5
三次四次方程的公式通解
source link: https://zhiqiang.org/math/root-formula-of-cubic-and-quartic-equations.html
Go to the source link to view the article. You can view the picture content, updated content and better typesetting reading experience. If the link is broken, please click the button below to view the snapshot at that time.
1、三次方程的公式通解
对于ax3+bx2+cx+d=0，有：
x_1=-\frac{b}{3 a}+ \sqrt[3]{\frac{bc}{6a^2}-\frac{b^3}{27a^3}-\frac{d}{2a}+\sqrt{{\color{red}\left(\frac{bc}{6a^2}-\frac{b^3}{27a^3}-\frac{d}{2a}\right)^2+ \left(\frac{c}{3a}-\frac{b^2}{9a^2}\right)^3}}}+\sqrt[3]{\frac{bc}{6a^2}-\frac{b^3}{27a^3}-\frac{d}{2a}-\sqrt{\left(\frac{bc}{6a^2}-\frac{b^3}{27a^3}-\frac{d}{2a}\right)^2+ \left(\frac{c}{3a}-\frac{b^2}{9a^2}\right)^3}}
x2=−b3a+−1+√3i23√bc6a2−b327a3−d2a+√(bc6a2−b327a3−d2a)2+(c3a−b29a2)3+−1−√3i23√bc6a2−b327a3−d2a−√(bc6a2−b327a3−d2a)2+(c3a−b29a2)3
x3=−b3a+−1−√3i23√bc6a2−b327a3−d2a+√(bc6a2−b327a3−d2a)2+(c3a−b29a2)3+−1+√3i23√bc6a2−b327a3−d2a−√(bc6a2−b327a3−d2a)2+(c3a−b29a2)3
其中判别式Δ为：
Δ = √(bc6a2−b327a3−d2a)2+(c3a−b29a2)3
2、四次方程的公式解
对于ax4+bx3+cx2+dx+e=0，有：
x1=−b4a+12√(b2a)2−2c3a+3√2(c2−3bd+12ae)3a3√2c3−9bcd+27ad2+27b2e−72ace+√(2c3−9bcd+27ad2+27b2e−72ace)2−4(c2−3bd+12ae)3+3√2c3−9bcd+27ad2+27b2e−72ace+√(2c3−9bcd+27ad2+27b2e−72ace)2−4(c2−3bd+12ae)333√2a−12√b22a2−4c3a−3√2(c2−3bd+12ae)3a3√2c3−9bcd+27ad2+27b2e−72ace+√(2c3−9bcd+27ad2+27b2e−72ace)2−4(c2−3bd+12ae)3−3√2c3−9bcd+27ad2+27b2e−72ace+√(2c3−9bcd+27ad2+27b2e−72ace)2−4(c2−3bd+12ae)333√2a+−b3+4abc−8a2d4a3√(b2a)2−2c3a+3√2(c2−3bd+12ae)3a3√2c3−9bcd+27ad2+27b2e−72ace+√(2c3−9bcd+27ad2+27b2e−72ace)2−4(c2−3bd+12ae)3+3√2c3−9bcd+27ad2+27b2e−72ace+√(2c3−9bcd+27ad2+27b2e−72ace)2−4(c2−3bd+12ae)333√2a
x2=−b4a+12√(b2a)2−2c3a+3√2(c2−3bd+12ae)3a3√2c3−9bcd+27ad2+27b2e−72ace+√(2c3−9bcd+27ad2+27b2e−72ace)2−4(c2−3bd+12ae)3+3√2c3−9bcd+27ad2+27b2e−72ace+√(2c3−9bcd+27ad2+27b2e−72ace)2−4(c2−3bd+12ae)333√2a+12√b22a2−4c3a−3√2(c2−3bd+12ae)3a3√2c3−9bcd+27ad2+27b2e−72ace+√(2c3−9bcd+27ad2+27b2e−72ace)2−4(c2−3bd+12ae)3−3√2c3−9bcd+27ad2+27b2e−72ace+√(2c3−9bcd+27ad2+27b2e−72ace)2−4(c2−3bd+12ae)333√2a−b3−4abc+8a2d4a3√(b2a)2−2c3a+3√2(c2−3bd+12ae)3a3√2c3−9bcd+27ad2+27b2e−72ace+√(2c3−9bcd+27ad2+27b2e−72ace)2−4(c2−3bd+12ae)3+3√2c3−9bcd+27ad2+27b2e−72ace+√(2c3−9bcd+27ad2+27b2e−72ace)2−4(c2−3bd+12ae)333√2a
x3=−b4a−12√(b2a)2−2c3a+3√2(c2−3bd+12ae)3a3√2c3−9bcd+27ad2+27b2e−72ace+√(2c3−9bcd+27ad2+27b2e−72ace)2−4(c2−3bd+12ae)3+3√2c3−9bcd+27ad2+27b2e−72ace+√(2c3−9bcd+27ad2+27b2e−72ace)2−4(c2−3bd+12ae)333√2a−12√b22a2−4c3a−3√2(c2−3bd+12ae)3a3√2c3−9bcd+27ad2+27b2e−72ace+√(2c3−9bcd+27ad2+27b2e−72ace)2−4(c2−3bd+12ae)3−3√2c3−9bcd+27ad2+27b2e−72ace+√(2c3−9bcd+27ad2+27b2e−72ace)2−4(c2−3bd+12ae)333√2a+b3−4abc+8a2d4a3√(b2a)2−2c3a+3√2(c2−3bd+12ae)3a3√2c3−9bcd+27ad2+27b2e−72ace+√(2c3−9bcd+27ad2+27b2e−72ace)2−4(c2−3bd+12ae)3+3√2c3−9bcd+27ad2+27b2e−72ace+√(2c3−9bcd+27ad2+27b2e−72ace)2−4(c2−3bd+12ae)333√2a
x4=−b4a−12√(b2a)2−2c3a+3√2(c2−3bd+12ae)3a3√2c3−9bcd+27ad2+27b2e−72ace+√(2c3−9bcd+27ad2+27b2e−72ace)2−4(c2−3bd+12ae)3+3√2c3−9bcd+27ad2+27b2e−72ace+√(2c3−9bcd+27ad2+27b2e−72ace)2−4(c2−3bd+12ae)333√2a+12√b22a2−4c3a−3√2(c2−3bd+12ae)3a3√2c3−9bcd+27ad2+27b2e−72ace+√(2c3−9bcd+27ad2+27b2e−72ace)2−4(c2−3bd+12ae)3−3√2c3−9bcd+27ad2+27b2e−72ace+√(2c3−9bcd+27ad2+27b2e−72ace)2−4(c2−3bd+12ae)333√2a+b3−4abc+8a2d4a3√(b2a)2−2c3a+3√2(c2−3bd+12ae)3a3√2c3−9bcd+27ad2+27b2e−72ace+√(2c3−9bcd+27ad2+27b2e−72ace)2−4(c2−3bd+12ae)3+3√2c3−9bcd+27ad2+27b2e−72ace+√(2c3−9bcd+27ad2+27b2e−72ace)2−4(c2−3bd+12ae)333√2a
其中判别式Δ为：
Δ = 256a3e3−192a2bde2−128a2c2e2+144a2cd2e−27a2d4+144ab2ce2−6ab2d2e−80abc2de+18abcd3+16ac4e−4ac3d2−27b4e2+18b3cde−4b3d3−4b2c3e+b2c2d2
### 五次及以上方程的公式通解
根据伽罗瓦理论，五次及以上方程无公式通解。
Q. E. D.
Recommend
About Joyk
Aggregate valuable and interesting links.
Joyk means Joy of geeK