Chapter Fourteen — Quadratic Equations and Inequalities
The Humongous Book of Algebra Problems
316
One-Variable Quadratic Inequalities
Inequalities that contain x
2
Note: Problems 14.35–14.37 refer to the inequality x
2
– x – 12 ≤ 0.
14.35 Identify the critical numbers of the quadratic expression.
The critical numbers of an expression are the values that make the expression
equal zero or cause it to be undefined. No real number, when substituted into
a quadratic expression, can cause a quadratic to be undefined, so calculate the
x-values for which x
2
– x – 12 = 0. Solve the equation by factoring,
using the method outlined in Problems 14.1–14.10.
The critical numbers of x
2
– x – 12 are x = –3 and x = 4.
Note: Problems 14.35–14.37 refer to the inequality x
2
– x – 12 ≤ 0.
14.36 Solve the inequality.
According to Problem 14.35, the critical numbers of the quadratic are x = –3
and x = 4. When graphed, those two values split the number line into three
intervals, as illustrated by Figure 14-1.
Figure 14-1: The critical numbers x = –3 and x = 4 divide the number line into three
intervals. Note that the critical numbers are plotted as closed points
due to the inequality sign ≤ in the original problem.
To decide whether to include the intervals pictured in Figure 14-1 as part of the
solution to the inequality, choose one “test value” from each, such as x = –4,
x = 0, and x = 5. Substitute each of the test values into the inequality. If a test
value satisfies the inequality (that is, results in a true inequality statement),
then the entire interval from which that test value was taken represents valid
solutions to the inequality.
The right side
of the inequality
needs to equal 0
before you do
anything else.
Algebraic
expressions
run the risk of
being undened
either when
they’re fractions
(because the
denominator could be
0) or when they contain
radicals with an even
index (because you
could have a negative
inside). Neither of
those things occur in a
quadratic expression,
so don’t worry about
quadratics being
undened. Focus
on when they
equal 0.
For more
information on closed
(solid) and open (hollow)
points on a graph—and
how they’re related to
inequality signs—look at
Problem 7.13.
x = –4 represents the left interval
of Figure 14.1 (because –4 ≤ –3), x = 0
represents the middle interval (because 0 is
between –3 and 4), and x = 5 represents the
right interval (because 5 ≥ 4).
Chapter Fourteen — Quadratic Equations and Inequalities
The Humongous Book of Algebra Problems
317
Of the three test values, only x = 0 satisfies the inequality. Therefore, the
solution is –3 ≤ x ≤ 4, the interval that contains x = 0.
Note: Problems 14.35–14.37 refer to the inequality x
2
– x – 12 ≤ 0.
14.37 Graph the solution to the inequality.
According to Problem 14.36, the solution to the inequality is –3 ≤ x ≤ 4. Graph
the solution by darkening that interval on the number line and using solid
endpoints, as illustrated by Figure 14-2.
Figure 14-2: The graph of the inequality x
2
– x – 12 ≤ 0.
Note: Problems 14.38–14.39 refer to the inequality 12x – 12x
2
≥ 3.
14.38 Solve the inequality.
Rewrite the inequality in form ax
2
+ bx + c ≥ 0.
–12x
2
+ 12x – 3 ≥ 0
Divide each term of the inequality by –3.
Calculate critical numbers by setting the quadratic expression equal to 0 and
solving for x.
Plot the critical number on a number line using a solid point, as illustrated by
Figure 14-3.
Use solid dots
on the number line
when the inequality
contains ≤ or ≥.
You could
divide by +3,
or you could choose
not to divide by
anything at all. The
nal answer isn’t
affected.
Remember
to reverse the
inequality symbol
when you multiply
or divide by a
negative.
Chapter Fourteen — Quadratic Equations and Inequalities
The Humongous Book of Algebra Problems
318
Figure 14-3: The quadratic equation 12x – 12x
2
≥ 3 has only one critical point, .
The critical point splits the number line in Figure 14-3 into two intervals:
and . Choose test points from both intervals (such as x = 0 and x = 1) to
identify the solution to the inequality.
Neither interval contains solutions to the inequality. The only valid solution is
.
Note: Problems 14.38–14.39 refer to the inequality 12x – 12x
2
≥ 3.
14.39 Graph the solution to the inequality.
According to Problem 14.38, the solution to the inequality is the single value
. To graph the solution, plot a single solid point at , as illustrated
by Figure 14-4.
Figure 14-4: The solution to the inequality 12x – 12x
2
≥ 3 is a single point on the
number line.
is a critical
number, so when you plug it
into 4x
2
– 4x + 1, you get 0.
Chapter Fourteen — Quadratic Equations and Inequalities
The Humongous Book of Algebra Problems
319
Note: Problems 14.40–14.41 refer to the inequality 2x
2
+ 3x – 1 > 0.
14.40 Solve the inequality.
Identify critical numbers using the quadratic formula, as the quadratic cannot
be factored.
Plot the critical numbers, and ,
on a number line, as illustrated by Figure 14-5, and choose test values (such as
x = –2, x = 0, and x = 2) from the three resulting intervals.
Figure 14-5: Plot the critical numbers of the quadratic using open points, as the
statement contains the > inequality symbol.
The solution to the equation is or , the intervals
that contain x = –2 and x = 2.
Type the roots
into a calculator
to get approximate
decimal values so you
can gure out where
to plot the points.
Chapter Fourteen — Quadratic Equations and Inequalities
The Humongous Book of Algebra Problems
320
Note: Problems 14.40–14.41 refer to the inequality 2x
2
+ 3x – 1 > 0.
14.41 Graph the solution to the inequality.
To graph the solution, graph the inequalities and
on the same number line, as illustrated by Figure 14-6.
Figure 14-6: The graph of 2x
2
+ 3x – 1 > 0.
Note: Problems 14.42–14.43 refer to the inequality 4x
3
+ 5x
2
– 6x < 0.
14.42 Solve the inequality.
Calculate the critical numbers by factoring the expression.
Factor by decomposition.
Set each factor equal to zero and solve for x to get critical numbers x = 0, x = –2,
and . As illustrated by Figure 14-7, those critical numbers separate the
number line into four intervals: x < –2, –2 < x < 0, , and .
Figure 14-7: Three different critical numbers will separate a number line into four
intervals.
This isn’t
a quadratic
equation, but you’ll
get a quadratic
expression once you
factor out the
greatest common
factor, x.
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