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This is the Mathematics notes from 2007. It is a free notes for revision.
§4-2 QUADRATIC INEQUALITIES
Definition
Quadratic inequalities in one variable are inequalities which can be written in one of the following forms: $a x^{2}+b x+c>0$,
\[
\begin{array}{l}
a x^{2}+b x+c<0
a x^{2}+b x+c \geq 0 \text { or }
a x^{2}+b x+c \leq 0 \text { where } a, b \text { and } c \text { are real numbers. }
\end{array}
\]
Procedure
Solving Quadratic Inequalities
1. Move all terms to one side.
2. Simplify and factor the quadratic expression.
3. Find the roots of the corresponding quadratic equation.
4. Use the roots to divide the number line into regions.
5. Test each region using theinequality.
Example 1 Solve the inequality, $x^{2}>x+2$.
Solution
\[
\begin{aligned}
x^{2} | >x+2
x^{2}-x-2 | >0
(x-2)(x+1) | >0
\end{aligned}
\]
The corresponding equation is $(x-2)(x+1)=0$ so...
\[
\begin{aligned}
x-2 | =0
x | =2
\end{aligned}
\]
or
\[
\begin{aligned}
x+1 | =0
x | =-1
\end{aligned}
\]
Now we test one point in each region.
Region | Test Point | \multicolumn{1}{c}{ Inequality } | Status
I | $x=-2$ | $(x-2)(x+1)=(-2-2)(-2+1)=4>0$ | True
II | $x=0$ | $(x-2)(x+1)=(0-2)(0+1)=-2>0$ | False
III | $x=3$ | $(x-2)(x+1)=(3-2)(3+1)=4>0$ | True
So the solution to this inequality is $x<-1$ or $x>2$.
Copyright©2007 by Lawrence Perez and Patrick Quigley