Wikibooks:Abstellraum: Strukturwissenschaften: Lösung von Gleichungen

${\displaystyle a\cdot x=b}$ mit ${\displaystyle a\in \mathbb {R} \setminus \{0\},b\in \mathbb {R} }$

${\displaystyle x={\frac {b}{a}}}$

${\displaystyle ax^{2}+bx+c=0,\quad a\neq 0}$

Man benutze die sog. „Mitternachtsformel“:

${\displaystyle x_{1,2}={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.}$

Kubische Gleichung: ${\displaystyle ax^{3}+bx^{2}+cx+d=0}$

Durch Substitution mit

${\displaystyle p={\frac {c}{a}}-{\frac {b^{2}}{3a^{2}}};\qquad q={\frac {d}{a}}+{\frac {2b^{3}}{27a^{3}}}-{\frac {bc}{3a^{2}}}}$

in die Form ${\displaystyle x^{3}+px+q=0}$ bringen.

Lösung

• Fall 1: ${\displaystyle {\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}>0}$

Eine reelle Lösung:

${\displaystyle x={\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}+{\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}-{\frac {b}{3a}}}$

• Fall 2: ${\displaystyle {\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}=0}$

Zwei reelle Lösungen:

${\displaystyle x_{1}={\sqrt[{3}]{\frac {q}{2}}}-{\frac {b}{3a}}}$

${\displaystyle x_{2}=-{\sqrt[{3}]{4q}}-{\frac {b}{3a}}}$

• Fall 3: ${\displaystyle {\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}<0}$ (Casus irreducibilis)

Drei reelle Lösungen:

${\displaystyle x_{1}=2{\sqrt {-{\frac {p}{3}}}}\cos \left({\frac {1}{3}}\arccos \left(-{\frac {q}{2}}{\sqrt {-{\frac {27}{p^{3}}}}}\right)\right)-{\frac {b}{3a}}}$

${\displaystyle x_{2,3}=-2{\sqrt {-{\frac {p}{3}}}}\cos \left({\frac {1}{3}}\arccos \left(-{\frac {q}{2}}{\sqrt {-{\frac {27}{p^{3}}}}}\right)\pm {\frac {\pi }{3}}\right)-{\frac {b}{3a}}}$

${\displaystyle ax^{4}+bx^{3}+cx^{2}+dx+e=0}$

Substitution mit ${\displaystyle p={\frac {3bd-c^{2}}{12a^{2}}}-{\frac {e}{a}};\qquad q={\frac {8ce-3d^{2}}{24a^{2}}}-{\frac {27b^{2}e-9bcd+2c^{3}}{216a^{3}}}}$

z ist eine beliebige Lösung der Gleichung ${\displaystyle z^{3}+pz+q=0}$.

Mit ${\displaystyle y=z+{\frac {c}{6a}}}$ ergibt sich:

• Fall 1: ${\displaystyle {\frac {b}{a}}y-{\frac {d}{a}}>0}$

${\displaystyle x_{1,2,3,4}=-{\frac {b}{4a}}\pm {\frac {1}{2}}{\sqrt {2y+{\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}}}+{\sqrt {{\frac {b^{2}}{8a^{2}}}-{\frac {1}{2}}y-{\frac {c}{4a}}\pm \left({\frac {b}{4a}}{\sqrt {2y+{\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}}}-{\sqrt {y^{2}-{\frac {e}{a}}}}\right)}}}$

• Fall 2: ${\displaystyle {\frac {b}{a}}y-{\frac {d}{a}}<0}$

${\displaystyle x_{1,2}=-{\frac {b}{4a}}\pm {\frac {1}{2}}{\sqrt {2y+{\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}}}-{\sqrt {{\frac {b^{2}}{8a^{2}}}-{\frac {1}{2}}y-{\frac {c}{4a}}\pm \left({\frac {b}{4a}}{\sqrt {2y+{\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}}}+{\sqrt {y^{2}-{\frac {e}{a}}}}\right)}}}$