Now try 0 from Interval B.
Original inequality
Let
False
The numbers in Interval B are notsolutions. Verify that the test number 5 from Inter-
val C satisfies the inequality, so all numbers there are also solutions.
Based on these results (shown by the colored letters in FIGURE 19), the solution set
includes the numbers in Intervals A and C, as shown on the graph in FIGURE 20. The
solution set is written in interval notation as
1 - q, - 32 ́ 1 4, q 2.
- 1270
02 - 0 - 127 x=0.
?
0
x^2 - x- 1270
554 CHAPTER 9 Quadratic Equations, Inequalities, and Functions
NOW TRY
EXERCISE 2
Solve and graph the solution
set.
x^2 + 2 x- 870
–3 04
FIGURE 20
NOW TRY
EXERCISE 3
Solve each inequality.
(a)
(b) 14 x- 122 6- 3
14 x- 122 7- 3
Solving Special Cases
Solve each inequality.
(a)
Because is never negative, it is always greater than Thus, the
solution set for is the set of all real numbers,
(b)
Using the same reasoning as in part (a), there is no solution for this inequality.
The solution set is 0.
12 x- 322 6- 1
12 x- 322 7- 1 1 - q, q 2.
12 x- 322 - 1.
12 x- 322 7- 1
EXAMPLE 3
Solving a Quadratic Inequality
Step 1 Write the inequality as an equation and solve it.
Step 2 Use the solutions from Step 1 to determine intervals.Graph the
numbers found in Step 1 on a number line. These numbers divide
the number line into intervals.
Step 3 Find the intervals that satisfy the inequality. Substitute a test
number from each interval into the original inequality to determine
the intervals that satisfy the inequality. All numbers in those intervals
are in the solution set. A graph of the solution set will usually look
like one of these. (Square brackets might be used instead of paren-
theses.)
or
Step 4 Consider the endpoints separately.The numbers from Step 1 are
included in the solution set if the inequality symbol is or. They
are not included if it is 6 or 7.
... Ú
This agrees with the solution set found in Example 1(a). NOW TRY
In summary, follow these steps to solve a quadratic inequality.
NOW TRY
NOW TRY ANSWERS
2.
- (a) 1 - q, q 2 (b) 0
1 - q, - 42 ́ 1 2, q 2
–4 0 2