2024年4月1日发(作者：可春琳)
四次函数
形如y=ax^4+bx^3+cx^2+dx+e(a≠0,b,c,d,e为常数)的函数叫做四次函数。
四次函数的图像
a*x^4+b*x^3+c*x^2+b*x+a=0的求解方法，对于一般的四次方程a*x^4+b*x^3
+c*x^2+d*x+e=0,先求解三次方程8y^3-4cy^2+(2bd-8e)y+e(4c-b^2)-d^2=0,得到
的y的任一实根分别代入下面两个方程：
x^2+(b+sqrt(8y+b^2-4c))x/2+(y+(by-d)/sqrt(8y+b^2-4c))=0
及x^2+(b-sqrt(8y+b^2-4c))x/2+(y-(by-d)/sqrt(8y+b^2-4c))=0
就可得到原方程的四个根。
在 数学中, 四次方程 是令一个 四次函数 等于零的结果.四次方程的一个例子如下
<math>2x^4+4x^3-26x^2-28x+48=0;</math>
它的通式是
<math>a_0x^4+a_1x^3+a_2x^2+a_3x+a_4=0,qquadmboxa_0ne0.<
/math>
代数基本定理 告诉我们, 一个四次方程总有四个解 (根). 它们可能是 复数 而且可能
有等根.
[编辑]解决四次方程
自然,人们为了找到这些根做了许多努力. 就像其它 多项式, 有时可能对一个四次方
程分解出因式;但更多的时候这样的工作是极困难的,尤其是当根是无理数或复数时.因此找
到一个通式解法或运算法则 (就像 二次方程那样, 能解所有的一元二次方程)是很有用的.
After much effort, such a formula was indeed found for quartics — but sinc
e then it has been proven (by Evariste Galois) that such an approach dead-e
nds with quartics; they are the highest-degree polynomial equations whose r
oots can be expressed in a formula using a finite number of arithmetic oper
ators and n-th roots. From quintics on up, one requires more powerful meth
ods if a general algebraic solution is sought, as explained under quintic equa
tions.
Given the complexity of the quartic formulae (see below), they are not o
ften used. If only the real rational roots are needed, they can be found (as i
s true for polynomials of any degree) via trial and error, using Ruffini's rule
(so long as all the polynomial coefficients are rational). In the modern age o
2024年4月1日发(作者：可春琳)
四次函数
形如y=ax^4+bx^3+cx^2+dx+e(a≠0,b,c,d,e为常数)的函数叫做四次函数。
四次函数的图像
a*x^4+b*x^3+c*x^2+b*x+a=0的求解方法，对于一般的四次方程a*x^4+b*x^3
+c*x^2+d*x+e=0,先求解三次方程8y^3-4cy^2+(2bd-8e)y+e(4c-b^2)-d^2=0,得到
的y的任一实根分别代入下面两个方程：
x^2+(b+sqrt(8y+b^2-4c))x/2+(y+(by-d)/sqrt(8y+b^2-4c))=0
及x^2+(b-sqrt(8y+b^2-4c))x/2+(y-(by-d)/sqrt(8y+b^2-4c))=0
就可得到原方程的四个根。
在 数学中, 四次方程 是令一个 四次函数 等于零的结果.四次方程的一个例子如下
<math>2x^4+4x^3-26x^2-28x+48=0;</math>
它的通式是
<math>a_0x^4+a_1x^3+a_2x^2+a_3x+a_4=0,qquadmboxa_0ne0.<
/math>
代数基本定理 告诉我们, 一个四次方程总有四个解 (根). 它们可能是 复数 而且可能
有等根.
[编辑]解决四次方程
自然,人们为了找到这些根做了许多努力. 就像其它 多项式, 有时可能对一个四次方
程分解出因式;但更多的时候这样的工作是极困难的,尤其是当根是无理数或复数时.因此找
到一个通式解法或运算法则 (就像 二次方程那样, 能解所有的一元二次方程)是很有用的.
After much effort, such a formula was indeed found for quartics — but sinc
e then it has been proven (by Evariste Galois) that such an approach dead-e
nds with quartics; they are the highest-degree polynomial equations whose r
oots can be expressed in a formula using a finite number of arithmetic oper
ators and n-th roots. From quintics on up, one requires more powerful meth
ods if a general algebraic solution is sought, as explained under quintic equa
tions.
Given the complexity of the quartic formulae (see below), they are not o
ften used. If only the real rational roots are needed, they can be found (as i
s true for polynomials of any degree) via trial and error, using Ruffini's rule
(so long as all the polynomial coefficients are rational). In the modern age o