1.6 equations and inequalities
DESCRIPTION
MATH 17 - COLLEGE ALGEBRA AND TRIGONOMETRYTRANSCRIPT
Chapter 1.6
Equations and
Inequalities
1
Algebraic Expressions
2
Algebraic expressions are symbolic
forms of numbers.
r hr
h
2
Algebraic Expressions
: number of hours you need
to finish a job
1: amount of work you can do
in an hour.
x
x
3
Algebraic Expressions
: number of hours your friend
needs to finish the same job
1 1: the amount of work you
can do together in an hour
y
x y
4
Example 1.6.1
Express the following as algebraic
expressions.
1. The perimeter of a rectangle two
sides of which are and .
2 2
a b
a b a
b
a
b 5
Example 1.6.1
2. Three consecutive odd integers
if is the smallest
2 4
x
x x x
6
Example 1.6.1
3. A number of two digits, if the
unit's digit is and the ten's
digit is .
25 2 10 5 73 7 10 3
10
x
y
y x
7
Equations and Inequalities
Equation
Inequality
is a statement that two
algebraic expressions are equal.
is a statement that two
algebraic expressions are not equal.
8
Example 1.6.2
2
2
The volume of a right circular cylinder
whose height is 3 times the radius is
cubic units.
radius: height: 3
volume radius height
3
r r
r r
9
Example 1.6.3
The distance of from 3 is not
greater than 4.
3 4
x
x
10
Example 1.6.4
Formulate the following problems.
1. The sum of 3 numbers is 34. The second
is 3 less than the first, and the third is 5
more than twice the first. What are the
numbers?
1st: 2nd: 3 3rd: 2 5
3 2 5 34
x x x
x x x
11
Example 1.6.4
2. A man has Php 10,000 to invest. He can
invest a part of it at 6% and the remainder
at 10%. How much should he invest at each
rate in order to realize an outcome of Php 760
on the two investments?
amo
unt invested at 6%:
amount invested at 10%: 10,000
0.06 0.10 10,000 760
x
x
x x
12
Example 1.6.4
3. One side of a rectangle exceeds 3 times the
other side by 2. Find the dimensions of the
smallest rectangle if the perimeter is at
least 36 m.
length of one side:
length of the other side: 3 2
perimeter o
x
x
f the rectangle: 2 2 3 2
2 2 3 2 36
x x
x x
13
Solution Set
The of an equation
or inequality is the set of all real
or complex values of the variables
that satisfy the given equation
or
solutio
inequal
n set
ity.
14
Solving
Equations/Inequalities
To an equation/inequality
means to find the solution set
of the equation/inequal
solve
ity.
15
Linear Equations
0ax b
bx
a
bSS
a
16
Example 1.6.5
Solve the following equations for
the indicated variable.
1. , for 4
4
4 4
4
bhV b
V bh
V Vb b
h h
VSS
h
17
Example 1.6.5
2. for y b m x a x
y bx a
m
y b y ba x x a
m m
y bSS a
m
18
Quadratic Equations
2
2
2
2
0
9 3, 3 3, 3
12 2 3, 2 3 2 3, 2 3
16 4 , 4 4 , 4
ax bx c
x x SS
x x SS
x x i i SS i i
19
Quadratic Equations
2
2
2
222 2
2
2 2
2 2 2
2 2
0
completing the square:
4 4 2 2
4
2 4 4
ax bx c
ax bx c
b cx x
a a
bb b c b bax xa a a a a
b b ac bx
a a a
20
Quadratic Equations
2 2
2
2
2
2
2
4
2 4
4
2 4
4
2 2
4
2
b b acx
a a
b b acx
a a
b b acx
a a
b b acx
a
21
Quadratic Equations
2
2
2 2
0
4
2
4 4,
2 2
ax bx c
b b acx
a
b b ac b b acSS
a a
Quadratic Formula
22
Solving Quadratic Equations
2
2
1. Write the equation as 0
2. . Use the quadratic formula.
b. Factor and apply
the theorem:
If 0 then 0 or 0
ax bx c
a
ax bx c
mn m n
23
Example 1.6.6
2
2
Solve the following.
1. 2 8
2 8 0
4 2 0
4 0 2 0
4 2
4,2
x x
x x
x x
x x
x x
SS
24
Example 1.6.6
2
2
2. 2 5 4
2 4 5 0
2 4 5
4 16 4 2 5 4 24
4 4
2 2 64 2 6 2 6
4 4 2
x x
x x
a b c
x
ii i
2 4
2
b b acx
a
25
Example 1.6.6
2
2
2 5 4
2 4 5 0
2 6
2
2 6 2 6,
2 2
x x
x x
ix
i iSS
26
Example 1.6.6
2
2
2
1 1 2 33.
3 3 9
1 1 2 3
3 3 3 3
: 3 3
x x
x x x
x x
x x x x
LCD x x
27
Example 1.6.6
2
2
2
2
2
2
1 1 2 3
3 3 3 3
: 3 3
1 1 2 33 3 3 3
3 3 3 3
3 3 3 32 3
3 3
3 3 2 3, 3, 3
3 3 2 3, 3, 3
2 3 0, 3, 3
x x
x x x x
LCD x x
x xx x x x
x x x x
x x x xx x
x x
x x x x x
x x x x x
x x x
28
Example 1.6.6
2 2 3 0, 3, 3
3 1 0
3 0 1 0
3 1
3 is an extraneous solution.
1
x x x
x x
x x
x x
SS
29
Example 1.6.6
2
2
1 1 2 3
3 3 3 3
1 3
Check:
1 :
1 1 1 1 3
1 3 1 3 4 2 4
1 2 1 3 6 3
1 3 1 3 8 4
x x
x x x x
x x
x
30
Example 1.6.6
21 1 2 3
3 3 3 3
If 3,
1 is undefined
3
3 is an extraneous solution
1
x x
x x x x
x
x
SS
31
: no. of boys
7 : no. of girls
x
x
Example 1.6.7
A man gave Php 120M to 7 children,giving Php 60M to the boys and thesame amount to the girls. In this way,each boy received Php 5M more thaneach girl. Find the number of boys andgirls.
32
: no. of boys
7 : no. of girls
60: money in M received by each boy
60: money in M received by each girl
7
60 605
7
x
x
x
x
x x
Example 1.6.7
33
2
2
60 605
7
: 7
60 607 7 5
7
420 60 60 5 7 , 0,7
420 60 60 35 5 , 0,7
5 155 420 0, 0,7
x x
LCD x x
x x x xx x
x x x x x
x x x x x
x x x
Example 1.6.7
34
2
2
5 155 420 0, 0,7
31 84 0
28 3 0
28 3
There are 3 boys and 4 girls.
x x x
x x
x x
x x
Example 1.6.7
35
Discriminant
2
2
4
2
Discriminant: 4
b b acx
a
D b ac
36
Discriminant
2 4
2
0 : only one real solution
0 : two distinct real solutions
0 : two distinct complex solutions
b b acx
a
D
D
D
37
Example 1.6.8
2
2
Find the nature of the solutions
of 4 8 3 0.
4 8 3
4 64 4 4 3 16 0
The equation has 2 distinct
real solutions.
x x
a b c
b ac
38
Sum/Product of Roots
2 2
Sum:
4 4
2 2
2
2
b b ac b b ac
a a
b
a
b
a
39
Sum/Product of Roots
2 2
22 2
2
2 2
2
Product:
4 4
2 2
4
4
4
4
b b ac b b ac
a a
b b ac
a
b b ac
a
40
Sum/Product of Roots
2 2
2
2 2
2
2
4
4
4
4
4
4
b b ac
a
b b ac
a
ac
a
c
a
41
Sum/Product of Roots
2 0
Sum of Roots:
Product of Roots:
ax bx c
b
a
c
a
42
Example 1.6.9
2
2
2
Find the sum and product of the
roots of 2 3 4.
2 3 4
2 3 4 0
2 3 4
3 3 4Sum: Product: 2
2 2 2
x x
x x
x x
a b c
43
Cubic Equations
3 2 0
To solve:
Factoring
Using quadratic formula
ax bx cx d
44
Example 1.6.10
3
3
3
2
2
Solve 1.
1
1 0
1 1 0
1 1 0
1 1 1
1 1 4 1 3
2 2
x
x
x
x x x
x x x
a b c
x45
Example 1.6.10
3 1
1
1
x
x
SS
46
Division of Polynomials
Long Division
Synthetic Division
47
Example 1.6.11
3Divide 3 2 by 2 using
1. long division
2. synthetic division
x x x
48
2
3 2
3 2
2
2
2 1
1. 2 0 3 2
2
2 3
2 4
2
2
4
x x
x x x x
x x
x x
x x
x
x
49
2
3 2
3 2
2
2
2 1
1. 2 0 3 2
2
2 3
2 4
2
2
4
x x
x x x x
x x
x x
x x
x
x
Quotient
Remainder50
3 2
2
2. 2 0 3 2
2 1 0 3 2
2 4 2
1 2 1 4
2 1 4
x x x x
x x r
51
Division of Polynomials
If a polynomial is divided
by we get a quotient
and a remainder .
P x
x a Q x
r
P x rQ x
x a x a
P x x a Q x r52
3
2
32
3 2
3 2 divided by 2
2 1
4
3 2 42 1
2 2
3 2 2 2 1 4
P x x x x
Q x x x
r
x xx x
x x
x x x x x
P x rQ x
x a x a
P x x a Q x r
53
Remainder Theorem
If a polynomial is divided by ,
the remainder is equal to .
P x x a
P a
54
Example 1.6.11
3
3
3
Use the remainder theorem to
determine the remainder when
3 2 is divided by 2.
3 2
2 2 3 2 2 8 6 2 4
4 is the remainder.
x x x
P x x x
P
55
2
3 2
3 2
2
2
2 1
1. 2 0 3 2
2
2 3
2 4
2
2
4
x x
x x x x
x x
x x
x x
x
x
Quotient
Remainder56
Factor Theorem
is a root of the equation 0
if and only if is a factor of .
a P x
x a P x
57
Example 1.6.11
4 3
4 3
4 3
4 3
Use the factor theorem to determine
if 1 is a factor of 2 4 1.
1 1
Is 1 a solution to 2 4 1 0
1 2 1 1 4 1 1 2 1 4 1 0
1 is a factor of 2 4 1.
x x x x
x x
P x x x x
P
x x x x
58
Example 1.6.11
3 2
3 2
3 2
3 2
3 2
23
Using the Factor Theorem, solve the equation
6 11 6 (Hint: Show that 1 is a factor.)
6 11 6
6 11 6 0
Is 1 a factor of 6 11 6?
6 11 6
1 1 6 1 11 1 6 0
1 is a factor of
x x x x
x x x
x x x
x x x x
P x x x x
P
x
3 2 6 11 6.x x x 59
3 2
3 2
2
3 2
2
6 11 6 1
1 6 11 6
1 1 6 11 6
1 5 6
1 5 6 0
5 6 0
6 11 6 0
1 5 6 0
x x x x
x x x x
x x r
x x x
x x x
60
3 2
2
6 11 6 0
1 5 6 0
1 2 3 0
1 0 2 0 3 0
1 2 3
1,2,3
x x x
x x x
x x x
x x x
x x x
SS
61
Fundamental Theorem of
Algebra
Every polynomial equation 0
with complex coefficients has at least
one root.
P x
62
Theorem
2
3 2
Every polynomial of degree can be
expressed as product of linear factors.
3 2 degree: 2
= 2 1
6 11 6 degree: 3
= 1 2 3
n
n
x x
x x
x x x
x x x
63
Theorem
1 2
1 2
1 2
1 2
Every polynomial equation 0
of degree has at most distinct roots.
In general, a polynomial equation can be
written as
0
, ,..., and are the distinct roots and
...
mk k k
m
m
P x
n n
P x a x r x r x r
r r r
k k
mk n 64
1 2
1 2
if
1, is a simple root.
2, is a double root.
, is a root of multiplicity .
mk k k
m
i i
i i
i i
P x a x r x r x r
k r
k r
k m r m
0
65
Example 1.6.12
2 4
Determine the roots of
1 3 5 0
distinct roots: 1, 3, 5
1 is a simple root.
3 is a double root.
5 is a root of multiplicity 4.
P x x x x
66
Theorem
A polynomial equation 0 of degree
has exactly roots, a root of multiplicity
being counted as roots.
P x
n n
k k
67
Example 1.6.13
2
Form an equation which has
1 as a double root
2 and 4 as simple roots
and no others.
1 2 4 0x x x
68
Theorem
The roots of 0 are precisely
the additive inverses of the roots of
0.
2 is a root of 0
2 is a root of 0
P x
P x
P x
P x
69
Example 1.6.15
5 3 2
5 3 2
5 3 2
5 2
Obtain an equation whose roots are
the negatives of the roots of
2 3 4 2 0
2 3 4 2 0
2 3 4 2 0
2 3 4 2 0
x x x x
P x x x x x
P x x x x x
x x x x
70
Variation of Signs
descending powers
variation of sign
If the terms of are arranged in
of , we say that
a occurs when two
successive terms have different signs.
P x
x
71
Example 1.6.16
5 4 2
5 3 2
Determine the number of variation of
signs for each polynomial.
1. 2 3 4
variation of signs: 3
2. 2 3 4 2
variation of signs: 4
x x x x
x x x x
72
Descartes Rule of Signs
The of the
polynomial equation 0 with
real coefficients is
number of positive roots
number of variat
equal to the
in
or less than that by
ion of signs
an even number.
P x
P x
73
Descartes Rule of Signs
The of 0
is
number of negative roots
number of positive roots the of 0.
P x
P x
74
Example 1.6.17
7 4 3
7 4 3
Determine the possible number of positive,
negative, and complex roots of
2 4 2 5 0
2 4 2 5 0
positive roots: 2 or 0
negative roots: 3 or 1
complex roots: 6 or 4 or 2
P x x x x x
P x x x x x
75
Rational Root Theorem
2
0 1 2
0
Consider
... 0, 0
with integral coefficients.
If is a root, where and are
relatively prime integers, then is a
factor of and is a factor of .
n
n n
n
a a x a x a x a
pp q
q
p
a q a
76
Example 1.6.18
3 2
3 2
3 2
Solve 2 3 7 3 0
: 1, 3 : 1, 2
1 3: 1, 3, ,
2 2
2 3 7 3 0
2 3 7 3 0
positive roots: 3 or 1 negative roots: 0
x x x
p q
p
q
x x x
x x x
77
3 2
3 2
2 3 7 3 0
1 3: 1,3, ,
2 2
2 3 7 3
1 2 3 7 3
2 1 6
2 1 6
1 is not a root
3
x x x
p
q
x x x
78
3 2
2
2 3 7 3
12 3 7 3
2
1 1 3
2 2 6
1 1 is a root and is a factor.
2 2
12 2 6 0
2
0
x x x
x
x x x
79
2
2
2
12 2 6 0
2
12 3 0
2
10 3 0
2
1 1 11
2 2
1 1 11 1 11, ,
2 2 2
x x x
x x x
x x x
ix x
i iSS
80
Example 1.6.19
4 3 2
4 3 2
4 3 2
Solve 8 14 13 6 0
: 1, 2, 3, 6 : 1
: 1, 2, 3, 6
8 14 13 6 0
8 14 13 6 0
positive roots: 0
negative roots: 4 or 2 or 0
x x x x
p q
pq
x x x x
x x x x
81
4 3 2
4 3 2
3 2
8 14 13 6 0
: 1, 2, 3, 6
8 14 13 6
1 1 8 14 13 6
1 7 7 6
1 7 7 6 0
1 is a root, 1 is a factor.
1 7 7 6 0
x x x x
pq
x x x x
x
x x x x
82
3 2
3 2
2
1 7 7 6 0
: 1, 2, 3, 6
7 7 6
6 1 7 7 6
6 6 6
1 1 1 0
6 is a root, 6 is a factor
1 6 1 0
x x x x
pq
x x x
x
x x x x
83
2
2
1 6 1 0
1 0 6 0 1 0
1 6 1 1 1
1 1 4 1 1
2
1 3 1 3
2 2
1 3 1 31, 6, ,
2 2
x x x x
x x x x
x x a b c
x
i
i iSS
84
Example 1.6.20
Solution: The LCD of the RE is
Multiplying both sides by the LCD:
2
1 2 7
2 1 2x x x x
2 1 .x x
2
1 2 72 1 2 1
2 1 2
x x x xx x x x
85
Checking the results shows that the LCD
≠ 0 for .
Therefore, the solution set is .
2 x
2 SS
1 2 2 7 x x 2 2 x x
2
1 2 72 1 2 1
2 1 2
x x x xx x x x
86
Example 1.6.21
Solution:
The LCD of the fractions is .
Multiplying both sides by the LCD yields:
2
3 2 3
3 2 5 6x x x x
3 2 x x
3 2 2 3 3 x x
3 6 2 6 3
3
x x
x87
Checking the results shows that the LCD
= 0 for . Thus, 3 is NOT a solution,
hence, there is NO SOLUTION to the
equation.
Therefore, the solution set is .
3x
SS
88
Example 1.6.22
3 3 1x x
2 2
3 3 1x x
square both sides of the
equation not the equation
23 9 6 1x x x
29 7 2 0x x 9 2 1 0x x
9 2 0 1 0x x
2
19
x x
3 1 3x x
89
Checking:
2:
9x
1 :x
22 8
3 19 3
2 23
9 3
2 is an extraneous solution
9
2
1 3 1 3 3 1 3
1SS
90
Example 1.6.23
2 5 1 2x x
2
2 5 1 4x x
22 5 2 2 3 5 1 4x x x x
23 4 2 2 3 5 4x x x
23 2 2 3 5x x x91
2 29 4 2 3 5x x x
23 2 2 3 5x x x
2 29 8 12 20x x x
2 12 20 0x x
10 2 0x x
10 0 2 0
10 2
x x
x x92
Checking:
10 :x 2 10 5 10 1 8 4
10 is an extraneous root.
2 :x 2 2 5 2 1 4 2
2 is an extraneous solution
SS
93
Example 1.6.24
4 21 2x x 2
2 22 1 0x x
2
2 1 0x 2 1 0x
2 1x
x i
,SS i i94
Example 1.6.25
22 2
6 1 5 1 6 0x x
2
Let 1ux
26 5 6 0u u
2 3 3 2 0u u
3
2u
2
3u
95
2 31
2x
4, 6
5SS
3
2u
2
3u
2 2
13x
2 4 3x x
5 4x
4
5x
3 6 2x x
6x
96
1. 2 7 1x x
4 22. 3 2 0x x
Try this at home!
97
End of section 1.6.1 EQUATIONS
98
Inequalities
A statement that one mathematical
expression is greater than or less than
another is called an .
Goal: Find solutions and solution sets
for ineq
ineq
uali
u
t
ality
ies.
99
Interval Notation
, ,
, ,
, ,
, ,
x a x b a b x a x b a b
x a x b a b x a x b a b
x x a a x x a a
x x a a x x a a
100
Example 1.6.26
Find the solutions of the following
linear inequalities.
1. 5 3 7
3 7 5
4 12
4 12
4 4
3
3 ,3
x x
x x
x
x
x
SS x x
101
4 52. 1
2
4 52 2 1
2
4 5 2 2
4 2 2 5
2 7
7
2
7 7,
2 2
xx
xx
x x
x x
x
x
SS x x
102
3. 4 3 5 10
9 3 15
3 5
3,5
x
x
x
SS
103
4. 6 2 2 9
8 2 7
74
2
74
2
7,4
2
x
x
x
x
SS
104
Example 1.6.28
2
2
Find the solution set for the following.
1. 2 15
2 15 0
3 5 0
. . : 5,3
x x
x x
x x
C N
, 5 3,SS 105
22. 3 2 0
2 1 0
. . : 2, 1
x x
x x
C N
106
0
0
0 0
107
. . : 2, 1C N
2, 1SS 2 1 0x x
2
2
2
2
3. 6 9 0
3 0
3
4. 2 1 0
1 0
1
x x
x
SS
x x
x
SS
108
Example 1.6.29
Find the solution set for the following
rational inequalities.
3 1 3 1 2 81. 2 0
4 4
3 1 92 0 0
4 4
3 1 2 40
4
. . : 4,9
x x x
x x
x x
x x
x x
x
C N
109
90 . . : 4,9
4
xC N
x
0
0
0
4,9SS 110
4 32.
2 1 1
4 30
2 1 1
4 1 3 2 10
2 1 1
4 4 6 30
2 1 1
7 20
2 1 1
1 7. . : 1, ,
2 2
x x
x x
x x
x x
x x
x x
x
x x
C N
111
0
0
0
0
1 7
, 1 ,2 2
SS
7 2 1 7
0 . . : 1, ,2 1 1 2 2
xC N
x x
112
Equations Involving
Absolute Values
0
or
2
2 or 2
x a a
x a x a
x
x x
113
Example 1.6.30
Solve the following equations.
1. 2 5 3
2 5 3 or 2 5 3
2 2 2 8
1 4
1, 4
x
x x
x x
x x
SS
114
2
2 2
2 2
2
2. 3 4 6
3 4 6 or 3 4 6
6 0 3 4 6
3 2 0 7 6 0
3 6 1 0
2 6 1
3,1 , 2 and 6 are extr. soln.
x x x
x x x x x x
x x x x x
x x x x
x x x
x x x
SS
115
3. 41
4 or 41 1
4 4 4 4
3 4 5 4
4 4
3 5
4 4,
3 5
x
x
x x
x x
x x x x
x x
x x
SS
116
2
2
2
4. 1 2 1
12
1
12
1
1 12
1
1 2
1 2 or 1 2
3 1
1,3
x x
x
x
x
x
x x
x
x
x x
x x
SS 117
Inequalities Involving
Absolute Values
If 0
and
or
a
x a a x a
a x x a
x a x a x a
118
Example 1.6.31
Solve the following inequalities.
1. 1 4 8
8 1 4 8
9 4 7
9 7
4 4
7 9 7 9,
4 4 4 4
x
x
x
x
x SS
119
2. 6 7
6 7 or 6 7
1 13
1, , 13
x
x x
x x
SS
120
3. 2 1 1
1 2 1 and 2 1 1
1 2 1 0
3 2
2
3
2, ,0
3
x x
x x x x
x x x
x
x
SS
│ │
0 2/3
│ │
0 2/3
121
2
2 2
2
2
4. 4 2
4 2 or 4 2
4 2
6 0
2 3 0 . . : 3,2
x x
x x x x
x x
x x
x x C N
122
2 3 0 . . : 3,2x x C N
0
0
0 0
1, 3 2,SS
123
2
2 2
1
2
4 2
4 2 or 4 2
, 3 2,
2,1
, 3 2,1 2,
x x
x x x x
SS
SS
SS
124
15. 1
4
1 11 and 1
4 4
11 0
4
4 10
4
50
4
50
4
50 . . : 5, 4
4
x
x x
x
x
x
x
x
x
x
xC N
x
125
50 . . : 5, 4
4
xC N
x
0
0
0
1, 5 4,SS
126
1
2
15. 1
4
1 11 and 1
4 4
, 5 4,
, 4 3,
, 5 4, , 4 3,
x
x x
SS
SS
SS
│ │ │
-5 -4 -3
│ │ │
-5 -4 -3
, 5 3,SS
127
Sample Problems related to
Inequalities
128
Example 1.6.27
A van can be rented from Company A
for Php 1,800 per day with no extra charge
for mileage. A similar van can be rented
from company B for Php 1,000 per day
plus Php 20 per km driven. How many
kilometers must you drive in a day in
order for the rental fee for Company B to
be more than that for Company A?129
Company A: 1,800
Company B: 1,000 + 20 per km
Let be the number of kilometers
to drive in a day
1800 is the amount of money to
pay Company A after driving
km.
1000 20 is the amount of money to
pay Company B a
x
x
x
fter driving
km.x 130
1000 20 1800
20 800
800
20
40
Conclusion: You should drive at least 40 km
in a day.
x
x
x
x
131