>0\qquad \qquad (3)}$

${\displaystyle \Delta :=\left({\frac {q}{2}}\right)^{2}+\left({\frac {p}{3}}\right)^{3}}$

Complex coefficients

The procedure is largely the same for complex coefficients, but there are only two cases:

• ${\displaystyle \Delta =0}$: This is the criterion for multiple zeros even in complexes. The formulas given above for this case apply unchanged.
• ${\displaystyle \Delta \neq 0}$: The formulas given above for the case apply analogously; the two third roots are to be chosen so that their product results. Extracting the complex third roots in a trigonometric way leads to a solution that corresponds to that given for the case , the casus irreducibilis . The angle must be adapted to the complex radicands . ${\displaystyle \Delta >0}$ ${\displaystyle -{\tfrac {p}{3}}}$ ${\displaystyle \Delta <0}$ ${\displaystyle 3\alpha }$ ${\displaystyle -{\tfrac {q}{2}}\pm {\sqrt {\Delta }}}$

literature

• Jörg Bewersdorff : Algebra for Beginners : From Equation Resolution to Galois Theory, Wiesbaden 2004, ISBN 3-528-13192-6 , introduction (PDF; 319 kB)
• Heinrich Dörrie : Cubic and biquadratic equations , Munich 1948
• Ludwig Matthiessen : Fundamentals of the ancient and modern algebra of the litteral equations , Leipzig 1896, document server
• Peter Pesic: Abel's proof , Springer 2005, ISBN 3-540-22285-5 . The story about the solution formulas from degree 2 to 4 and the complete proof of Abel.

Web links

• Cardano formulas for solving the equation of the third degree on mathematik.ch