四次方程求根公式
对于 ax4+bx3+cx2+dx+e=0 (a≠0),其通解为:
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
你不如直接5层for
如果是计算几何double大概率也过不去,long double也基本被卡时间(
不愧是被认为又臭又长的四次求根公式啊