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 Depdendent Variable

 Number of equations to solve: 23456789
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 Dependent Variable

 Number of inequalities to solve: 23456789
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 Solve for:

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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) = ax2+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: You can use the graph of a quadratic function to solve quadratic inequalities.

Example. Solve the quadratic inequality x2 − 2x − 3 < 0.

The graph of f(x) = x2 − 2x − 3 opens upward, because the coefficient of x2 is +1. Since

x2− 2x − 3 = (x − 3)(x + 1),

the roots are x = 3 and x = −1.
Therefore, the graph looks like this: The original inequality asks for the values of x for which the parabola is below the x-axis:

 x2 − 2x − 3 < 0 parabola below x-axis ? The parabola is below the x-axis for −1 < x < 3.

Example. Solve the quadratic inequality −6 − 5x − x2 ≥ 0.

The graph of f(x) = −6 − 5x − x2 opens downard, because the coefficient of x2 is −1. Since

−6 − 5x − x2 = −(x2 + 5x + 6) = −(x + 2)(x + 3),

the roots are x = −2 and x = −3.
Therefore, the graph looks like this: The original inequality asks for the values of x for which the parabola is below or on the x-axis. The
solution is x ≤ −3 or x ≥ −2.

Warning! You can’t put the inequalities x ≤ −3 and x ≥ −2 together by writing “−2 ≤ x ≤ 3”. This
says that “−2 ≤ −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 x2 − 4x + 5 ≤ 0.

The graph of f(x) = x2 − 4x + 5 opens upward, because the coefficient of x2 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: (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 x2 − 4x + 5 ≤ 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

JUNK > 0 or JUNK < 0.

(≥ 0 or ≤ 0 are also okay.) If there are terms on both sides, add or subtract terms to move everything
to one side.

2. Combine the terms in JUNK into one piece. For example, add or subtract fractions over a common
denominator.

3. Factor as much as you can. For example,
write as 4. Find the values of x for which JUNK = 0 and the values of x for which JUNK is undefined. These are
the break points for your sign chart.

5. 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.

6. Solve the inequality by examining the sign chart.
Example. Solve for x = −1; is undefined for x = 3. Set up a sign chart with x = 3 and x = −1 Let 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, The values I get determine the sign (+ or −) of on each interval. I’ve marked the signs
above the sign chart.

The original inequality asks where 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 First, move everything to one side and combine terms over a common denominator: for x = 2 and is undefined for x = −3. These are the break points on the sign
chart: The inequality asks where is greater than or equal to 0. The solution is x ≤ −3 or
x ≥ 2.
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