Quadratic Inequalities

In this section, I'll consider quadratic inequalities. I'll solve them using the graph of the quadratic function. I'll also look at other inequalities, which I'll solve using sign charts.

A quadratic function is a function of the form $f(x) = ax^2 + bx + c$ . The graph of a quadratic function is a parabola.

If $a > 0$ , the parabola opens up; if $a < 0$ , the parabola opens down:

$$\matrix{\hbox{\epsfxsize=2.5in \epsffile{quadratic-inequalities1.eps}} & \hskip0.5in & \hbox{\epsfxsize=2.5in \epsffile{quadratic-inequalities2.eps}} \cr a > 0 & & a < 0 \cr}$$

You can use the graph of a quadratic function to solve quadratic inequalities.


Example. Solve the quadratic inequality $x^2 - 2x - 3 < 0$ .

The graph of $f(x) = x^2
   - 2x - 3$ opens upward, because the coefficient of $x^2$ is $+1$ . Since

$$x^2 - 2x - 3 = (x - 3)(x + 1),$$

the roots are $x = 3$ and $x = -1$ .

Therefore, the graph looks like this:

$$\hbox{\epsfysize=1.75in \epsffile{quadratic-inequalities3.eps}}$$

The original inequality asks for the values of x for which the parabola is below the x-axis:

$$\matrix{x^2 - 2x - 3 & < & 0 & \cr \hbox{parabola} & \hbox{below} & x\hbox{-axis} & \hbox{?} \cr}$$

$$\hbox{\epsfysize=1.75in \epsffile{quadratic-inequalities4.eps}}$$

The parabola is below the x-axis for $-1 < x < 3$ .


Example. Solve the quadratic inequality $-6 -5x - x^2 \ge 0$ .

The graph of $f(x) = -6
   -5x - x^2$ opens downard, because the coefficient of $x^2$ is -1. Since

$$-6 -5x - x^2 = -(x^2 + 5x + 6) = -(x + 2)(x + 3),$$

the roots are $x = -2$ and $x = -3$ .

Therefore, the graph looks like this:

$$\hbox{\epsfysize=1.75in \epsffile{quadratic-inequalities5.eps}}$$

The original inequality asks for the values of x for which the parabola is below or on the x-axis. The solution is $x \le -3$ or $x \ge -2$ .

Warning! You can't put the inequalities $x \le -3$ and $x \ge -2$ together by writing "$-2 \le x
   \le 3$ ". This says that "$-2 \le -3$ ", which is absurd. Rule of thumb: The solution set occupied two shaded pieces on the number line, so two inequalities are required to write the answer.


Example. Solve the quadratic inequality $x^2 - 4x + 5 \le 0$ .

The graph of $f(x) = x^2
   - 4x + 5$ opens upward, because the coefficient of $x^2$ is $+1$ . The quadratic formula shows that the roots are complex numbers; this means that the graph does not intersect the x-axis. It must look like this:

$$\hbox{\epsfysize=1.75in \epsffile{quadratic-inequalities6.eps}}$$

(I've located the vertex of the parabola --- it's at $x = 2$ --- for reference, but it doesn't come into this problem.)

The inequality $x^2 - 4x
   + 5 \le 0$ asks for what values of x the parabola is below or on the x-axis. Since the parabola lies entirely above the x-axis, there are no solutions.


You can also solve inequalities using sign charts.

  1. Make sure the inequality has the form

$$\hbox{JUNK} > 0 \quad\quad\hbox{or}\quad\quad \hbox{JUNK} < 0.$$

($\ge 0$ or $\le 0$ are also okay.) If there are terms on both sides, add or subtract terms to move everything to one side.

  1. Combine the terms in JUNK into one piece. For example, add or subtract fractions over a common denominator.
  2. Factor as much as you can. For example,

$$\hbox{write} \dfrac{x^2 - 4}{x^2 + 5x} \quad\hbox{as}\quad \dfrac{(x - 2)(x + 2)}{x(x^2 + 5)}.$$

  1. Find the values of x for which $\hbox{JUNK} = 0$ and the values of x for which JUNK is undefined. These are the break points for your sign chart.
  2. Pick points at random in the intervals determined by the break points. Plug them into JUNK to determine whether JUNK is positive or negative on each interval. You'll see how to do this in the examples below.
  3. Solve the inequality by examining the sign chart.

Example. Solve $\dfrac{x + 1}{x - 3} < 0$ .

$\dfrac{x + 1}{x - 3} =
   0$ for $x = -1$ ; $\dfrac{x + 1}{x - 3}$ is undefined for $x = 3$ . Set up a sign chart with $x = 3$ and $x =
   -1$

$$\hbox{\epsfxsize=4in \epsffile{quadratic-inequalities7.eps}}$$

Let $f(x) = \dfrac{x +
   1}{x - 3}$ . I pick points at random in each of the three intervals: -2, 0, 4. (Pick points which make the calculations simple!) I plug the points into f; for example,

$$f(-2) = \dfrac{-2 + 1}{-2 - 3} = \dfrac{-1}{-5} = \dfrac{1}{5}.$$

The values I get determine the sign (+ or -) of $f(x) = \dfrac{x + 1}{x -
   3}$ on each interval. I've marked the signs above the sign chart.

The original inequality asks where $\dfrac{x + 1}{x - 3}$ is negative. From the sign chart, the solution is $-1 < x
   < 3$ .

By the way, it's purely coincidental that the +'s and -'s alternate --- they don't have to do that!


Example. Solve $2 \ge \dfrac{x + 8}{x + 3}$ .

First, move everything to one side and combine terms over a common denominator:

$$2 \ge \dfrac{x + 8}{x + 3}, \quad 2 - \dfrac{x + 8}{x + 3} \ge \dfrac{x + 8}{x + 3} - \dfrac{x + 8}{x + 3}, \quad 2 - \dfrac{x + 8}{x + 3} \ge 0, \quad 2\cdot \dfrac{x + 3}{x + 3} - \dfrac{x + 8}{x + 3} \ge 0,$$

$$\dfrac{2x + 6}{x + 3} - \dfrac{x + 8}{x + 3} \ge 0, \quad \dfrac{(2x + 6) - (x + 8)}{x + 3} \ge 0, \quad \dfrac{x - 2}{x + 3} \ge 0.$$

$f(x) = \dfrac{x - 2}{x
   + 3} = 0$ for $x = 2$ and $\dfrac{x - 2}{x + 3}$ is undefined for $x = -3$ . These are the break points on the sign chart:

$$\hbox{\epsfxsize=4in \epsffile{quadratic-inequalities8.eps}}$$

The inequality $\dfrac{x
   - 2}{x + 3} \ge 0$ asks where $\dfrac{x - 2}{x + 3}$ is greater than or equal to 0. The solution is $x \le -3$ or $x \ge 2$ .


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