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lectures:三次方程式

三次方程式

$$x^3+ax^2+bx+c=0 $$

標準化

$$(x-r)^3+a(x-r)^2+b(x-r)+c=0$$ $$x^3-3rx^2+3r^2x-r^3+ax^2-2arx+ar^2+bx-br+c=0$$ $$x^3+(a-3r)x^2+(3r^2-2ar+b)x+ar^2-r^3-br+c=0$$

1実根

$$ x^3+3px-2q=0 $$

$$ x=(q+\sqrt{D})^{\frac{1}{3}}+(q-\sqrt{D})^{\frac{1}{3}} $$

$$ \begin{align} x^3&=(q+\sqrt{D})+3(q+\sqrt{D})^{\frac{2}{3}}(q-\sqrt{D})^{\frac{1}{3}}+3(q+\sqrt{D})^{\frac{1}{3}}(q-\sqrt{D})^{\frac{2}{3}}+(q-\sqrt{D})\\ &=2q+3(q+\sqrt{D})^{\frac{1}{3}}(q-\sqrt{D})^{\frac{1}{3}}\left((q+\sqrt{D})^{\frac{1}{3}}+(q-\sqrt{D})^{\frac{1}{3}}\right)\\ &=2q+3(q^2-D)^{\frac{1}{3}}\left((q+\sqrt{D})^{\frac{1}{3}}+(q-\sqrt{D})^{\frac{1}{3}}\right)\\ \end{align} $$ $$(q^2-D)^{\frac{1}{3}}=-p$$ $$D=q^2+p^3$$ ===== 因数分解 ===== $$x^3+3px-2q=(x-\alpha)(x-\beta)(x-\gamma)$$ $$ \begin{align} \alpha+\beta+\gamma=0\\ \alpha\beta+\beta\gamma+\gamma\alpha=3p\\ \alpha\beta\gamma=2q\\ \end{align} $$ $$\alpha+\beta=-\gamma$$ $$\alpha\beta=3p-(\alpha+\beta)\gamma=3p+\gamma^2$$ $$x^3+3px-2q=(x^2+\gamma x+3p+\gamma^2)(x-\gamma)$$

lectures/三次方程式.txt · 最終更新: 2021/12/25 14:09 by kimi

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