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DAOYI PENG
Solution:
\begin{equation} x=-b/a. \end{equation}
Solution:
\begin{equation} x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}, \quad x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}. \end{equation}
Vieta‘s theorem:
\begin{equation}x_1+x_2=-b/a, \quad x_1x_2=c/a. \end{equation}
Cardano‘s solution. The roots of the incomplete cubic equation
\begin{equation}\label{eq:3.1} y^3+py+q=0 \end{equation}
are given by
\[y_1=A+B, \quad y_{2,3}=-\frac{1}{2}(A+B)\pm i\frac{\sqrt{3}}{2}(A-B),\]
where
\[A=\left(-\frac{q}{2}+\sqrt{D}\right)^{1/3},\ B=\left(-\frac{q}{2}-\sqrt{D}\right)^{1/3},\ D=\left(\frac{p}{3}\right)^3+\left(\frac{q}{2}\right)^2,\ i^2=-1,\]
with $A$ and $B$ being any of the values of the respective cubic roots such that $AB=-p/3.$
The number of real roots of the cubic equation \eqref{eq:3.1} depends on the sign of the discriminant $D$:
\begin{align*}
D>0 &\quad \text{one real and two complex conjugate roots}, \\
D<0 &\quad \text{three real roots}, \\
D=0 &\quad \text{one simple real and one twofold real roots} \\
&\quad \text{or, if $p=q=0$, one threefold real root}.
\end{align*}
Trigonometric solution. If the coefficients $p$ and $q$ of the incomplete cubic equation \eqref{eq:3.1} are real, then its roots can also be expressed with trigonometric functions as shown below.
i) Let $p<0$ and $D<0$. Then
\[y_1=2\sqrt{-\frac{p}{3}}\cos\frac{\alpha}{3}, \quad y_{2,3}=-2\sqrt{-\frac{p}{3}}\left(\frac{\alpha}{3}\pm
\frac{\pi}{3}\right),\]
where the trigonometric functions are evaluated taking into account the formula
\[\cos\alpha=-\frac{q}{2\sqrt{-(p/3)^3}}.\]
ii) Let $p>0$ and $D\geqslant0$. Then
\[y_1=2\sqrt{\frac{p}{3}}\cot(2\alpha), \quad y_{2,3}=\sqrt{\frac{p}{3}}\Big[\cot(2\alpha)\pm i\frac{\sqrt{3}}{\sin(2\alpha)}\Big],\]
where the trigonometric functions are evaluated are using the formulas
\[\tan\alpha=\left(\tan\frac{\beta}{2}\right)^{1/3},\quad \tan\beta=\frac{2}{q}\left(\frac{p}{3}\right)^{3/2}, \quad
|\alpha|\leqslant\frac{\pi}{4}, \quad |\beta|\leqslant\frac{\pi}{2}.\]
iii) Let $p<0$ and $D\geqslant0$. Then
\[y_1=-2\sqrt{-\frac{p}{3}}\frac{1}{\sin(2\alpha)}, \quad y_{2,3}=\sqrt{-\frac{p}{3}}\Big[\frac{1}{\sin(2\alpha)}\pm i\sqrt{3}\cot(2\alpha)\Big],\]
where the trigonometric functions are evaluated using the formulas
\[\tan\alpha=\left(\tan\frac{\beta}{2}\right)^{1/3},\quad \sin\beta=\frac{2}{q}\left(-\frac{p}{3}\right)^{3/2}, \quad
|\alpha|\leqslant\frac{\pi}{4}, \quad |\beta|\leqslant\frac{\pi}{2}.\]
In all three cases, the real value of the cubic root is taken.
The roots of the cubic complete equation
\begin{equation}\label{eq:3.2}
ax^3+bx^2+cx+d=0 \quad (a\neq0).
\end{equation}
are evaluated by the formulas
\[x_k=y_k-\frac{b}{3a}, \quad k=1,2,3,\]
where the $y_k$ are roots of the incomplete cubic equation \eqref{eq:3.1} with coefficients
\[p=-\frac{1}{3}\left(\frac{b}{a}\right)^{2}+\frac{c}{a}, \quad q=\frac{2}{27}\left(\frac{b}{a}\right)^3-\frac{bc}{3a^2}+\frac{d}{a}.\]
Vieta‘s theorem for the roots of the cubic equation \eqref{eq:3.2}:
\begin{align*}
x_1+x_2+x_3 & = -b/a, \\
x_1x_2+x_1x_3+x_2x_3 & = c/a, \\
x_1x_2x_3 & =-d/a.
\end{align*}
Reduction to an incomplete equation. The quartic equation in question is reduced to an incomplete equation
\begin{equation}\label{eq:4.1}y^4+py^2+qy+r=0. \end{equation}
with the change of variable
\[x=y-\frac{b}{4a}.\]
Decartes-Euler solution. The roots of the incomplete equation \eqref{eq:4.1} are given by
\begin{equation}\label{eq:4.2}
\begin{matrix}
y_1=\frac{1}{2}(\sqrt{z_1}+\sqrt{z_2}+\sqrt{z_3}), & y_2=\frac{1}{2}(\sqrt{z_1}-\sqrt{z_2}-\sqrt{z_3}), \\
y_3=\frac{1}{2}(-\sqrt{z_1}+\sqrt{z_2}-\sqrt{z_3}),& y_4=\frac{1}{2}(-\sqrt{z_1}-\sqrt{z_2}+\sqrt{z_3}).
\end{matrix}
\end{equation}
where $z_1,z_2,z_3$ are roots of the cubic equation
\begin{equation}\label{eq:4.3}
z^3+2pz^2+(p^2-4r)z-q^2=0,
\end{equation}
which is called cubic resolvent of \eqref{eq:4.1}. The signs of the roots in \eqref{eq:4.2} are chosen so that
\[\sqrt{z_1}\sqrt{z_2}\sqrt{z_3}=-q.\]
The roots of the incomplete quartic equation \eqref{eq:4.1} are determined by the roots of the cubic resolvent \eqref{eq:4.3}; see the table below.
TABLE
Relation between the roots of the incomplete quartic equation and the roots of its cubic resolvent
| Cubic resolvent \eqref{eq:4.3} | Quartic equation \eqref{eq:4.1} |
| All roots are real and positive* | Four real roots |
| All roots are real, one positive and two negative* | Two pairs of complex conjugate roots |
|
One root is positive and two roots are complex conjugate |
Two real and two complex conjugate roots |
*By Vieta‘s theorem, the product of the roots $z_1,z_2,z_3$ is equal to $q^2\geqslant0$.
Ferrari‘s solution. Suppose $z_0$ is any of the roots of the auxiliary cubic equation \eqref{eq:4.3}. Then the four roots of the
incomplete equation \eqref{eq:4.1} are found by solving two quadratic equations
\begin{align*}
y^2-\sqrt{z_0}y+\frac{p+z_0}{2}+\frac{q}{2\sqrt{z_0}}=0, \\
y^2+\sqrt{z_0}y+\frac{p+z_0}{2}-\frac{q}{2\sqrt{z_0}}=0.
\end{align*}
Algebraic equation of general form of degree $n$: $a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0,\,a\neq0$. the coefficients $a_k$ are real or complex numbers.
1. For brevity, denote the left-hand side of the equation, which is a polynomial of degree $n$, by
\begin{equation}\label{eq:5.1}
a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0,\quad a\neq0.
\end{equation}
A number $x=\xi$ is called a root of the equation in question, and also a root of the polynomial $P_n(x)$, if $P_n(\xi)=0$.
A number $x=\xi$ is called a root of multiplicity $m$ if $P_n(x)=(x-\xi)^{m}Q_{n-m}(x),$ where $m$ is a positive integer $(1\leqslant m\leqslant n)$ and $Q_{n-m}(x)$ is a polynomial of degree $n-m$ such that $Q_{n-m}(\xi)\neq0.$
2. Main theorem of polynomial algebra. An algebraic equation of degree $n$ has exactly $n$ roots (real or complex),
provided a root of multiplicity $m$ is counted $m$ times.
3. If an algebraic equation has roots $x_1,x_2,\cdots,x_s$ with respective multiplicities $k_1,k_2,\ldots, k_s\, (k_1+k_2+\cdots+k_s=n)$, then the left-hand side of the equation can be factorized so that
\[P_n(x)=a_n(x-x_1)^{k_1}(x-x_2)^{k_2}\cdots(x-x_s)^{k_s}.\]
4. An algebraic equation of odd degree with real coefficients has always at least one real root.
5. Suppose an algebraic equation with real coefficients has a complex root $\xi=\alpha+i\beta$. Then this equation has also the
complex conjugate root $\eta=\alpha-i\beta$, with the multiplicities of $\xi$ and $\eta$ being the same.
6. An algebraic equation of degree $n$ with integer coefficients $a_k$ cannot have rational roots other than irreducible
fractions $p/q$ with $p$ being a divisor of $a_0$ and $q$ a divisor of $a_n$. If $a_n=1$, then all rational roots of the algebraic
equation (if any) are integer divisors of the free term $a_0$ and can be found by simple search.
7. Any equation with $n\leqslant4$ is solvable in terms of radicals, which means that its roots can be expressed in terms of the equation coefficients using the operations of addition, subtraction, multiplication, division, and root extraction. For $n> 4$, an algebraic equation is generally unsolvable in terms of radicals; this fact is known as the Ruffini-Abel theorem.
8. Vieta‘s Theorem. The following relations between the roots on an algebraic equation (taking into account their
multiplicities) and its coefficients hold true:
\begin{align*}
x_1+x_2+\cdots+x_n &=\sum_{i=1}^{n}x_i=-a_{n-1}/a_n, \\
x_1x_2+x_1x_3+\cdots+x_{n-1}x_n &=\sum_{1\leqslant i<j}^nx_ix_j=a_{n-2}/a_n, \\
x_1x_2x_3+x_1x_2x_4+\cdots+x_{n-2}x_{n-1}x_n &=\sum_{1\leqslant i<j<k}^nx_ix_jx_k=-a_{n-3}/a_n, \\
\cdots\cdots\cdots\cdots\cdots\cdots\cdots&\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots, \\
x_1x_2\cdots x_n &=\prod_{i=1}^{n}x_i=(-1)^na_0/a_n.
\end{align*}
Root of a polynomial in one variable
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原文地址：http://www.cnblogs.com/pengdaoyi/p/4270758.html
