solving certain polynomial inequalitiesSolvingCertainPolynomialInequalities2007-04-20 13:29:322008-05-10 16:52:26Feature\usepackage{amssymb,amscd}
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\newcommand{\real}{\mathbb{R}}In this article, we discuss inequality of the form $p(x)\ge 0$ or $p(x)>0$, where $p(x)$ is a polynomial with real coefficients such that $p(x)$ can be expressed as product of linear factors:
$$p(x)=(x-a_1)(x-a_2)\cdots (x-a_n)$$
where $a_i$ are real numbers (in the \PMlinkescapetext{language} of field theory, this means that $p(x)$ splits in the field of real numbers).
When we plot the polynomial $y=p(x)$, whenever it crosses the $x$-axis, the crossing point is a root of $p(x)$. On either side of the crossing point, the values of $p(x)$ may be negative or positive. The way to solve the inequality $p(x)>0$ or $p(x)\ge 0$ is to look at how $p(x)$ crosses the $x$-axis, and to realize that when $x$ is very large, $p(x)$ will be positive. This idea can be illustrated in the following figure:
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If we start from the far right, the curve (graph of $p(x)$) is above the horizontal axis, and so the values of $p(x)$ are positive there. As it approaches $a_1$, its values decrease until it crosses $a_1$ and dips below the horizontal axis. The values of $p(x)$ are now negative. As it continues to travel along to the left, its values increase until it passes over another crossing point, $a_2$, and the values become positive as soon as it passes $a_2$. When $p(x)$ reaches $a_3$ however, it merely touches the $x$-axis (at $a_3$) and then rises again. So on either side of $a_3$ the values of $p(x)$ are positive. Nevertheless, an analysis of how $p(x)$ crosses the $x$-axis is enough to give us some clue on how to solve inequalities of the form $p(x)\ge 0$ or $p(x)>0$.
With this idea in mind, the steps are devised when solving inequalities of this type:
\begin{enumerate}
\item arrange $a_i$ so that they are in the ascending order: $a_1\ge a_2\ge \cdots \ge a_n$
\item plot $a_i$ on the number line (the real axis), so each $a_i$ is now a point on the line
\item label above the interval $i_1:=(a_1,\infty)$ to the right of $a_1$ positive
\item go to point $a_2$
\item if $a_2\ne a_1$, label above the interval $i_2:=(a_2,a_1)$ adjacent to $i_1$ negative
\item if $a_2=a_1$, label above $a_2$ negative
\item go to $a_3$, and iterate the labeling processes 4-6
\item stop when $(-\infty,a_n)$ is labeled.
\item if we are trying to solve $p(x)\ge 0$, then all intervals that are labeled positive, including the end points, are solutions to the inequality
\item if we are solving $p(x)>0$, then all intervals labeled positive, excluding the end points, are solutions to the inequality.
\end{enumerate}
\textbf{Remark}. After all the labeling is done, there should a total of $n+1$ labels, one over each interval, including the null intervals (the points).
To see how this works, let us look at some actual examples.
\begin{itemize}
\item Solve $(x-2)x(x+3)\ge 0$.
\begin{enumerate}
\item Plot $2$, $0$, $-3$ on the number line.
\item The intervals separated by these points are $(2,\infty),(0,2),(-3,0),(-\infty,-3)$.
\item Since no two points are the same, the intervals that are labeled positive are $(2,\infty)$ and $(-3,0)$.
\item The solutions to the inequality are $[2,\infty) \cup [-3,0]$, the square brackets signify that the end points are included in the solutions.
\end{enumerate}
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\item Solve $(x-8)^7>0$.
\begin{enumerate}
\item Plot $8$ on the number line.
\item The intervals separated by these points are $(8,\infty),(-\infty,8)$, since $8$ is repeated $7$ times.
\item Start with labeling $(8,\infty)$ positive (the first label)
\item The point $8$ is then labeled alternately $-(2), +(3), -(4), +(5), -(6), +(7)$.
\item The last label goes to $(-\infty,8)$, which is negative.
\item Therefore, the solution set is $(8,\infty)$ (excluding $8$).
\end{enumerate}
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\item Solve $(x-1)(x+1)^4\ge 0$.
\begin{enumerate}
\item Plot $1,-1$ on the number line.
\item The intervals separated by these points are $(1,\infty),(-1,1)$, and $(-\infty,-1)$, since $-1$ is repeated $4$ times.
\item Start with labeling $(1,\infty)$ positive (the first label), followed by $(-1,1)$ as negative.
\item The point $-1$ is then labeled alternately $+(3), -(4), +(5)$.
\item The last label goes to $(-\infty,-1)$, which is negative.
\item Therefore, the solution set is $[1,\infty) \cup \lbrace -1\rbrace$.
\end{enumerate}
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\end{itemize}
\textbf{Remarks}.
\begin{enumerate}
\item
In the last two examples, we observe that a simplification can be made when solving the inequality: whenever we have repeating roots ($a_i=a_{i+1}$). Depending on the parity of the number of repeating roots, we have two situations:
\begin{itemize}
\item If the number $n_i$ of repeating root, say $a_i$, is odd, then solving inequality involving $(x-a_i)^{n_i}$ is the same as solving the same inequality with $(x-a_i)^{n_i}$ replaced by $(x-a_i)$. In other words, their solution sets are the same. For example,
\begin{center} solving $(x-8)^7>0$ is the same as solving $(x-8)>0$. \end{center}
\item If the number $n_i$ of repeating root $a_i$ is even, then we look at whether the inequality is strict or not.
\begin{itemize}
\item If the inequality is strict, then we can completely eliminate $(x-a_i)^{n_i}$ from the inequality without altering the solution set. For example, \begin{center} solving $(x-1)(x+1)^4>0$ is the same as solving $(x-1)>0$. \end{center}
\item Otherwise, we need to remember the roots themselves as solutions. Therefore, the solution set for $(x-1)(x+1)^4\ge 0$ is the same as the solution set of $(x-1)\ge 0$ together with the root $-4$.
\end{itemize}
\end{itemize}
\item
Using the rules above, we may also solve inequalities $p(x)<0$ or $p(x)\le 0$. The solution set for $p(x)<0$ is the complement of the solution set for $p(x)\ge 0$, and the solution set for $p(x)\le 0$ is the complement of the solution set for $p(x)>0$.
\end{enumerate}