quadratic inequalityQuadraticInequality2005-07-16 18:08:302007-04-20 16:11:30Topicinequalities for real numbers% this is the default PlanetMath preamble. as your knowledge
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\newtheorem*{thmplain}{Theorem}The \PMlinkescapetext{{\em normal form}} of a {\em quadratic inequality} is
\begin{align}
ax^2\!+\!bx\!+\!c < 0
\end{align}
or
\begin{align}
ax^2\!+\!bx\!+\!c > 0
\end{align}
where $a$, $b$ and $c$ are known real numbers and\, $a \neq 0$.
Solving such an inequality, i.e. determining all real values of $x$ which satisfy it, is based on the fact that the graph of the quadratic polynomial function\, $x\mapsto ax^2\!+\!bx\!+\!c$\, is the parabola
$$y = ax^2\!+\!bx\!+\!c,$$
opening upwards if\, $a > 0$\, and downwards if\, $a < 0$.
For obtaining the solution we first have to determine the real zeroes of the polynomial $ax^2\!+\!bx\!+\!c$, i.e. solve the \PMlinkname{quadratic equation}{QuadraticFormula}\,
$ax^2\!+\!bx\!+\!c = 0$.
\begin{itemize}
\item If there is two distinct real zeroes $x_1$ and $x_2$ (say\, $x_1 < x_2$),\, then the parabola intersects the $x$-axis in these points.\, In the case\, $a > 0$\, the parabola opens upwards and thus\, $y < 0$\, in the interval\, $(x_1,\,x_2)$, but\, $y > 0$\, outside this interval.\, I.e., for positive $a$, the solution of (1) is
$$x_1 < x < x_2$$
and the solution of (2) is
$$x < x_1\,\,\,\mbox{or}\,\,\,x > x_2$$
(note that the latter solution-domain consists of two distinct portions of the $x$-axis and therefore must be expressed with two separate inequalities, not with a double inequality as the former).\, For negative $a$ we must swap those solutions for (1) and (2).
\begin{figure}[!htb]
\begin{center}
\includegraphics{parabola.1.eps}
\end{center}
\caption{Solving for $ax^2\! + \!bx\! + \!c < 0$ when $a > 0$ and the quadratic has two distinct roots}
\end{figure}
\item If there is only one real zero of the polynomial (we may say that\, $x_2 = x_1$), the parabola has $x$-axis as the \PMlinkname{tangent}{TangentLine} in its apex.\, For positive $a$ the other points of parabola are above the $x$-axis, i.e. they have\, $y > 0$\, always\, but\, $y < 0$\, never.\, So, (1) has no solutions, but (2) is true for all\, $x \neq x_1$ (i.e.\, $x < x_1$\, or\, $x > x_1$).\, For the case of negative $a$ we again must change those solutions for (1) and (2).
\begin{figure}[!htb]
\begin{center}
\includegraphics{parabola.2.eps}
\end{center}
\caption{Solving for $ax^2\! + \!bx\! + \!c > 0$ when $a > 0$ and the quadratic has only one root}
\end{figure}
\item There can still appear the possibility that the polynomial has no real zeroes (the roots of the equation are imaginary).\, Now the parabola does not intersect or touch the $x$-axis, but is totally above the axis when $a$ is positive ($y > 0$\, always) and totally below the axis when $a$ is negative
($y < 0$\, always).\, Thus we get no solutions at all (the inequality is impossible) or all real numbers $x$ as solutions, according to what the inequality (1) or (2) demands.
\begin{figure}[!htb]
\begin{center}
\includegraphics{parabola.3.eps}
\end{center}
\caption{$ax^2\! + \!bx\! + \!c > 0$ for all $x$ when $a > 0$ and the quadratic has no roots}
\end{figure}
\end{itemize}