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MathWorld
CRC Archive
对于四次方程 $ x^4+px^2+qx+r = 0$ 因为 $x_1+x_2+x_3+x_4= 0$ 可定义
\begin{align*}
\alpha &= (x_1+x_2)(x_3+x_4)=-(x_1+x_2)^2 \\
\beta &= (x_1+x_3)(x_2+x_4)=-(x_1+x_3)^2 \\
\gamma &= (x_1+x_4)(x_2+x_3)=-(x_2+x_3)^2
\end{align*}The resolvent cubic 定义为
\begin{align*}
h(x) &= (x-\alpha)(x-\beta)(x-\gamma)\\
&= x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\alpha\gamma+\beta\gamma)x-\alpha\beta\gamma.
\end{align*}比较
\begin{align*}
P(x) &= x^4+px^2+qx+r\\
&= (x-x_1)(x-x_2)(x-x_3)(x-x_4) \\
&= x^4+\left({\,\prod_{i\not=j}^4 x_ix_j}\right)x^2+(x_1+x_2)(x_1+x_3)(x_2+x_3)x \\
&\phantom{=}-x_1x_2x_3(x_1+x_2+x_3)
\end{align*}
得
\begin{align*}
h(x) &= x^3-2px^2=(p^2-r)z+q^2.
\end{align*}
解这个三次方程得 $\alpha$, $\beta$, 与 $\gamma$, 从而解出 $x_i$.
Faucette 1996 A Geometric Interpretation of the Solution of the General Quartic Polynomial JSTOR
Dave Auckly 2007 Solving the Quartic with a Pencil PDF
Degenerate conic
geometric solution to a quartic equation |
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