实系四次方程有四个实根的条件

zuijianqiugen

（1）已知实系四次方程：x⁴=ax²+bx+c有四个实根
（2）求系数a、b、c所满足的充要条件。

cn8888

画出函数图形,f(x)=x^4-ax^2-bx-c
求导f'(x)=4x^3-2ax-b
,三个实根是x1 x2 x3 (x1<=x2<=x3)
f(x1)<=0
f(x2)>=0
f(x1)<=0
这就是充要条件

cn8888

看你的样子是想求解四次方程
x4+dx2+d2/4=(a+d)x2+bx+(c+d2/4)
右边完全平方,判别式等于零
b2-4(a+d)(c+d2/4)=0
关于d的三次方程,求解出来就可以了,肯定至少有一个实根的
然后两边平方差就可以了

cn8888

我看很难有显式表达,还是不要浪费时间了吧

cn8888

Solving a quartic equation
http://en.wikipedia.org/wiki/Quartic_function

恭喜你了,原来还真的有显示表达式.

Nature of the roots[edit]
Given the general quartic equation
\(ax^4 + bx^3 + cx^2 + dx + e = 0\)
with real coefficients and \(a \ne 0\), the nature of its roots is mainly determined by the sign of its discriminant
\(\begin{split}\Delta\ =\ &256 a^3 e^3 - 192 a^2 b d e^2 - 128 a^2 c^2 e^2 + 144 a^2 c d^2 e - 27 a^2 d^4 \\
&+ 144 a b^2 c e^2 - 6 a b^2 d^2 e - 80 a b c^2 d e + 18 a b c d^3 + 16 a c^4 e \\
&- 4 a c^3 d^2 - 27 b^4 e^2 + 18 b^3 c d e - 4 b^3 d^3 - 4 b^2 c^3 e + b^2 c^2 d^2\end{split}\)
This may be refined by considering the signs of three other polynomials:
\(P = 8ac - 3b^2\)
such that \(\tfrac{P}{8a^2}\) is the second degree coefficient of the associated depressed quartic (see below);
\(\Delta_0 = c^2 - 3bd + 12ae\)
which is 0 if the quartic has a triple root; and
\(D = 64 a^3 e - 16 a^2 c^2 + 16 a b^2 c - 16 a^2 bd - 3 b^4\)
which is 0 if the quartic has two double roots.

The possible cases for the nature of the roots are as follows:[7]

• If \(\Delta < 0\) then the equation has two real roots and two complex conjugate roots.
• If \(\Delta > 0\) then the equation's four roots are either all real or all complex.
  • If \(P < 0\) and \(D < 0\) then all four roots are real and distinct.
  • If \(P > 0\) or \(D > 0\) then there are two pairs of complex conjugate roots.[8]
• If \(\Delta = 0\) then either the polynomial has a multiple root, or it is the square of a quadratic polynomial. Here are the different cases that can occur:
  • If \(P < 0\) and \(D < 0\) and \(\scriptstyle \Delta_0 \neq 0\), there is a real double root and two real simple roots.
  • If (\(P > 0\) and \(D \neq 0\)) or \(D > 0\), there is a real double root and two complex conjugate roots.
  • If \(\scriptstyle \Delta_0 = 0\) and\( D \neq 0\), there is a triple root and a simple root, all real.
  • If\( D = 0\), then:
    • If \(P < 0\), there are two real double roots.
    • If \(P > 0\), there are two complex conjugate double roots.
    • If \(\scriptstyle \Delta_0 = 0\), all four roots are equal to \(-\tfrac{b}{4a}\)

There are some cases that do not seem to be covered, but they can not occur. For example \(\scriptstyle \Delta > 0\), \(P = 0\) and \(D ≤ 0\) is not one of the cases. However if \(\scriptstyle \Delta > 0\) and \(P = 0\) then \(D > 0\) so this combination is not possible.

zuijianqiugen

cn8888 发表于 2014-5-8 18:30
Solving a quartic equation
http://en.wikipedia.org/wiki/Quartic_function

还有更漂亮的解答。参见：http://zuijianqiugen.blog.163.co ... 622011112903052939/

cn8888

zuijianqiugen 发表于 2014-5-8 19:17
还有更漂亮的解答。参见：http://zuijianqiugen.blog.163.com/blog/static/1265240622011112903052939/

不厚道呀,你自己都知道了答案然后到这里问,害我帮你找资料

zuijianqiugen

cn8888 发表于 2014-5-9 12:17
不厚道呀,你自己都知道了答案然后到这里问,害我帮你找资料

免费教你学学知识、长长见识，你应该感谢我才对呀！

wayne

类似题目论坛有提及过。
Sturm's_theorem
http://en.wikipedia.org/wiki/Sturm's_theorem
http://zh.wikipedia.org/wiki/施图姆定理

葡萄糖

§3  结式
12.3.1    两个一元多项式的结式的定义
考虑域 上的多项式
\[\begin{gathered}
f(x) = {a_0}{x^n} + {a_1}{x^{n - 1}} +  \cdots  + {a_n} \\
g(x) = {b_0}{x^m} + {b_1}{x^{m - 1}} +  \cdots  + {b_m}  \\
\end{gathered} \]
由给定 上多项式
\[\begin{gathered}
  {q_1}(x) = {y_1}{x^{n - 1}} + {y_2}{x^{n - 2}} +  \cdots  + {y_n} \\
  {q_2}(x) = {x_1}{x^{m - 1}} + {x_2}{x^{m - 2}} +  \cdots  + {x_m} \\
\end{gathered} \]       
这里系数 是待定的。那么， 的充分必要条件是下面等式成立：
\[\left\{ \begin{gathered}
  {a_0}{x_1}{\text{                                  = }}{b_0}{y_1},  \\
  {a_1}{x_1} + {a_0}{x_2}{\text{                        = }}{b_1}{y_1} + {b_0}{y_2}, \\
  {a_2}{x_1} + {a_1}{x_2} + {a_0}{x_3}{\text{              = }}{b_2}{y_1} + {b_1}{y_2} + {b_0}{y_3},  \\
   \cdots  \cdots  \cdots  \cdots  \cdots  \cdots  \cdots  \cdots  \cdots  \cdots  \cdots  \cdots  \cdots  \cdots  \cdots  \cdots  \cdots  \cdots   \\
  {\text{                 }}{a_n}{x_{m - 1}} + {a_{n - 1}}{x_m} = {\text{            }}{b_m}{y_{n - 1}} + {b_{m - 1}}{y_n}, \\
  {\text{                                 }}{a_n}{x_m} = {\text{                       }}{b_m}{y_n}.  \\
\end{gathered}  \right.\left( 1 \right)\]
令
\[R(f,g) = \left| {\begin{array}{*{20}{r}}
  {{a_0}}&{{a_1}}&{{a_2}}& \cdots &{{a_n}}&0& \cdots &0 \\
  0&{a{}_0}&{{a_1}}&{{a_2}}& \cdots &{{a_n}}& \cdots &0 \\
   \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots  \\
  0&0& \cdots &{{a_0}}&{{a_1}}&{{a_2}}& \cdots &{{a_n}} \\
  {{b_0}}&{{b_1}}&{{b_2}}& \cdots &{{b_m}}&0& \cdots &0 \\
  0&{{b_0}}&{{b_1}}&{{b_2}}& \cdots &{{b_m}}& \cdots &0 \\
   \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots  \\
  0&0& \cdots &{{b_0}}&{{b_1}}&{{b_2}}& \cdots &{{b_m}}
\end{array}} \right|\]
其中${a_i}$有$m$行，其中${b_i}$有$n$行，称$R(f,g)$为多项式$f,g$的结式。
命题  两个一元多项式的结式等于0当且仅当此二多项式不互素或首相系数都为零。
证明  设两个一元多项式为
\[\begin{gathered}
  f(x) = {a_0}{x^n} + {a_1}{x^{n - 1}} +  \cdots  + {a_n},{\text{    (}}n \geqslant 1) \\
  g(x) = {b_0}{x^m} + {b_1}{x^{m - 1}} +  \cdots  + {b_m}{\text{      (}}m \geqslant 1) \\
\end{gathered} \]
充分性     若${a_0} = {b_0} = 0$，则显见有$R(f,g) = 0$。今设${a_0},{b_0}$不全为零，不妨设${a_0} \ne 0$，且$f$与$g$不互素，即有公因式$d(x)$，$\deg d(x) \geq 1$。于是
$f = d{q_1},g = d{q_2}$。因$f \ne 0$，故${q_1} \ne 0$，且$\deg {q_1} < n$。若$g \ne 0$，则$\deg {q_2} < m$；若$g = 0$，则令${q_2} = 0{x^{m - 1}} + 0{x^{m - 2}} +  \cdots  + 0$。易知此时$f{q_2} = g{q_1}$，且${q_1} \ne 0$，故齐次线性方程组（1）有非零解，于是
$R(f,g) = 0$
必要性   若$R(f,g) = 0$，而${a_0},{b_0}$不全为零，我们来证明$f$与$g$不互素。因为此时齐次线性方程组（1）有非零解。故存在不全为零的${q_1}(x),{q_2}(x) \in K[x]$，使$f{q_2} = g{q_1}$，而且当${q_1}$（或${q_2}$）不为零时，其次数小于$n$（小于$m$）。不妨设${a_0} \ne 0$，即$f \ne 0$。若$g = 0$，则$f,g$显见不互素。今设$g \ne 0$。因$f|g{q_1}$，若$(f,g) = 1$，则有$f|{q_1}$，与$\deg {q_1} < n$矛盾。证毕。

12.3.3   用一个多项式与它的微商的\(\color{red}{结式}\)表达该多项式的\(\color{red}{判别式}\)
现在设
\[f(x) = {a_0}{x^n} + {a_1}{x^{n - 1}} +  \cdots  + {a_n}{\text{    }}({a_0} \ne 0)\]
根据前面对其判别式的定义，我们有
\[D(f) = a_0^{2n - 2}\prod\limits_{1 \leqslant i \leqslant j \leqslant n} {{{({\alpha _j} - {\alpha _i})}^2}} \]
因为
\[f(x) = {a_0}\prod\limits_{i = 1}^n {(x - {\alpha _i})} \]
故
\[f'(x) = {a_0}\sum\limits_{i = 1}^n {(x - {\alpha _1}) \cdots (x - {\alpha _{i - 1}})(x - {\alpha _{i + 1}}) \cdots (x - {\alpha _n})} \]
以$x = {\alpha _i}$代入上式，得
\[f'({\alpha _i}) = {a_0}\prod\limits_{\begin{gathered}
  j = 1 \\
  j \ne i \\
\end{gathered}
}^n {({\alpha _i} - {\alpha _j})} \]
\[\prod\limits_{i = 1}^n {f'({\alpha _i}) = a_0^n\prod\limits_{\begin{gathered}
  i,j = 1  \\
  i \ne j \\
\end{gathered}
}^n {({\alpha _i} - {\alpha _j}) = {{( - 1)}^{\frac{{n(n - 1)}}{2}}}a_0^n\prod\limits_{1 \leqslant i \leqslant j \leqslant n} {{{({\alpha _j} - {\alpha _i})}^2}} } } \]
从而有
\[\begin{gathered}
  R(f,f') = a_0^{n - 1}\prod\limits_{i = 1}^n {f'({\alpha _i})}  \\
  {\text{                = }}{( - 1)^{\frac{{n(n - 1)}}{2}}}a_0^{2n - 1}\prod\limits_{1 \leqslant i \leqslant j \leqslant n} {{{({\alpha _j} - {\alpha _i})}^2}}   \\
  {\text{                = }}{( - 1)^{\frac{{n(n - 1)}}{2}}}{a_0}D(f)  \\
\end{gathered} \]
\[\begin{gathered}
  R(f,f') = {( - 1)^{\frac{{n(n - 1)}}{2}}}{a_0}D(f)  \\
\Longrightarrow \color{red}{D(f)={( - 1)^{\frac{{n(n - 1)}}{2}}}{a_0^{-1}}R(f,f')}
\end{gathered} \]
这就是$f$的判别式与$R(f,f')$之间的关系式。
http://zh.wikipedia.org/wiki/判別式