mit ![{\displaystyle a\in \mathbb {R} \setminus \{0\},b\in \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/159e79bfdd6b64035df777d188b24588515bbdfd)
![{\displaystyle ax^{2}+bx+c=0,\quad a\neq 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/35e2b05d123e361ea9da24df2c14d436cfed934d)
Man benutze die sog. „Mitternachtsformel“:
![{\displaystyle x_{1,2}={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a0252e1df0e4d27a338a3725eccc16e5b803c11d)
Kubische Gleichung:
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}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/15a775331163e6d0c076833f18038d5ab3da0f0e)
in die Form
bringen.
Lösung
- Fall 1:
![{\displaystyle {\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}>0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/31d94a7668beea4b1e032b36952de62bf8cc276b)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e8062a0bbf133e2f4a27a7fb98aa6df146ebb54d)
- Fall 2:
![{\displaystyle {\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3bef996f26f4d75ba8965128c9d171580f3c332c)
Zwei reelle Lösungen:
![{\displaystyle x_{1}={\sqrt[{3}]{\frac {q}{2}}}-{\frac {b}{3a}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/215a3e7ec5e47a0d28ec1949000c873651cf0706)
![{\displaystyle x_{2}=-{\sqrt[{3}]{4q}}-{\frac {b}{3a}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/82ff4a9edebf265f3a109b3f71d987b25d92fc1f)
- Fall 3:
(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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aec84a3eccf61bd655487db9958564b09769540e)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/853f478ae6559b517ea31bff6a93195ecd787196)
Substitution mit
z ist eine beliebige Lösung der Gleichung
.
Mit
ergibt sich:
- Fall 1:
![{\displaystyle {\frac {b}{a}}y-{\frac {d}{a}}>0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c28d5820361d436b5f896e128e8116a507079c1c)
![{\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)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/20bc4ed953e615888f03c7b6a1b805727868814f)
- Fall 2:
![{\displaystyle {\frac {b}{a}}y-{\frac {d}{a}}<0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb57df8bfc59d7360fff4fb61d1ef2846a94e8fd)
![{\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)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/709381352a83ed072ddc9f58f3c3576dea539b44)