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一般一元四次方程的通解

2022-08-02 16:00 作者:げいしも_蕓  | 我要投稿


解題思路:

使用換元使原四次方程化為

%5Clambda%5E4%3Dp%5Clambda%5E2%2Bq%5Clambda%2Br%EF%BC%88p%2Cq%2Cr%5Cin%20%5Ctext%20R%EF%BC%89形式

后對兩邊配方,變形為

(%5Clambda%5E2%2By)%5E2%3D(u%5Clambda%2Bv)%5E2%EF%BC%88y%2Cu%2Cv%5Cin%20%5Ctext%20C%EF%BC%89

再簡化為關于%5Clambda的一元二次方程,解出%5Clambda,進而解出x

求解過程:

%E5%AF%B9%E4%BA%8Eax%5E4%2Bbx%5E3%2Bcx%5E2%2Bdx%2Be%3D0%EF%BC%88a%2Cb%2Cc%2Cd%2Ce%5Cin%20%5Ctext%20R%EF%BC%8Ca%E2%89%A00)

%E4%BB%A4B%3D%5Cfrac%20ba%EF%BC%8CC%3D%5Cfrac%20ca%EF%BC%8CD%3D%5Cfrac%20da%EF%BC%8CE%3D%5Cfrac%20ea

%E4%BA%8E%E6%98%AFx%5E4%2BBx%5E3%2BCx%5E2%2BDx%2BE%3D0

%E4%BB%A4x%3Dz%2Bt%EF%BC%8C%E4%BB%A3%E5%85%A5%E5%90%8E%E5%B1%95%E5%BC%80%EF%BC%8C%E5%BE%97%E5%88%B0%EF%BC%9A

(z%5E4%2B4z%5E3t%2B6z%5E2t%5E2%2B4zt%5E3%2Bt%5E4)%5C%5C%2BB(z%5E3%2B3z%5E2t%2B3zt%5E2%2Bt%5E3)%5C%5C%2BC(z%5E2%2B2zt%2Bt%5E2)%5C%5C%2BD(z%2Bt)%5C%5C%2BE%3D0


%E6%95%B4%E7%90%86%E5%BE%97%E5%88%B0%EF%BC%9A

z%5E4%5C%5C%2B(4t%2BB)z%5E3%5C%5C%2B(6t%5E2%2BBt%2BC)z%5E2%5Cnewline%2B(4t%5E3%2B3Bt%5E2%2B2Ct%2BD)z%5C%5C%2B(t%5E4%2BBt%5E3%2BCt%5E2%2BDt%2BE)%3D0

若要消掉三次項,則要使t%3D-%5Cfrac%20B4

%E4%BA%8E%E6%98%AF%EF%BC%9A

z%5E4%2B(%5Cfrac%20%7BB%5E2%7D8%2BC)z%5E2%2B(%5Cfrac%20%7BB%5E3%7D8-%5Cfrac%7BBC%7D2%2BD)z%2B(-%5Cfrac%7B3B%5E4%7D%7B256%7D%2B%5Cfrac%7BB%5E2C%7D%7B16%7D-%5Cfrac%7BBC%7D4%2BE)%3D0

%E4%BB%A4p%3D%5Cfrac%7BB%5E2%7D8%2BC%EF%BC%8Cq%3D%5Cfrac%7BB%5E3%7D8-%5Cfrac%7BBC%7D2%2BD%EF%BC%8Cr%3D-%5Cfrac%7B3B%5E4%7D%7B256%7D%2B%5Cfrac%7BB%5E2C%7D%7B16%7D-%5Cfrac%7BBD%7D4%2BE

%E6%95%85%E6%9C%89z%5E4%2Bpz%5E2%2Bqz%2Br%3D0

%E7%A7%BB%E9%A1%B9%E5%BE%97%EF%BC%9Az%5E4%3D-pz%5E2-qz-r

%E8%A6%81%E4%BD%BF%E5%B7%A6%E8%BE%B9%E9%85%8D%E6%96%B9%EF%BC%8C%E4%B8%A4%E8%BE%B9%E9%9C%80%E5%90%8C%E6%97%B6%E5%8A%A0%E4%B8%8A2z%5E2y%2By%5E2%EF%BC%88y%E4%B8%BA%E4%B8%80%E5%8F%82%E6%95%B0%EF%BC%89

%E5%BE%97%E5%88%B0%EF%BC%9Az%5E4%2B2z%5E2y%2By%5E2%3D(-p%2B2y)z%5E2-qz-(r-y%5E2)

%E4%BA%8E%E6%98%AF%EF%BC%9A(z%5E2%2By)%5E2%3D(-p%2B2y)z%5E2-qz-(r-y%5E2)

這時等式左邊完成配方

我們知道,若一個關于x的一元二次多項式ax%5E2%2Bbx%2Bc是完全平方式,則其各項系數(shù)滿足

b%5E2-4ac%3D0

而在這個例子中,要使得等式右邊為完全平方式,則有:

%5CDelta%3Dq%5E2-4(-p%2B2y)(-r%2By%5E2)%3D-8y%5E3%2B4py%5E2%2B8ry-4pr%2Bq%5E2%3D0

這是一個關于y的一元三次方程,取其任意一個解y_1

并且可以推出:%5Cfrac%7Bq%5E2%7D%7B4p-8y_1%7D%3Dr-y_1%5E2

對于一元三次方程的解法,可以查看這篇專欄:

下面對等式右邊進行配方:

(-p%2B2y_1)z%5E2-qz-(r-y_1%5E2)%3D-%EF%BC%BB(p-2y_1)z%5E2%2Bqz%2B(r-y_1%5E2)%5D

%3D-%5B(p-2y_1)z%5E2%2Bqz%2B%5Cfrac%7Bq%5E2%7D%7B4p-8y_1%7D%5D%3D-(p-2y_1)(z%5E2%2B%5Cfrac%7Bq%7D%7Bp-2y_1%7D%2B%5Cfrac%7Bq%5E2%7D%7B4(p-2y_1)%5E2%7D)

%3D-(p-2y_1)(z%2B%5Cfrac%20q%7B2p-4y_1%7D)%5E2%3D(-p%2B2y_1)(z%2B%5Cfrac%20q%7B2p-4y_1%7D)%5E2

于是:

(z%5E2%2By_1)%5E2%3D(-p%2B2y_1)(z%2B%5Cfrac%20q%7B2p-4y_1%7D)%5E2

兩邊同時開方,得到兩組等式:

1.%5Ctext%7B%20%20%7Dz%5E2-%5Csqrt%7B-p%2B2y_1%7D%5Ctext%7B%20%7Dz%2By_1-%5Cfrac%20q%7B2p-4y_1%7D%5Csqrt%7B-p%2B2y_1%7D%3D0

2.%5Ctext%7B%20%20%7Dz%5E2%2B%5Csqrt%7B-p%2B2y_1%7D%5Ctext%7B%20%7Dz%2By_1%2B%5Cfrac%20q%7B2p-4y_1%7D%5Csqrt%7B-p%2B2y_1%7D%3D0

解這兩組關于z的一元二次方程得到z_1~z_4,即可得到x_1~x_4

通解:

%E5%AF%B9%E4%BA%8Eax%5E4%2Bbx%5E3%2Bcx%5E2%2Bdx%2Be%3D0%EF%BC%88a%2Cb%2Cc%2Cd%2Ce%5Cin%20%5Ctext%20R%EF%BC%8Ca%E2%89%A00%EF%BC%89

其解為:

x_%7B1%2C2%2C3%2C4%7D%3D-%5Cfrac%2014B%5Cpm%5Cfrac%2012%5Csqrt%7B-p%2B2y%7D%5Ctext%7B%20%7D%5Cpm%20%5Cfrac%2012%5Csqrt%7B-p-2y%5Ctext%7B%20%7D%5Cpm%20%5Cfrac%7B2q%7D%7Bp-2y%7D%5Csqrt%7B-p%2B2y%7D%7D%5Ctext%7B%20%7D

其中,

y%3D%5Cfrac%20p6%2B%5Csqrt%5B3%5D%7B-%5Cfrac%7Bpr%7D6%2B%5Cfrac%7Bp%5E3%7D%7B216%7D%2B%5Cfrac%7Bq%5E2%7D%7B16%7D%2B%5Csqrt%7B(-%5Cfrac%7Bpr%7D6%2B%5Cfrac%7Bp%5E3%7D%7B216%7D%2B%5Cfrac%7Bq%5E2%7D%7B16%7D)%5E2%2B(-%5Cfrac%20r3-%5Cfrac%7Bp%5E2%7D%7B36%7D)%5E3%7D%7D%5Cnewline%2B%5Csqrt%5B3%5D%7B-%5Cfrac%7Bpr%7D6%2B%5Cfrac%7Bp%5E3%7D%7B216%7D%2B%5Cfrac%7Bq%5E2%7D%7B16%7D-%5Csqrt%7B(-%5Cfrac%7Bpr%7D6%2B%5Cfrac%7Bp%5E3%7D%7B216%7D%2B%5Cfrac%7Bq%5E2%7D%7B16%7D)%5E2%2B(-%5Cfrac%20r3-%5Cfrac%7Bp%5E2%7D%7B36%7D)%5E3%7D%7D

p%3D%5Cfrac%20%7BB%5E2%7D8%2BC%EF%BC%8Cq%3D%5Cfrac%7BB%5E3%7D8-%5Cfrac%7BBC%7D2%2BD%EF%BC%8Cr%3D-%5Cfrac%7B3B%5E4%7D%7B256%7D%2B%5Cfrac%7BB%5E2C%7D%7B16%7D-%5Cfrac%7BBD%7D4%2BE

B%3D%5Cfrac%20ba%EF%BC%8CC%3D%5Cfrac%20ca%EF%BC%8CD%3D%5Cfrac%20da%EF%BC%8CE%3D%5Cfrac%20ea

(注:后兩個±可同時取不同符號,第一和第三個±號同號,以此得到四個解x_1~x_4)


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