Chabot Mathematics
§8.5 PolyNom
InEqualities
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Review § 8.4
MTH 55
 Any QUESTIONS About
• §8.4 → Equations in Quadratic Form
 Any QUESTIONS About HomeWork
• §8.3 → HW-40
Chabot College Mathematics
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
PolyNomial InEquality
 PolyNomial InEqualities:
3
2 2
5x  8  3, 3x  2 x  x , x  5x  4  0.
 Second-Degree polynomial inequalities
in one variable are called Quadratic
inequalities.
 To solve polynomial inequalities, focus
attention on where the outputs of a
polynomial function are positive and
where they are negative
Chabot College Mathematics
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
2
Example  Solve x  2 x  8  0.
 SOLUTION: Consider the related
function y = f(x) = x2 + 2x − 8
y
and its graph.
8
 Since the graph positive
y-values
opens upward,
-5
the y-values are
positive outside
the interval formed
by the x-Intercepts
Chabot College Mathematics
4
6
positive
y-values
4
2
-4 -3 -2 -1
-2
-4
-6
-8
1
2 3 4 5
x
y = x2 + 2x – 8
-10
-12
-14
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
2
Example  Solve x  2 x  8  0.
 SOLUTION: Thus y > 0 for x-values
OUTSIDE of the region described
y
the x-intercepts
8
 It follows that the
solution set of the
inequality is
x | x  4 or x  2,
or  , 4    2,  .
6
positive
y-values
positive
y-values
4
2
-5 -4 -3 -2 -1
-2
-4
-6
-8
1
2 3 4 5
y = x2 + 2x – 8
-10
-12
-14
Chabot College Mathematics
5
x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
StreamLine Solution
 In the next example, simplify the
process by concentrating on the sign of
a polynomial function over each
interval formed by the x-intercepts.
 We will do this by tracking the sign of
each factor.
• By looking at how many positive or
negative factors are being multiplied, we
will be able to determine the sign of the
polynomial function.
Chabot College Mathematics
6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Solve f x   x  9 x  0
3
 SOLUTION - first solve the related
equation
f (x) = 0
x3 − 9x = 0
x(x − 3)(x + 3) = 0
Factoring
Using the principle
x = 0 or x − 3 = 0 or x + 3 = 0 of zero products
x = 0 or x = 3 or x = −3.
Chabot College Mathematics
7
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Solve f x   x  9 x  0
3
 The function f(x) has zeros at −3, 0 & 3.
 We will use the factorization
f(x) = x(x − 3)(x + 3).
 The product x(x − 3)(x + 3) is positive or
negative, depending on the signs of x,
x − 3, and x + 3.
 This is easily determined using a chart
Chabot College Mathematics
8
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Solve f x   x  9 x  0
3
 The Sign Chart for f(x) = x(x – 3)(x + 3)
Interval:
(, 3)
Sign of x:
Sign of x – 3:
Sign of x + 3:
Sign of product
x(x – 3)(x + 3)
–
–
–
–
(3,0)
(0,3)
(3, )
–
–
+
+
–
+
+
+
+
–3
0
+
–
3
+
 Note that ANY Negative No. < 0
Chabot College Mathematics
9
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Solve f x   x  9 x  0
3
 A product is negative when it has an
odd number of negative factors.
 Since the sign allows for EQuality, the
ENDpoints −3, 0, and 3 ARE solutions.
 The chart then displays solution set:
(, 3]  [0, 3], or  x | x  3 or 0  x  3.
Chabot College Mathematics
10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Solve PolyNom InEq by Factoring
1. Add or subtract to get 0 on one side of
a “f(x) Greater/Less Than” InEquality

e.g.; 3x2 < 7 − 5x  3x2 + 5x − 7 < 0
2. Next solve the related f(x) = 0
polynomial equation by factoring.
3. Use the numbers found in step (1) to
divide the number line into intervals.
4. Using a test value (or Test-Pt) from
each interval, determine the sign of
each factor over that interval.
Chabot College Mathematics
11
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Solve PolyNom InEq by Factoring
5. Determine the sign of the product of
the factors over each interval.
6. Select the interval(s) for which the
inequality is satisfied and write
set-builder notation or interval notation
for the solution set.
7. Include the endpoints of the intervals
whenever ≤ or ≥ describes the InEquality
Chabot College Mathematics
12
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example Solve x  2x  7.
2
 SOLUTION – Obtain Zero on One Side
the InEquality
x 2  2x ─ 7  2x  7  2x ─ 7
x 2  2x ─ 7  0
 Next Solve the
2
x

2
x

7

0
2
Associated
x  2x  7  0
Equation by
2
Quadratic x   2   2   4 17 
Formula
2 1
Chabot College Mathematics
13
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example Solve x  2x  7.
2
 Isolating x from Quadratic Formula
x  1.8
2  32
so
x
 1 2 2
x  3.8
2
 This divides the No. line into 3 intervals
0
0
– 4 –3 –2 –1 0
1 2 2
1
2
3 4
5
1 2 2
 Pick “Round Numbers as Test Points
• −3, 0, and 4
Chabot College Mathematics
14
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example Solve x  2x  7.
2
 Constructing the Sign Table: x2−2x−7 > 0
Interval
Point
Value
Result
x 2  2x  7
,1  2 2 
1  2 2,1  2 2 
1  2 2, 
−3
8
+
0
−7
–
4
1
+
x  2x  7  0
2
Chabot College Mathematics
15
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example Solve x  2x  7.
2
 The Sign Intervals on the Number Line
++++++++ 0 ––––––––– 0+++++
– 4 –3 –2 –1 0
1 2 2
1
2
3 4
5
1 2 2
 Thus for x2 > 2x +7  x2 − 2x − 7 > 0 is
POSITIVE (exceeds Zero) in intervals
,1  2 2 
)
– 4 –3 –2 –1 0
1 2 2
Chabot College Mathematics
16
1  2
1
2

2,  .
(
3 4
1 2 2
5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Solve x2 + x ≤ 12
 Solution: write the related equation
x2 + x − 12 = 0 Set the quadratic expression equal to 0
(x − 3)(x + 4) = 0 Factor
x = 3 or x = −4 Use the zero-products theorem
 Plot the Break-Points” on a number line
to create boundaries between the three
intervals
I
-10 -9
-8
II
-7
Chabot College Mathematics
17
-6
-5
-4
-3
-2
-1
III
0
1
2
3
4
5
6
7
8
9
10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Solve x2 + x ≤ 12
 Choose a test number (or Test-Pt) from
each interval and substitute that value
into x2 + x − 12 ≤ 0
I
-10 -9
Test

-8
-7
-6
-5
-4
-3
-2
-1
III
0
1
2
3
4
5
6
7
8
9
10
For [−4, 3]
choose 0
For [3, ) we
choose 4
(–5)2 + (–5) – 12 (0)2 + (0) – 12
= 25 – 5 – 12
= 0 – 12
(4)2 + (4) – 12
= 16 + 4 – 12
For (−, −4],
choose −5
=8
= –12
=8
False
True
False
Chabot College Mathematics
18
II
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Solve x2 + x ≤ 12
 The Table Reveals the at the CENTRAL
interval is the Solution
• Last time the solution was the
EXTERIOR Intervals
 The solution is the interval [−4, 3]
[
-10 -9
-8
Chabot College Mathematics
19
-7
-6
-5
-4
-3
-2
-1
0
1
2
]
3
4
5
6
7
8
9
10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Solve x2 + 5x > 0
 Solution: write & Solve the related
2
equation
x  5x  0
x( x  5)  0
x  0 or x  5
 Plot the Break-Pts on the No. line to
form boundaries between the 3 intervals
I
-10 -9
II
-8
-7
Chabot College Mathematics
20
-6
)
-5
-4
-3
III
-2
-1
(0
1
2
3
4
5
6
7
8
9
10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Solve x2 + 5x > 0
 Choose a Test-Point from each interval
and substitue that value into The
InEquality x2 + 5x > 0
For (−, −5),
choose −6
For (0, )
choose 1
(–6)2 + 5(–6) > 0 (–1)2 + 5(–1) > 0
36 – 30 > 0
1–5>0
(1)2 + 5(1) > 0
1+5>0
–4 > 0
False
6>0
True
6>0
True
Chabot College Mathematics
21
For (−5, 0)
choose −1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Solve x2 + 5x > 0
 The Sign Table Reveals that the
EXTERIOR intervals are the Solution
 The solution in interval Notation
(−∞, −5) U (0, ∞)
 The Solution on the Number Line
(
)
-10 -9
-8
Chabot College Mathematics
22
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  (x + 3)(x + 1)(x – 2) < 0
 Solution: write the related equation
(x + 3)(x + 1)(x – 2) = 0 Write Related Eqn
x + 3 = 0 or x + 1 = 0 or x − 2 = 0
Using Zero-Products Theorem
x = −3 or x = −1 or x = 2 Solving for x
 Plot the Break-Points” on a number line
to create FOUR intervals
-10 -9
-8
Chabot College Mathematics
23
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  (x + 3)(x + 1)(x – 2) < 0
 Solution: Create Sign/Truth Chart
-10 -9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
(, 3) (3, 1)
Interval
5
6
7
8
9
10
(1, 2)
(2, )
Test Number
4
2
0
3
Test Results
True or False
18 < 0
True
4<0
False
6 < 0
True
24 < 0
False
 Thus the Soln Set: (−∞, −3) U (−1, 2)
• On Number Line:
)
-10 -9
-8
-7
Chabot College Mathematics
24
-6
-5
-4
-3
(
-2
-1
)
0
1
2
3
4
5
6
7
8
9
10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Skid Marks
 A car involved in an accident left skid
marks over 75 feet long. Under the road
conditions at the accident, the distance
d (in feet) it takes a car traveling v miles
per hour to stop is given by the equation
2
d  0.05v  v.
 The accident occurred in a 25 mph zone.
 Was the driver Speeding?
Chabot College Mathematics
25
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Skid Marks
 SOLUTION: Solve the InEquality:
(stopping distance) > 75 feet, or 0.05v  v  75.
2
0.05v 2  v  75
0.05v 2  v  75  0
0.05v 2  v  75  0
5v 2  100v  7500  0
v  20v  1500  0
2
Chabot College Mathematics
26
v  50 v  30   0
v  50  0 or v  30  0
v  50
or v  30
 The Solns divide
the number line
into 3 intervals
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Skid Marks
 Make Sign-Chart, and Divide No. Line
Interval
Point
Result
– 60
Value
2
0.05v  v
45
(–∞, –50)
(–50, 30)
0
–75
–
(30, ∞)
40
45
+
+
++++++ 0 ––––––––––––––– 0 ++++++
– 60 –50
Chabot College Mathematics
27
0
30 40
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Skid Marks
 ++++ on No. Line indicates Speeding
++++++ 0 ––––––––––––––– 0 ++++++
– 60 –50
0
30 40
 For this situation, we look at only the
positive values of v. Note that the
numbers corresponding to speeds
between 0 and 30 mph (i.e., 0 ≤ v ≤ 30 )
are not solutions of 0.05v2+v
 Thus, the car was traveling more than 30
miles per hour  Excessive Speed
Chabot College Mathematics
28
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Solve x4 ≤ 1
 SOLUTION: Solve Related Eqn for x
x 1
4
x  1  0, x  1  0, x  1  0
2
x4  1  0
x

 These Breakx  1x  1x  1 0 Points divide the
number line into
x  1x  1x  1 0 3 intervals
2

x  1, x  1
1 x 1  0
2
2
2
Chabot College Mathematics
29
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Solve x4 ≤ 1
 Make Sign Chart & Associated No. Line
Value of

 Result
Interval Point x  1x  1 x 2  1
(−∞, –1)
−2
15
+
(–1, 1)
0
−1
–
(1, ∞)
2
15
+
++++++++0 –––0 +++++++
–3 –2 –1 0
Chabot College Mathematics
30
1
2
3 4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  Solve x4 ≤ 1
 The Sign-Regions on No. Line
++++++++0 –––0 +++++++
–3 –2 –1 0
1
2
3 4
 The solution set consists of all x
between −1 & 1, including both −1 & 1.
–2
[
–1
0
]
1
2
{−1 ≤ x ≤ 1} or [−1, 1]
Chabot College Mathematics
31
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
ONE-Sign Theorem
 If a polynomial equation has NO
REAL solution, then the polynomial
is either ALWAYS positive or
always negative
 Example  Use One-Sign Theorem to
Solve InEquality:
x – 2x  2  0.
2
Chabot College Mathematics
32
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  1-Sign Theorem x 2 – 2x  2  0.
 SOLUTION:
Solve Related Eqn
x – 2x  2  0.
2
 The Expression does NOT factor so use
Quadratic Formula
• First Test Discriminant for Real Solutions
b – 4ac  2   4 12    4
2
2
 Discriminant is NEGATIVE →
NO Real-Number Solution
Chabot College Mathematics
33
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Example  1-Sign Theorem x 2 – 2x  2  0.
b – 4ac  2   4 12    4
2
2
 Since the discriminant is negative, there
are no real roots.
 Use 0 as a test point, which yields 2.
 The inequality is always positive, the
solution set is ALL REAL Numbers
• Using Interval Notion: (−∞, ∞).
Chabot College Mathematics
34
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
WhiteBoard Work
 Problems From §8.5 Exercise Set
• 16, 28, 34, 68
 x2 − x − 6 < 0
Graphically
Chabot College Mathematics
35
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
All Done for Today
Acapulco
Cliff
Diving
Chabot College Mathematics
36
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
37
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
Graph y = |x|
6
 Make T-table
x
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
Chabot College Mathematics
38
5
y = |x |
6
5
4
3
2
1
0
1
2
3
4
5
6
y
4
3
2
1
x
0
-6
-5
-4
-3
-2
-1
0
1
2
3
-1
-2
-3
-4
-5
file =XY_Plot_0211.xls
-6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
4
5
6
5
5
y
4
4
3
3
2
2
1
1
0
-10
-8
-6
-4
-2
-2
-1
0
2
4
6
-1
0
-3
x
0
1
2
3
4
5
-2
-1
-3
-2
M55_§JBerland_Graphs_0806.xls
-3
Chabot College Mathematics
39
-4
M55_§JBerland_Graphs_0806.xls
-5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt
8
10