1.6 equations and inequalities

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