note[4]
例:解方程 \[ x^4 - 2x^3 +x^2 - 15 = 0 \]
解:
令 \(y=x-\dfrac{1}{2}\) ,则 \[ -15 + \left(y + \dfrac{1}{2}\right)^2 - 2\left(y+\dfrac{1}{2}\right)^3 + \left(y + \dfrac{1}{2}\right)^4 = 0 \]
目的在于构造无奇次项的多项式。
构造方式如下:
设 \(y = x + a\)
然后代入原式,找到 \(y^3\) 的系数,这题是 \(-4a-2\)
故令 \(-4a-2=0\) 解得 \(a = -\dfrac{1}{2}\)
全部展开并整理得到 \[ y^4 - \dfrac{y^2}{2} - \dfrac{239}{16} = 0 \] 令 \(z = y^2\) ,则 \[ z^2 - \dfrac{z}{2} - \dfrac{239}{16} = 0 \] 解得 \[ z=\dfrac{1}{4} + \sqrt{15} \quad\mathtt{or}\quad z=\dfrac{1}{4} - \sqrt{15} \] 则 \[ y = \sqrt{\dfrac{1}{4} + \sqrt{15}}\quad\mathtt{or}\quad y=-\sqrt{\dfrac{1}{4} + \sqrt{15}}\quad\mathtt{or}\quad z=\dfrac{1}{4} - \sqrt{15} \]
\[ x = \dfrac{1}{2} + \sqrt{\dfrac{1}{4} + \sqrt{15}}\quad\mathtt{or}\quad x= \dfrac{1}{2} -\sqrt{\dfrac{1}{4} + \sqrt{15}}\quad\mathtt{or}\quad z=\dfrac{1}{4} - \sqrt{15} \]
同理,可以解出 \(x\) 的共轭复根。整理得 \[ \begin{aligned} x_1 &= \dfrac{1}{2} + \sqrt{\dfrac{1}{4} + \sqrt{15}} \\x_2&= \dfrac{1}{2} - \sqrt{\dfrac{1}{4} + \sqrt{15}} \\x_3&= \dfrac{1}{2} + i\sqrt{-\dfrac{1}{4}+\sqrt{15}} \\x_4&= \dfrac{1}{2} - i\sqrt{-\dfrac{1}{4}+\sqrt{15}} \end{aligned} \]
参考文献:
[1] https://www.wolframalpha.com/input?i=x%5E4+-+2x%5E3+%2Bx%5E2+-+15+%3D+0