طريقة ديكارت

في كتابه الهندسة (1637)، طبق ديكارت هذه الطريقة لحل المعادلات من الدرجة الرابعة بمجهول واحد:

نعتبر معادلة رباعية : ${\displaystyle ax^{4}+bx^{3}+cx^{2}+dx+e=0~}$

نضع

${\displaystyle x=z-{\frac {b}{4a}}~}$

فنجد:

${\displaystyle z^{4}+pz^{2}+qz+r=0~}$

${\displaystyle z^{4}+pz^{2}+qz+r=(z^{2}+az+b)(z^{2}-az+c)~}$

ننشر فنجد:

${\displaystyle z^{4}+pz^{2}+qz+r=z^{4}+(b+c-a^{2})z^{2}-a(b-c)z+bc~}$

حسب خاصية تساوي حدوديتين نستنتج أن:

${\displaystyle {\begin{cases}b+c-a^{2}=p\\a(b-c)=-q\\bc=r\end{cases}}}$

${\displaystyle {\begin{cases}b+c=a^{2}+p\\b-c=-{\frac {q}{a}}\\bc=r\end{cases}}}$

من (أ)+(ب) و (أ)-(ب) نستنتج أن${\displaystyle {\begin{cases}b={\frac {1}{2}}\left(a^{2}+p-{\frac {q}{a}}\right)\\c={\frac {1}{2}}\left(a^{2}+p+{\frac {q}{a}}\right)\\bc=r.\end{cases}}}$

${\displaystyle {\begin{cases}b={\frac {1}{2}}\left(a^{2}+p-{\frac {q}{a}}\right)\\c={\frac {1}{2}}\left(a^{2}+p+{\frac {q}{a}}\right)\\{\frac {1}{4}}\left(a^{2}+p-{\frac {q}{a}}\right)\left(a^{2}+p+{\frac {q}{a}}\right)=r\end{cases}}}$

من (ت) نستنتج أن

${\displaystyle (a^{2}+p)^{2}-{\frac {q^{2}}{a^{2}}}=4r}$.

${\displaystyle a^{6}+2pa^{4}+(p^{2}-4r)a^{2}-q^{2}=0}$.

نضع a²=y

${\displaystyle y^{3}+2py^{2}+(p^{2}-4r)y-q^{2}=0}$.

${\displaystyle a={\sqrt {y_{1}}}}$.

${\displaystyle {\begin{cases}b={\frac {1}{2}}\left(y_{1}+p-{\frac {q}{\sqrt {y_{1}}}}\right)\\c={\frac {1}{2}}\left(y_{1}+p+{\frac {q}{\sqrt {y_{1}}}}\right)\\a={\sqrt {y_{1}}}.\end{cases}}}$

إذن ${\displaystyle z^{4}+pz^{2}+qz+r=0}$ ${\displaystyle \left(z^{2}+z{\sqrt {y_{1}}}+{\frac {1}{2}}\left(y_{1}+p-{\frac {q}{\sqrt {y_{1}}}}\right)\right)\left(z^{2}-z{\sqrt {y_{1}}}+{\frac {1}{2}}\left(y_{1}+p+{\frac {q}{\sqrt {y_{1}}}}\right)\right)=0}$.

• ${\displaystyle z^{2}+z{\sqrt {y_{1}}}+{\frac {1}{2}}\left(y_{1}+p-{\frac {q}{\sqrt {y_{1}}}}\right)=0}$
• ${\displaystyle z^{2}-z{\sqrt {y_{1}}}+{\frac {1}{2}}\left(y_{1}+p+{\frac {q}{\sqrt {y_{1}}}}\right)=0}$

${\displaystyle z_{1}={\frac {1}{2}}\left(-{\sqrt {y_{1}}}+{\sqrt {-y_{1}-2p+{\frac {2q}{\sqrt {y_{1}}}}}}\right)}$

${\displaystyle z_{2}={\frac {1}{2}}\left(-{\sqrt {y_{1}}}-{\sqrt {-y_{1}-2p+{\frac {2q}{\sqrt {y_{1}}}}}}\right)}$

${\displaystyle z_{3}={\frac {1}{2}}\left({\sqrt {y_{1}}}+{\sqrt {-y_{1}-2p-{\frac {2q}{\sqrt {y_{1}}}}}}\right)}$

${\displaystyle z_{4}={\frac {1}{2}}\left({\sqrt {y_{1}}}-{\sqrt {-y_{1}-2p-{\frac {2q}{\sqrt {y_{1}}}}}}\right)}$.

https://fr.wikiversity.org/wiki/%C3%89quation_du_quatri%C3%A8me_degr%C3%A9/M%C3%A9thode_de_Descartes

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