CHAPTER 1

Elementary Set theory

A set is defined as a collection of well-defined objects. Each object in the set is called the
element or member of the set. Sets are normally denoted by capital letters like A, B, C ,..., X , Y , Z
while elements of sets are denoted by lower-case letters like a, b,..., y, z. For example "V" is the
set of english vowels. The set V may be written as V = {a, e, o, i, u}, "N" is set of counting
nmbers between 1 and 12. The set N can be written as N = {2, 3, 4, 5, ..., 11}. The elements that
made up a set have to be well-defined and easily identifiable.

1.1 Representing sets and basic concepts

Sets are commonly represented using braces {}. There are other ways of representing sets. Some
of these ways are described as follows.
➢ List or Roster Method. This involves listing all the elements of a given set explicitly. All
the members of the set are laid bare by displaying all its objects. Ex: {x1 , x2 ,..., xn } is the
set consisting of the elements x1 , x2 ,..., xn . A set containing exactly one element or
consisting of a single element is denoted {x} and is known as a singleton set.
➢ Specifier Method. This is a set defined by specification, that is, identifying a given set by
describing property that are possessed by its members. Let X be a given set, if P(x) is a
property which every element x X possesses, then one writes {x  X : P( x)} to
describe the set that consists of all objects x for which the property P(x) is valid.

Example
List the elements of the following sets:
• A = {x : x is odd and x is greater than 0 and less than 12}.
• B = {x : x is even and x is greater than 19 and less than 31}.
• C = {x : x is an odd integers, x > 0}.

Membership of a set-the symbol 

If an object x is an element of (or belongs to) a set A , one writes x  A . When x is not
a member of A , one can write x  A . When both x and y are elements of A , we write
x, y  A .

Subsets and supersets

If a A then it follows that {a}  A . Let A and B be given sets. Set A is said to be a
subset of set B , written A  B or B  A if every element of A is also an element of B . That
is, if a  A  a  B . The set B is also called a superset of A . If set A is not a subset of B ,
we write A  B. The notation A  B denotes a proper subset.

Equality of sets
Two sets A and B are said to be equal if they consist of same elements. That is, A
and B are equal if every element of A is an element of B and every element of B is an

1
element of B . Symbolically, A = B if and only if for all x, x  A  x  B and write A  B and
B  A . If A is not equal to B , one writes A  B. Example, {1,2,3} and {2,1,3} are equal;
and {1, 1, 2, 2, 3} and {1,2,3} are equal. We note that the repetition of elements of the set is
irrelevant.

Theorem
Let A, B ,C be any sets. Then the following hold
• A A
• If A  B and B  A , then A = B
• If A  B and B  C , then A  C .

Example
State whether each statement is true or false:
i. {1,4,3} = {3,4,1}
ii. {1,3,1,2,3,2}  {1,2,3}
iii. {1,2} = {2,1,1,2,1}
iv. {4} {{4}}
v. {4}  {{4}}
vi.   {{4}}.
Solution
The order of elements of a set is irrelevant, so we see that the two sets are equal. The answer to
(i ) is TRUE. The set {1,3,1,2,3,2} = {1,2,3} since repetition of elements of a set does not
increase its size, also every set is a subset of itself and therefore (ii ) is TRUE. Number (iii) is
also TRUE since repetition is irrelevant. We see that {4} is an element or a point in the set
{{4}} and so it belongs to the set, the answer is TRUE. The answer to (v ) is FALSE since {4} is
a member of the set {{4}} and cannot be a subset. Number (vi) is TRUE since an empty set is a
subset of every set.

Universal and Empty sets

An empty set or null set is a set that contains no elements or has no elements as its
members. Using a word we are yet to define, one defines an empty set as a set with zero
’cardinality’. It is denoted by  or {} or {x : x  x} . An empty set is not the same as {} or
{0} . On the other hand, a Universal set is the set containing all the objects of particular interest
or the largest set of interest.

Theorem 1.2 For any set A , we have that   A  X .

Practice Exercise 1.1

Which of the following sets are equal?
A = {x : x 2 − 4 x + 3 = 0}, B = {x : x 2 − 3x + 2 = 0}, C = {x : x  N, x < 3},
D = {x : x  N, x is odd x < 5}, E = {1,2}, F = {1,2,1}, G = {3,1}, H = {1,1,3}.

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1.2 Set operations
Intersection of sets
. The intersection of sets A and B is a set which contains all elements that are common
to both sets.
A  B = {x : x  Aand x  B}.

Disjoint Sets
Given that sets A and B have no elements in common, then their intersection will have zero
number of elements and we say that A and B are disjoint and write A  B =  .

Union of sets
The union of sets A and B is another set C which contains all the elements of set A together
with all the elements of set B . That is:
A  B = {x : x  A or x  B or x  A  B} = {x : x  A or x  B}.

Theorem
• For any sets A and B , we have the following
A B  A  A B
A B  B  A B

• Given that A  B then A  B = A and A  B = B .

Example
If the universal set X = {x : 0  x  6} , A = {x : x  3} , B = {x : 1  x  5} , and C = {x : x > 6} .
Find (i) A  B (ii) A C (iii) A  B (iv) ( A  B) and (v) Ac  C c .
c

Solution
A = [0,3], B = [1,5], C = .
Thus
(i) A  B = [0,5] = {x : 0  x  5} ,
(ii) A  C = A = {x : x  3} ,
(iii) A  B = [1,3] = {x : 1  x  3}
(iv) ( A  B) c = [0, 1)  (3, 6] = {x : 0  x < 1}  {x : 3 < x  6} ,
(v) Ac  C c = (3, 6]  X = X = {x : 0  x  6}.

Complement of a set
The complement of a set A denoted by A or Ac is a set that contains all the elements of the
universal set that are not in A : Ac = {x : x  X , x  A}.

Relative Complement
Relative complement or set difference A \ B of two sets A and B is the set of those elements
of A which do not belong to B and is given by given by
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A \ B = A − B = A  B c = {x : x  A, x  B}. There is no assumption that B  A .

Symmetric difference
Is defined as A  B = ( A  B) \ ( A  B) = ( A \ B)  ( B \ A).

Example
Given that A = {−5,−3,−1,0,1,2,3}, B = {−4,−3,0,3,5,8}. Find A  B.

Solution
We will use both identities to give the answer. Recall that A \ B are those elements in set
A that are not in set B . So, A \ B = {−5,−1,1,2} , B \ A = {−4,5,8} and
A  B = ( A \ B)  ( B \ A) = {−5,−4,−1,1,2,5,8}.
Using the second identity: A  B = {−5,−4,−3,−1,0,1,2,3,5,8} ,
A  B = {−3,0,3} and AB = ( A  B) \ ( A  B) = {−5,−4,−1,1,2,5,8} which gives the same
answer.

Practice Exercise 1.2

Given that A = {5,6,7,9}, B = {0,2,4,6,8} ,
X = {0,1,2,3,4,5,6,7,8,9}. List the elements of each of the following sets
(a) A c (b) B c (c) A c  B c (d ) Ac  B c (e) A  B ( f ) ( A  B) c ( g ) ( A  B) c (h) ( A c  B) c
(i) ( B c  A) c
What do you notice about your answers to (i) (c ) and ( g ) ; (ii). (d ), ( f ) ?

1.3 Laws of the Algebra of sets

Theorem
Let X be a given set and A, B,C  X then we have the following identities:
1. Idempotent laws: A  A = A, A  A = A.
c c
2. Involution law: ( A ) = A.
3. Commutative laws: A  B = B  A, A  B = B  A.
4. Associative laws: ( A  B)  C = A  ( B  C ), ( A  B)  C = A  ( B  C ) .
5. Distributive law
A  ( B  C ) = ( A  B)  ( A  C ), A  ( B  C ) = ( A  B)  ( A  C ) .
6. Identity laws: A   = A, A  X = A, A  X = X , A   = .
7. Complement laws: A  Ac = X , A  Ac = , X c = ,  c = X .
8. DeMorgan’s laws: ( A  B)c = Ac  B c , ( A  B)c = Ac  B c .

Proof:
We will give proofs to some of the Laws.
•Involution law: ( Ac )c = A. Let
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 x  ( Ac ) c  ( x  X ) and x  Ac  ( x  X ) and x  A  x  X  A  x  A since A  X .
a n d ( x  X • DeMorgan’s
 Bc ) 
laws: ( A (B)x
c
=
Ac  BA
c c
. Wea nwith
start dxthe
c
B implication.
forward , s i n Let
ce A .
c
, Bc
Conversely, let, since ( A  B)  X .
c

To prove the second part of the DeMorgan’s law, we make use of the first part and the
( Ac  B c )c = ( Ac )c  ( B c )c = A  B by the Involution law. Then take
Involution law.
complement of both sides: ( A  B) = (( A  B ) ) = ( A  B ) by the Involution law. Hence
c c c c c c c c

Example
Prove that A \ ( A \ B) = A  B for any sets A and B .

Solution
By definition we have that
A \ ( A \ B) = A  ( A  B c ) c = A  ( Ac  B) = ( A  Ac )  A  B =   A  B = A  B.

Example
Prove that A  B if and only if A \ B =  .
Solution
First assume that A  B then A \ B  B \ B = B  B c = . Conversely, assume that
A \ B = A  B c =  , then it follows that A  B   . If A  B   , it follows that one of the
following is true: A  B , B  A or A = B . Thus A  B .

Practice Exercise 1.3

Let X be given and A, B,C  X . Show the following identities:
(a). A \ B = B c \ Ac (b). A \ ( B  C ) = ( A \ B)  ( A \ C ) (c). ( A \ B) \ C = A \ ( B  C ).

1.4 Finite and infinite sets

A set is said to be finite if it has a finite number of objects as its members or elements. That is, a
set is said to be finite if it contains as its members exactly n elements. If a set has an infinite
number of elements then it is said to be infinite, eg: (0,1) .

Example
State whether each statement is true or false:
• Every subset of a finite set is finite.
• Every subset of an infinite set is infinite.

Solution
Recall that a subset of a given set is usually less than or equal to its superset. Since the superset
is finite, automatically the subset will be a finite set. Therefore, the answer to (i ) is TRUE. The
second one is FALSE because we can always find a finite subset of every infinite set.

Cardinality of a set
The cardinality of a set A is the number of elements contained in the given set and is denoted by
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| A | or card S or n( A) or # ( A) . It is a positive integer number.
Lemma
Let A and B be finite sets. Given that A and B are disjoint sets. Then we have that
• A  B is finite and n( A  B) = n( A) + n( B).
• n( A \ B) = n( A) − n( A  B) , this gives us the number of elements in set A only.
• n( A ) = n( X ) − n( A) where X is the universal set and it gives the number of
c

elements not in A .

Theorem (Inclusion-Exclusion Principle).

Suppose that A and B are given finite sets. Then A  B and A  B are finite and
• n( A  B) = n( A) + n( B) − n( A  B)

• n( A  B  C ) = n( A) + n( B) + n(C ) − n( A  B) − n( A  C ) − n( B  C ) + n( A  B  C )
where C is a finite set.

Example
Suppose that in a MTH 101 class, the number of students in group B is 120 students and 58 of
them are in Medical Biochemistry. How many students in group B are not in Medical
Biochemistry, using the cardinality of sets?

Solution
Let n( X ) = 120 , the total number of students in MTH 101 group B class and n( A) = 58 , the
number of students in Medical Biochemistry programme. Then applying the above Lemma 4.3
number (3) we have that n ( A ) = n ( X ) − n ( A) = 120 − 58 = 62 . Therefore, 62 are in other
c

programmes other than Medical Biochemistry.

Example
Suppose a list A contains 30 students in a MTH 101 class, and a list B contains 40 students
in Phy 101 class, and suppose that there are 25 names on both lists. Find the number of students
(a). Only on list A , (b). Only on list B (c). On list A or B (or both), (d). On exactly one list.

Solution
Given that n( A) = 30,n( B) = 40,n( A  B) = 25 .
(a)The number on list A only is given by n( A \ B) = n( A) − n( A  B) = 30 − 25 = 5 and
(b) Those on list B only is n( B \ A) = n( B) − n( A  B) = 40 − 25 = 15.
(c) The number on list A or B (or both) is
n( A  B) = n( A) + n( B) − n( A  B) = 30 + 40 − 25 = 45.
(d) The number on exactly one list is the sum of those on list A only and those on list B only
and it is equal to 20 .

Direct product
The direct product of two finite sets A and B is given by
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 A  B = {( a, b) : x  A, b  B}.
Generally, the direct product of n finite sets is defined as follows:
A1  A2  ...  An = {(a1 , a2 , ..., an ) | a1  S1 , a2  S 2 , ..., an  S n }.

Theorem
(Multiplication rule) Let A1 , A2 ,..., An be finite sets. Then
| A1  A2  ...  An |=| A1 |  | A2 | ... | An | .

Example
Given that A = {1,2,3,4} and B = {5,6} . Find the direct product A B of sets A and B and
justify the multiplication rule.

Solution
Applying the definition, we have that
A B = {(1,5),(1,6),(2,5),(2,6),(3,5),(3,6),(4,5),(4,6)}.
Next, n( A) = 4, n( B) = 2 and n( A  B) = 8 .
Thus 8 =| A  B |=| A |  | B |= 4  2 = 8 and the multiplication rule holds.

Power set
A set that contains all possible subsets of a given set is known as the Power set. It is defined by
P( X ) = { A : A  X } = 2 X . The cardinality of a Power set is given by | 2 X |= 2| X | which is finite if
| X | is finite.

Example
Find the power set P ( A) of A = {1,2,3,4} and the power set P (B ) of B = {1,{2,3},4}.
Solution
First calculate the cardinality of the sets to know the number of elements that will be
contained in the power sets. The cardinalities are, n( A) = 4, n( B) = 3 , so the power sets will have
2 4 = 16 elements for A and 2 3 = 8 elements for B . Therefore
P( A) = { A, {1},{2},{3},{4},{1,2},{1,3},{1,4},
 {2,3},{2,4},{3,4},{1,2,3},{1,2,4},{1,3,4},{2,3,4}, }.
P( B) = {B, {1},{{2,3}},{4},{1,{2,3}},{1,4},{{2,3},4}, }.
Now do this. Exercise. Given that A = {1,2} , find the power set P ( A) of A .

Practice Exercise 1.4

1. What is {} ?. How many elements does it have?

2. Which of the following formulas/inclusions are correct?
(a ).   {, {}}, (b).{}  {{}}, (c).   {{}} .
3. Is {{}} a singleton? Justify your ’Yes’ or ’No’ answer!
4. How many elements do the following sets contain?
( a ) {1,2,1} (b) {1,2,{1,2}}, (c ) {{2}} ( d ) {{1},1} (e) {1, } ( f ) {{}, } ( g ) {{}, {}}
( h). {x, 5 x − 10}, for x  R.
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1.5 Venn diagram
A pictorial representation of sets. Below are diagram illustration of Union, Intersection,
complement and disjoint sets.

A A B
B

A B A B
C
A

A\B A Only

A B

A\(B C)
A B
A B

C
A C
C
B

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 (A B) (B C)
C
A B\ C =A B C
A B A B

C C

A B=
A B C
B
A A B

C
Disjoint set

Example
In an election involving three parties for the chairmanship and gubernatorial elections of Lagos
State, voters cast their votes as follows: 190voted for party A , 200 for party B and 250 for
party C . 80 voted for A and B , 60 voted for A and C , 100 voted for B and C and 40
voted for B alone. If 500people voted during the election, find:
a. The number of voters who voted for all the three parties.
b. The number of voters who voted for A and B but not C .
c. The number of voters who did not vote for any party.

Solution
Representing the question using Venn diagram, we have

90
n(

=1
B)

)
=2

n (A 80-x 40
00

x
60 - x 100 - x

9
n(C)=250
 Recall that the number that voted for B alone is
40 , therefore,
40 = 200 − (80 − x) − x − (100 − x)
= 200 − 80 − 100 + x − x + x
40 = 20 + x
x = 20
Substituting the value of x = 20 in the diagram
above
90

n(
=1

B)
A)

=2
n( 60 40

0
70

0
20
40 80

110

80 n(C)=250

C only = 250 − 80 − 40 = 250 − 140 = 110

A only = 190 − 60 − 20 − 40 = 190 − 120 = 70

a. The number of voters that voted for all the three parties is 20
b. The number of voters who voted for A and B but not C is
n(( A  B ) \ C ) = n( A  B ) − n( A  B  C )
= 60 − 20 = 40
c. The number of voters who did not vote for any party
= 500 − (70 + 60 + 20 + 40 + 110 + 80 + 40)
= 500 − 420
= 80

Practice Exercise 1.5

1. Each first year student in in Faculty of Science, FUNAI has a Maths requirement A and
a Science requirement B . A pool of 100 students show that 50 students have completed
A , 35 students completed B , and 25 students completed both requirements A and
B . Use a Venn diagram to find the number of students who have completed: (a). At
least one of A and B (b). Exactly one of A or B (c) neither A nor B .
2. In a survey about Newspaper reading culture of 120 FUNAI staff, it was found that 65
read Sun News, 45 read Vanguard, 42 read Daily Times, 20 read both the Sun and
Vanguard, 25 read both the Sun and Daily Times, and 15 read both Vanguard and Daily
Times while 8 read all the three Newspapers. (a). Find the number that read at least one
of the three Newspapers. (b). Find the number that read exactly one Paper. Use Venn
diagram to illustrate your answer.
3. Given that A and B are non-disjoint sets, show on a Venn diagram the following sets.
(a). Ac (b). B c (c). A  B c (d ). Ac  B c (e). Ac  B c .

1.6 Set of Numbers

A special example of set is the set of numbers used often in mathematics. These include
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the set of natural or counting numbers, usually denoted by ℕ; set of integers, usually denoted by
ℤ; set of rational numbers, usually denoted by Q; set of irrational numbers, usually denoted by
IQ; set of real numbers, usually denoted by R and set of complex numbers, usually denoted by
ℂ.

The Natural Number

The set of natural numbers, ℕ is the set of positive counting numbers defined as the
smallest subset of ℝ containing 1 and containing n + 1 whenever it contains n. It is also known as
the set of positive integers, not including zero.

Peano’s Postulates
The set of Natural numbers ℕ has the following properties known as Peano’s postulates.
1. ℕ is non-empty.
2. Associated with each natural number n, there is a unique natural number say n
called the successor of n .
3. There is a natural number say n that is not a successor of any natural number.
4. Distinct natural numbers have distinct successor, that is, if n  m then n  m .
5. The only subset of ℕ that contains n and the successors of all its elements is ℕ
itself.
Remark
We note that n = 1 and n = n +1 . The above axioms are known as Peano’s Postulates.

The Integers
The set of integer numbers denoted by ℤ is the set containing zero, positive and negative
numbers. It is given by Z = {n : n = 0 or n  N or − n  N} = {..., −1,0,1,...} .

The rational numbers

p
The set of rational numbers is defined as follows ℚ = {x : x = ; p,q  0  Z}.
q
The irrational numbers
The set of irrational numbers IQ , is the set of numbers that cannot be expressed as rational
numbers. Examples are  , e, 2 , etc.

Repeating decimals
A repeating decimal is a decimal that has a digit, or a block of digits that keep repeating over and
over and over again without ever ending. Examples of repeating decimals, their equivalent
fractions and the digit or block of digits repeating:, etc.

Example
What fraction is equal to 0.5555...?

Solution
We solve using the following steps:
• Let x = 0.5555...
• After quick examination, one observes that the repeating digit is 5 .

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 • To place the repeating digit (5) to the right of the decimal point, one needs to move
the decimal one place to the right, which is multiplying by 10 : 10x = 5.555...
• Replace the repeating digit (5) to the right of the decimal point: x = 0.5555...
• Then solve the two equations for x :
10x = 5.555
x = 0.5555
and x = 5 .
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Example
Find the equivalent fraction of 1.04242424242....
Solution
We will follow the same steps as above. Let x = 1.04242424242 . Observe that the repeating
block of digits is 42 , so we multiply by 1000 to place the decimal point to the left of the
repeating digits. Thus, 1000x = 1042.42424242 . Next we place the repeating digit(s) to the right
of the decimal point by multiplying by 10 : 10x = 10.4242424242 . Now solving the two
equations in x , we have that x = 1032 .
990

Existence of the real numbers

There exists a complete ordered field, called the set of real numbers and denoted by R .

Field Properties
The real number system is a set {a, b, c,...} on which the operations of addition and
multiplication are defined so that every pair of real numbers has a unique sum and product, both
real numbers, with the following properties:
1. a + b = b + a and ab = ba (commutative laws).
2. (a + b) + c = a + (b + c) and (ab)c = a(bc) (associative laws)
3. a (b + c) = ab + ac (distributive law)
4. there exist distinct real numbers 0 and 1 such that a + 0 = a and a1 = a  a
5. for each a , there exists a number − a such that a + (−a) = 0 , and if a  0 , there is
a real number 1 such that a( 1 ) = 1.
a a

The order relation

The real number system is ordered by the relation < , which has the following
properties:
1. For each pair of real numbers a and b , exactly one of the following holds:
a = b, a < b, or b < a.
2. If a < b and b < c , then a < c (transitivity)
3. If a < b , then a + c < b + c for any c , and if c > 0 , then ac < bc .
A field with an order relation satisfying (1) − (3) is an ordered field. Therefore a real number
forms an ordered field.

Different types of intervals on ℝ

Let a,b  R and a < b then we have the following

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 1. [a, b] = {x : a  x  b} , bounded closed interval.
2. ( a, b] = {x : a < x  b} , bounded half open and half closed interval.
3. [a, b) = {x : a  x < b} , bounded half closed and half open interval.
4. (a, b) = {x : a < x < b} , bounded open interval.
5. (a, ) = {x : a < x < } , unbounded open interval.
6. [a, ) = {x : a  x < } , unbounded half closed interval.
7. (−, b) = {x : − < x < b} , unbounded open interval.
8. (−, b] = {x : − < x  b} , unbounded half closed interval.
The symbol  is called infinity.

13
 CHAPTER 2
The Number System

2.1 Indices
Given a real number x and a positive integer n . Then, by x n , "read as x raise to power n ", we
mean x  x  ...  x .
n −times

Here, x is called the base and n is called the index or the exponent or the power. For example:

53 = 5  5  5 = 125
7 5 = 7  7  7  7  7 = 16807
27 = 2  2  2  2  2  2  2 = 128
x1 = x
( x + y ) 2 = ( x + y )  ( x + y ) = x 2 + 2 xy + y 2

Basic rules of indices.

There are a few basic laws guiding operations on exponents. These are summarized below.
Let a, b, m, n be real numbers with a, b > 0 . Then:
(a) a m  a n = a m + n .
Example

26  24 = 26+4 = 210
(0.2)5  0.27 = 0.25+7 = 0.212
3−0.6  3−0.9 = 3−0.6−0.9 = 3−1.5

(b) a n  b n = (ab) n
Example

34  7 4 = (3 7)4 = 214

(1.6)3  53 = (1.6  5)3 = 83

an
(c) m
= a n−m
a
Example

2 8  2 2 = 2 8−2 = 2 6 (not 2 4 )
45
−2
= 45− −2 = 47
4
2
x2 2 −3 −1
= x = x (not x 3 )
x3

14
 an a
(d) n
= ( )n
b b
Example

311 3
11
= ( )11
4 4
1 1 2
=( )
52 5

(e) (a n ) m = a nm

Example
1
(34 ) 2 = 32
(23 ) 7 = 221
Special cases of these laws are:
an
(a) a 0 = a n − n = n = 1
a
a0 1
(b) a − n = a 0− n = n = n
a a
1 1
(c) Fractional indices: a n = n a (In particular, a 2 = a )
m 1
(d) Mixed fractional indices: a = (a ) = ( n a ) m .
n n m

Example
1
42 = 4 = 2
2
27 3 = (3 27 ) 2 = 32 = 9
−1
2
1 1 1
16 = 1
= =
2 16 4
16
Algebra involves combinations of one or more of these laws together with other known basic
properties of real numbers.

Example
Evaluate:
x
(a) 2a 3  3a 4 (b) x 4 x −6 (c ) ( ) − 3 (d ) (4 x 5 )(3x −9 )
y
(3x 5 y −6 z 5 )(4x 3 y 2 z ) 4
x5 y
(e) (f)
2 x 4 y 4 z −4 4
xy −3

Solution:
15
 • 2a 3  3a 4 = 2  3  a 3  a 4 = 6a 3+ 4 = 6a 7
1
• x 4  x −6 = x 4−6 = x − 2 = 2
x
x −3 1 y 3 y3
•( ) = =( ) = 3
y x x x
( )3
y
12
• 4 x 5  3x −9 = 4  3  x 5  x −9 = 12x 5−9 = 12x − 4 =
x4
(3x 5 y −6 z 5 )(4x 3 y 2 z ) 12x 5+3 y −6+ 2 z 5+1 8− 4 − 4 − 4 6 + 4 4 −8 10 6 x 4 z10
• = = 6x y z = 6x y z =
2 x 4 y 4 z −4 2 x 4 y 4 z −4 y8
4
x5 y x5 y 4 4 4
• =4 = x y = xy
4
xy−3 xy−3

Remark: Some Common Errors in Algebra

The student is asked to be careful of these most frequent errors in algebraic
manipulations.
0
(A) Division by Zero. Division by Zero is forbidden. Everyone knows that = 0 . What
7
7
about ? You simply cannot divide by zero so don’t ever do it. Remember that division by zero
0
is undefined.

Example
Spot the mistake in the work below.

• Suppose that : b = a.
• Multiply both sides by a : ab = a 2 .
• Subtract b 2 from both sides: ab − b 2 = a 2 − b 2 .
• Factor both sides: (a − b)b = (a − b)(a + b) .
• Divide both sides a − b : b = a + b .
• Since b = a , then a = 2a
• Divide both sides by a : 1 = 2
Answer It is obvious that 1  2 . A closer look at the steps show that since a = b , then
a − b = 0 . Therefore, it is very wrong to divide both sides by a − b .
(B) Bad Use of Parenthesis Bad use of parenthesis or ommission of parenthesis may lead to
complications in algebra. Examples of correct and incorrect uses of parenthesis are given below:
Example
Correct: (3x) 2 = 32 x 2 = 9 x 2 . Incorrect: 3x 2
Correct: (−4)2 = (−4)(−4) = 16 . Incorrect: − 4 2 = −16

Example
Subtract 4x − 7 from x 2 + 3 x − 7 .
Correct: x 2 + 3x − 7 − (4x − 7) = x 2 + 3x − 7 − 4 x + 7 = x 2 − x .
16
 Incorrect: x 2 + 3x − 7 − 4 x − 7 = x 2 − x − 14 .
Example
Convert 5 x to fractional indices.
1
2
Correct: 5 x = (5x)
1
Incorrect: 5 x = 5 x 2
(C) Improper Distribution. Be careful when using the distribution property! These two
errors are most frequent.
Example
Simplify 4(2x 2 − 3) .
Correct: 4(2x 2 − 3) = 8 x 2 − 12 .
Incorrect: 4(2x 2 − 3) = 8 x 2 − 3 . Make sure that you distribute 4 to all the terms in the
parenthesis. Often people just multiply the first term by 4 and ignore the other terms.
Example
Simplify 2(3x − 5)2 .
Correct: 2(3x − 5)2 = 2(3x − 5)(3x − 5) = 2(x 2 − 30x + 25) = 2 x 2 − 60x + 50 .
Incorrect: 2(3x − 5)2 = (6x − 10)2 = 36x 2 − 120x + 100 .
The issue here is that the exponentiation should have been performed before distribution.
(D) Cancellation Errors
Example
x − 12
Simplify
x
x − 12 x 12 12
Correct: = − = 1−
x x x x
x − 12 x
Incorrect: = − 12 = 1 − 12 = −11
x x
(E) False Assumptions These assumption are not true.
( x + y) 2  x 2 + y 2 .
x+ y  x + y
1 1 1
 +
x+ y x y

Practice Exercise 2.1

Simplify:
1 1
x 22 1 −
(a) (b) (x 3 ) 9 (c) ( x −2 ) 4 , (d) (4x 3 )(7x 4 )
x11 16
x2
(e) ( − 4 ) −3 (f) (4x 2 y 3 z −10 )(2x −5 y −1 z12 )
y

2.2 Exponential and Logarithmic Functions

17
 We have studied the basic laws governing the operation of indices.
An exponential function with base a is a function f : R → R + such that
f ( x) = a x , a > 0 x  R.

Example
Convert the following to the indicated base

(a) (27)− x to base 3.

1
(b) ( ) 4 x to base 5
25
Solution (a) 27 = 33 . Therefore (27)− x = (33 ) − x = 3−3 x
1 1
(b) ( ) = 5 − 2. Therefore ( ) 4 x = (5− 2 ) 4 x = 5−8 x
25 25
The exponential function has the following properties.
(a) f ( x + y ) = a x+ y = a x .a y = f ( x). f ( y )
(b) ( f ( x)) y = (a x ) y = a xy = f ( xy)
ax f ( x)
(c) f ( x − y ) = a x − y = y
=
a f ( y)
1 1
From (a ) , f (0) = a 0 = 1 , and f (− x) = a − x =x
= . Any exponential function can
a f ( x)
be written equivalently in the form f ( x) = e kx where k R and the irrational number
e = 2.7182... is the Euler number. If 0 < a < 1, then k < 0 . On the other hand, for a > 1 , k > 0 .
If a > 0, a  1 , then the exponential function f a : R → R + given by f a ( x) = a x e has an
inverse function. The inverse function f a−1 : R + → R is called the logarithmic function with base
a and denoted by loga .
Observe that
y = a x if and only if x = log a y.
log a y
Thus, for any y > 0 , y = a . In particular, a x = e xlna where lna = loge a .

Logarithmic Functions

log a y is that index (power) to which a must be raised to obtain y.

1
1
For example log3 9 = 2 because 9 = 32 . But log3 3 = because 3 = 3 2 . In general, loga 1 = 0
2
because a 0 = 1 and loga a = 1 because a1 = a .
The basic properties of the logarithm function are obtained by reversing the laws of
indices.
(a) a m  a n = a m + n , therefore loga x + loga y = loga xy
logb x
(b) a n  b n = (ab) n therefore loga x =
logb a
18
 an x
(c) m
= a n − m , therefore loga x − loga y = loga
a y
(d)) (a n ) m = a nm therefore mloga x = loga x m

Example
Solve:
7 2 x − 7 x +1 + 10 = 0 .

Solution
The expression 7 2 x − 7 x +1 + 10 = (7 x ) 2 − 7(7x ) + 10 . Thus the equation is of the form
y 2 − 7 y + 10 = 0 where y = 7 x . The solution this quadratic equation is y = 2 or y = 5 . So
ln2 ln5
7 x = 2 or 7 x = 5 . Taking log7 on both sides gives x = log7 2 = or log7 5 =
ln7 ln7
Example
Evaluate log 10 0.0001
Solution
1 1 −8
0.0001= = 4 = 10− 4 = 10 .
10000 10
−8
Therefore log 10
0.0001= log 10
10 = −8

Example
1
Write ( ) x in the form e kx for an appropriate k .
27
Solution
1 1
( ) x = ( 3 ) x = 3−3 x = (eln3 ) ( −3 x ) = e ( −3ln3) x . Thus k = −3ln3.
27 3
Logarithmic and exponential functions have many important applications in sciences and
engineering. They are often used interchangeably, as shown below.

Example
A bacteria colony starts with 40 bacteria and doubles every minute. Determine the estimated
number of bacteria in the colony after 15 minutes.
Solution: The number of bacteria can be shown in a table below.

Time (mins) 0 1 2 ... t

40 80 160 ... 40 2 t

Number of
Bacteria

The number of bacteria at any given time is given by N (t ) = 40  2t . Since 2 = e ln 2 , we

write N (t ) = 40etln2  40e0.693t . Thus N (15)  40e 0.69315 = 40e10.395 .

19
Practice Exercise 2.2
(1) Convert the following to the indicated base
1
(a) ( ) − x to base 4
4
4 25
(b) 2( ) 4 x to base
5 16
1
(c) (64)x to base 2.
3
(2) Solve:
(a) log4 x = 2
(b) log5 x = 3
(c) 2(32 x ) − 7(3x ) − 15 = 0 .
(3) If log7 2, log7 (2 x − 1) and log7 (2x + 3) are in A.P., find x .
(4) Evaluate each of the following
(a) log10 0.0001
(b) log 5 125
1
(c) log
2 
1
2
(d) loge e .
(5) Write the following exponential functions in the form e kx for an appropriate k .
(a) 2 x
x
1
(b)
3
x
1
(c)
e
(d) 10 x
(e) 0.1x

2.3 Surds
We have noted that if n is not a perfect square, then n is not a rational number. In that
case, n is a quadratic surd (or simply a surd). The general form of a quadratic surd is a + b c
where a, b, c are rational numbers. Addition or subtraction of surds are done component wise
Given a surd a + b c , the conjugate is simply a − b c . This has the intersting property
that (a + b c )(a − b c ) = a 2 − b2c.
Example
Evaluate (a) (3 − 2 3) + (2 + 3) (b) 4 − (3 − 4 11) .

(a) (3 − 2 3) + (2 + 3) = (3 + 2) + (−2 + 1) 3 = 5 − 3
(b) 4 − (3 − 4 11) = (4 + 0 11) − (3 − 4 11) = (4 − 3) + (0 − −4) 11 = 1 + 4 11 .

20
Example
Evaluate: (3 + 5 2 )(3 − 5 2 ).
Solution: (3 + 5 2 )(3 − 5 2 ) = 32 − 3 5 2 + 3 5 2 − 52  2 = 9 − 50 = −41.
1
To convert an expression of the form to a surd, we multiply both the numerator and the
a+b c
denominator by the conjugate a − b c .
1 1 a −b c a −b c a b
So =  = 2 = 2 − 2 c
a +b c a +b c a −b c a −b c a −b c a −b c
2 2 2

Example
Write the following in the general form of a quadratic surd:
5 6 1 1
(a) (b) (c)
3 3− 5 3 21  7
Solution
5 6 5 6 3 5 12 10 3
(a) =  = = 0+
3 3 3 3 3
1 1 3+ 5 3+ 5 3 1
(b) =  = = + 5
3− 5 3− 5 3+ 5 9−5 4 4
1 1 1 7 1
(c) = = = = 0+ 7
3 21  7 3 147 21 7 147 147

Practice Exercise 2.3

2
Simplify: (i) 8 3 (ii) 6  12 (iii) 2 5  3 5 . (iv) (v) 5 45
2
5 3 2
(vi) (vii)
12 2 − 18

2.4 Polynomials
A polynomial function f is any function that has expression of the form
f ( x) = a0 + a1x + a2 x2 + ... + an xn ;an  0
where a0 , a1 , a2 ,..., an are real numbers and n a non nonnegative integer.

Example
The function f : R → R given by f ( x) = 27 + 3x − 13x 4 + 4 x 5
is a polynomial function because 27 + 3x − 13x 4 + 4 x 5 = 27 + 3 x + 0 x 2 + 0 x 3 − 13x 4 + 4 x 5 .
1
These functions: g ( x) = x + 4x + 1 and h( x) = + x 2 + 3 x 7 are not polynomials because of
x
1
1
x = x 2 and = x −1
x
The domain of any polynomial is the set of real numbers. The degree of a polynomial is the
21
highest index n such that an  0 . For example, the polynomial f ( x) = 27 + 3x − 13x 4 + 4 x 5 is of
degree 5 . A polynomial of degree 1 is called linear, degree 2 quadratic, degree 3 cubic and
degree 4 quartic.
Polynomials can be combined to get new ones through addition, subtraction, multiplication but
not division.

Addition/Substraction of Polynomials
Addition and subtraction are done term wise.
Example 2.31
Given p( x) = 2 x 2 − 3x + 4 and q( x) = 7 x 4 − 4 x 2 + 2 x + 8 , then
• p( x) + q( x) = (2x 2 − 3x + 4) + (7 x 4 − 4 x 2 + 2 x + 8)
= (0x 4 + 0 x 3 + 2 x 2 − 3x + 4) + (7 x 4 + 0 x 3 − 4 x 2 + 2 x + 8)
= (0 + 7)x 4 + (0 + 0)x 3 + (2 − 4) x 2 + (−3 + 2)x + 4 + 8
= 7 x 4 − 2 x 2 − x + 12

• p( x) − q( x) = (2x 2 − 3x + 4) − (7 x 4 − 4 x 2 + 2 x + 8)
= (0x 4 + 0 x 3 + 2 x 2 − 3x + 4) − (7 x 4 + 0 x 3 − 4 x 2 + 2 x + 8)
= (0 − 7) x 4 + (0 − 0)x 3 + (2 + 4) x 2 + (−3 − 2)x + 4 − 8
= −7 x 4 + 6 x 2 − 5 x − 4

Multiplication of Polynomials
If p is a polynomial of degree n and q of degree m , then pq is another polynomial of degree
m + n .Multiplication of two polynomials may be achieved by setting up a multiplication table
and using the relevant law of indices.
Example
If f ( x) = 3x 3 − 2 x + 4 and g ( x) = 2 x 4 + 3x 2 + 7 x − 5 , then we set up the multiplication table as
follows:

 2x 4 + 3x 2 + 7x −5
3x 3
6x 7 9x 5 21x 4
− 15x 3
− 2x − 4x 5 − 6x 3 − 14x 2 + 10x
+4 8x 4 12x 2 28x − 20

Gathering the like terms:

6 x 7 + 9 x 5 − 4 x 5 + 21x 4 + 8 x 4 − 15x 3 − 6 x 3 − 14x62 + 12x 2 + 28x + 10x − 20
= 6 x 7 + 5 x 5 + 29x 4 − 21x 3 − 2 x 2 + 38x − 20

Factorisation of Polynomials
Factorisation is the opposite of multiplication of polynomials. In factorisation, we seek for
polynomials of smaller degrees whose products give the original polynomial.
Example
Fill in the empty bracket:

22
 x 2 − 9 x + 20 = ( x − 4)( )
Solution: To obtain x 2 , x must be multiplied by another x . So we must have
x 2 − 9 x + 20 = ( x − 4)(x... )
To get + 20 , we must multiply − 4 by − 5 . So
x 2 − 9 x + 20 = ( x − 4)(x − 5)

Example
Fill in the empty bracket:
x 3 + 3x 2 − 12x + 4 = ( x − 2)(... )
Solution:
To get x 3 , x must be multiplied by x 2 . So we must have
x 3 + 3x 2 − 12x + 4 = ( x − 2)(x 2 ... ) = ( x 3 − 2 x 2 ) + (...)
To find (...), we subtract the term on the right hand side from the left:
( x 3 + 3x 2 − 12x + 4) − ( x 3 − 2 x 2 ) = 5 x 2 − 12x + 4 .Repeat this process: 5 x 2 − 12x + 4 = ( x − 2)(?) To
get 5x 2 we must multiply x by 5x and to get + 4 we must multiply − 2 by − 2 . Thus
5 x 2 − 12x + 4 = ( x − 2)(5x − 2) and then x 3 + 3x 2 − 12x + 4 = (x - 2)(x2 + 5x - 2) .

Long Division
The method used in the example above is the technique of long division. This principle is
illustrated below:

Example
Find the quotient and the remainder when p(x)=x 3 +2x 2 +4 is divided by d(x) = x+4.

Solution:
The recurring step is:
1), use x to divide the highest power of the polynomial. i.e. x 3  x = x 2
2) Use x 2 to multiply (x+4) to obtain x 3 + 4x 2
3) Subtract x 3 + 4x 2 from the polynomial x 3 + 2 x 2 + 4 to get − 2 x 2 + 4 .
Next repeat the steps with − 2 x 2 + 4

x 2 − 2 x − 8 = q ( x)
(x+4) x3 + 2x 2 + 0x + 4
x3 + 4x 2
0 − 2x 2 + 0x + 4
0 − 2x 2 − 8x
0 − 8x − 4
0 − 8 x − 32
−28 = r

Example
Divide the polynomial expression x 4 − 3 x 2 + 1 by x-2.

23
Solution:
First write x 4 − 3 x 2 + 1 as x 4 + 0 x 3 − 3x 2 + 0 x + 1 .Then the division is done below:

x 3 + 2 x 2 + x + 2 = q( x)
(x-2) x 4 + 0 x 3 − 3x 2 + 0 x + 1
x 4 − 2x3
0 + 2 x 3 − 3x 2 + 0 x + 1
0 + 2x3 − 4x 2
0 + x 2 + 0x + 1
0 + x 2 − 2x
2x +1
2x − 4
5=r

Theorem
Given a polynomial p of degree n, and a real number a, there exists a unique polynomial q and a
unique real number r such that p(x)=(x-a)q(x)+r.

The polynomial q is called the quotient of division of p(x) by x-a and r the remainder. If r=0, we
say that x-a is a factor of p. A quick check shows that r=p(a). The quotient q is obtained through
long division.

Theorem
(The Remainder Theorem)
If p(x) is divided by x-a, the remainder is p(a).

Theorem (The Factor Theorem)

For a given polynomial p, if p(a)=0, then (x-a) is a factor of p.

Example
Given p( x) = x 3 − 5 x 2 + x + 15 . For what values of x is p(x)=0?

Solution:
If we calculate: 𝑝(0) = (0)3 − 5(0)2 + (0) + 15
𝑝(1) = (1)3 − 5(1)2 + (1) + 15
p(2) = (2)3 − 5(2) 2 + (2) − 12 = −22
𝑝(3) = (3)3 − 5(3)2 + (3) + 15 = 0.
Thus by the factor theorem, x-3 is not a factor of p.

Practice Exercise 2.4

(1) What is the remainder when (a) 11 is divided by 3 (b) 15 is divided by 4 (c) 12 is divided by
5
24
(2) Is (a) 5 a factor of 30? (b) 7 a factor of 29?

(3) If f ( x) = x 3 + 3x 2 − 6 x − 8
(a) Find f(2).
(b) Use the factor theorem to write a factor of f.
(c) Express f(x) as a product of 3 linear factors.
(4) Is x-3 a factor of x 3 + 5 x 2 − 12x − 36 .
(5) What is the remainder when 3x 5 − 5 x 3 + 57 is divided by x+2?
(6) Divide the left hand side by the right hand side:
(a) x 3 − 2 x 2 + 3 (x-1)
(b) x + 4 x − 12
2
(x+2)
(c) x + 3x + 4
3
(x+1)
(d) 2 x 3 − 15x 2 − 2 x + 120 (2x+5)
(e) 6 x 4 + 13x 3 − 36x 2 − 43x + 30 (x-2)
(f) x + 1
4
(x+1)

(7) If x+2 is a factor of x 3 + ax 2 + bx − 6 and if the remainder on division by x-1 is 3, find the
values of a and b.

2.5 Equations
A polynomial equation is any expression of the form p ( x) = q( x) where p, q are
polynomials. This can be simplified as P (x) = 0 where P = p − q .
Given a polynomial equation, we seek to find the roots of the polynomial. The roots or
the zeros of a polynomial p are all the values of x for which p (x ) = 0 . For example, if
p( x) = x 2 − 10x + 24 , then the zeros of p are x = 6 and x = 4 .

Linear Equations
A polynomial equation is called linear if none of the variables that appear in it is raised
to any other power except 1. For example, x + 3 = 9. On the other hand, the equation y = 3x − 5
is also linear, but in two variables x and y .
Any linear equation with one unknown has the general form ax + b = c , a  0 , where x
c−b
is the variable and a, b, c are real parameters. The equation has a solution given by .
a
Example
2
Solve: x + 8 = 16.
9
Solution:
2 9 9 2 9
Subtract 8 from both sides: x = 8. Multiply both sides by :  x =  8 = 36. Thus the
9 2 2 9 2
solution is x = 36 .

Example
Find the root of 5( x − 2) = −4(2x + 7) + x .

25
 Solution:
Open the brackets: 5x −10 = −8x − 28 + x = −7x − 28
Add 7 x to both sides: 12x −10 = −28
Add 10 to both sides: 12x = −18 .
−3
Divide both sides by 12 : x = .
2
Remark: Observe that each operation is carried out on both sides. This is in order to
main the balance of the equation.
Example
2 1 3
Solve for x , given that x + = 2 x − .
3 5 10
Solution:
The LCM of the denominators is 30 . Multiply both sides by 30 .
2 1 3
30( x + ) = 30(2x − ) . So 20x + 6 = 60x − 9 .
3 5 10
Subtract 60x from both sides: − 40x + 6 = −9
Subtract 6 from both sides: − 40x = −15 .
− 15 3
Divide both sides by − 40 : x = = .
− 40 8
Remark: A linear polynomial eqaution may have no solution. In some other cases, it may
have infinitely many solutions.

Example
Solve for x if 5( x − 4) = 5 x + 12.

Solution:
Open the brackets: 5x − 20 = 5x + 12
Subtarct: 5x from both sides:-20=12
This is not true. So the equation has no solution.

Example
Solve for x if 7x +14 − 3x = 4x +14 .
Solution:
Add 3x to both sides: 7x +14 = 7x +14.
Subtract 7 x from both sides: 14 = 14
This is always true no matter the value of x . So this is true for all real numbers x .
Remark: A linear polynomial equation in more than one variable is solved accordingly.

Example
Given p = 2l + 3w, solve w .
Solution:
Subtract 2l from both sides p − 2l = 3w .
p − 2l
Divide both sides by 3 : w = .
3

Practice Exercise 2.4 B

(1) Solve for x : 2x + 3 = − x + 9 .
26
 (2) Solve for s : cx + 3t = 2s − t .
(3) Solve for d : 3d +14 = 7e − 4d .

Quadratic Equations
A quadratic equation is a polynomial equation where the degree of the polynomial is 2 .
The general form of a quadratic equation is ax 2 + bx + c = 0 where a  0 .
Not all quadratic equations can be factored into linear equations in R . Thus, not all quadratic
equations have roots. For example, p( x) = x 2 − 2 x + 5 = 0 cannot be factored in R and therefore
does not have a root in R .
The following rules may help in recognizing the factor of some quadratic expressions.
(a) X 2 + 2 AX + A2 = ( X + A) 2 , Perfect square
(b) X 2 − 2 AX + A2 = ( X − A) 2 Perfect square
(c) X 2 − A2 = ( X − A)( X + A) Difference of two squares
In general, X n − An = ( X − A)( X n −1 + AX n −2 + A2 X n−3 + ... + An−1 ) Given any quadratic
equation ax 2 + bx + c = 0 a  0 , we seek to rewrite it as perfect square or at least as a difference
of two perfect squares. This process is called completing the square. The steps are shown below:
b c
Step1. Set ax 2 + bx + c = a ( x 2 + x + )
a a
b 2 2 b c b b b c
Step 2: Add and subtract: ( ) : x + x + = x 2 + x + ( ) 2 − ( ) 2 +
2a a a a 2a 2a a
b b b
Since x 2 + x + ( ) 2 is of the form (a), with A = , X = x , write
a 2a 2a
b c b b b c b b 2 − 4ac
x 2 + x + = x 2 + x + ( ) 2 − [( ) 2 − ] = ( x + ) 2 − .
a a a 2a 2a a a (2a) 2
If b 2 − 4ac  0 , then this is the difference of two squares and we can factor ax 2 + bx + c
b b 2 − 4ac b b 2 − 4ac
as a( x + − )( x + + )
a 2a a 2a
The number b 2 − 4ac is called the discriminant of the quadratic equation. If b 2 − 4ac < 0 , then
we cannot write the quadratic expression as a difference of two squares and the quadratic
expression ax 2 + bx + c cannot be factored in R (but can be factored in C ).
b
The case b 2 − 4ac = 0 reduces to ax2 + bx + c = a( x + ) 2 . In this case, the two roots of the
2a
−b
quadratic equation are the same and it is given by x = .
2a
On the other hand, if b 2 − 4ac > 0 , then the equation has two distinct roots given by
− b + b 2 − 4ac − b − b 2 − 4ac
x= and x = .
2a 2a
− b + b 2 − 4ac − b − b 2 − 4ac
If we set  = and  = , then
2a 2a
b c
x2 + x + = ( x −  )( x −  ) = x 2 − ( +  ) x +  ) . Comparing the coefficients gives
a a

27
 −b c
= + and = 
a a
.

Example
Solve the equation: 3x 2 − 7 x + 4 = 0 .

Solution:
The discriminant is (−7)2 − 4(3)(4) = 49 − 48 = 1 > 0. Therefore, the equation has two distinct
7+ 1 4 7− 1
roots. The roots are x1 = = and x2 = =1
2(3) 3 2(3)

Example
Find the roots of the polynomial p( x) = x 4 − 13x 2 + 36 .
Solution:
The polynomial is of the form y 2 − 13y + 36 where y = x 2 . For the quadratic equation
y 2 − 13 y + 36 = 0, the discriminant is (−13)2 − 4(1)(36)= 169 − 144 = 25 > 0 . Therefore, the
13 + 25 18 13 − 25 8
equation has two distinct roots given by y1 = = = 9 and y2 = = = 4. The
2 2 2 2
polynomial p therefore has four distinct roots namely  y1 =  9 = 3 and  y2 = 2 .
Example
Solve for t if 16t − 4t − 12 = 0 .

Solution:
The equation is of the form x 2 − x − 12 = 0 where x = 4t . The discriminant is 1 − 4(−12) = 49 > 0
1 + 49 1 − 49
. Solving this yields x1 = = 4 and x2 = = −3 . Thus 4t = 4 or 4t = −3 . But
2 2
4t = −3 is impossible because 4 > 0, t  R. Thus 4t = 4 = 41 and t = 1 .
t

Example
Solve for a if 32 a +1 − 28.3a + 9 = 0 .

Solution:
We can rewrite this equation as 3 x 2 − 28x + 9 = 0 where x = 3a . The discriminant of this
equation is (−28)2 − 4(3)(9) = 784 − 108 = 676 . Therefore the roots are
28 − 676 28 − 26 1 28 + 676 28 + 26 1
x1 = = = and x2 = = = 9. Thus 3a = = 3−1 or
2(3) 6 3 2(3) 6 3
3a = 9 = 32 . This yields a = −1 or a = 2 .

Example
A surveyor errorneously measured the dimensions of a rectangulaer portion of land as Area
= 10m 2 and perimeter 4m . Why is the measurement erroneous?

28
Solution
Let l be the length and b the breath. Then
lb = 10.................. ...(1)
2(l + b) = 4......... .......(2) .
Thus l = 2 − b and lb = (2 − b)b = 10 . Thus b 2 − 2b + 10 = 0 . The discriminant is
(−2)2 − 4(1)(10)= −36 < 0 . Thus the equation has no(real) root and therefore such a plot of land
does not exist in real life.

Example
Find a quadratic equation whose roots are the squares of the roots of the equation
x 2 − ax − b = 0 .

Solution
Let the root of x 2 − ax − b = 0 be  and  . Then the desired quadratic equation is
( x −  2 )( x −  2 ) = 0 . i.e x 2 − ( 2 +  2 ) x +  2  2 = 0.
Recall that  + =a and  = −b. Thus  2  2 = ( ) 2 = (−b) 2 and
 +  = ( +  ) − 2 = (a) + 2b .
2 2 2 2
Thus the equation is
x − ( +  ) x +   = x + (a + 2b) x + b = 0.
2 2 2 2 2 2 2 2

Example
If  and  are the roots roots of x 2 − 8 x + 7 , find:
(i)  2 +  2 (ii)  −  (iii)  3 +  3 .
Solution:
Observe that ( +  ) = 8 and  = 7 .
(i)  2 +  2 = ( +  ) 2 − 2 = (8)2 − 2(7) = 50 .
(ii) ( −  ) 2 =  2 − 2 +  2 =  2 +  2 − 2 = 50 − 14 = 36 . Therefore
 −  = ( −  ) 2 = 6 .
(iii)  3 +  3 = ( +  )( 2 −  +  2 ) = 8(50 − 7) = 344 .

Example
1 1
If the equation x 2 + ax + 1 has  and as roots, find the equation whose roots are  2 and
 2
Solution:
1 1
The equation is ( x −  2 )( x − )=0 or x 2 − ( 2 + ) x + 1 = 0. But
 2
2
1 1
2 + = ( + ) 2 − 2 = (−a ) 2 − 2 . Thus the equation is x 2 + (2 − a 2 ) x + 1 = 0 .
 2


Example
Suppose the equations x 2 + px + 4 = 0 and x 2 + qx + 3 = 0 have a common root, write this root in
terms of the other two roots.
Solution:
29
Let  be the common root. Then  2 + p + 4 = 0 and  2 + q + 3 = 0 . Therefore
−1
( p − q ) = −1 or  = .
p−q
Suppose that  is the second root of x 2 + px + 4 = 0 and  the second root of x 2 + qx + 3 = 0 .
Then  +  = − p , and  +  = −q .
−1
Thus  −  = p − q and then  =
 −

Practice Exercise 2.5

(1) Write the following equations in the form ax 2 + bx + c = 0 , and give the value of a, b , and c .
(a) x 2 = 6 x (b) 2 x 2 = 32 (c) 3 x 2 = 5 x + 1 (d) 10 = 3x − x 2 (e) ( x + 2)2 = 9 (f) 4 x 2 = 64
(2) By examining the discriminants, determine which of these quadratic equations have (real)
roots.
(a) x 2 + 2 x − 63 = 0
(b) x 2 + 8 x − 4 = 0
(c) x 2 − 5 x − 6 = 0
(d) x 2 + 7 x − 13 = 0
(e) 3x 2 + 5 x + 6 = 0
(f) 4 x 2 + 6 x + 10 = 0

(3) Solve each of these equations.

(a) x 2 + 4 = 0
(b) 2 x 2 − 18 = 0
(c) 2 x 2 + 18 = 0
(d) x 2 − 3 = 0
(e) 3x 2 + 21 = 0
(f) x 2 − 12x + 35 = 0
(g) x 2 − 3x − 40 = 0
(h) 2 x 2 − 5 x = 3
(i) 3 x 2 + 25x = 18
(j) 15x 2 − 2 x − 8 = 0
(k) x 2 + 8 x + 15 = 0
(4) If  and  are the roots of 7 x 2 − 8 x + 1 , find
(a)  +  , (b)  (c)  + 
(5) If  and  are the roots of x 2 − 9 x + 5 , show that
(a)  3 +  3 = 594 (b) ( +  ) 2 = 61 .

2.6 Polynomial Equations

Theorem (Fundamental Theorem of Arithmetic)
If p(x) is a polynomial of degree n , n  N with complex coefficients , then p(x) has exactly n
complex roots when counting according to multiplicity of the roots.

30
This theorem tells us the maximum number of roots to expect from a given polymial. For
example, a polynomial of degree 3 must have three roots, though, some of the roots may be the
same. It does not, however, tell us how to find the roots.
When trying to find all roots of a polynomial, it usually helps to find an easy root first.
For example, with the polynomial p( x) = x 3 + x 2 + 7 x − 9 , we may observe that p(x) has a root at
x = 1 , then we can factor x − 1 to get p( x) = ( x − 1)(x 2 + 2 x + 9) through long division. We may
then use the quadratic formula to find the roots of x 2 + 2 x + 9 = 0 .
Finding the roots of some other polynomial eqautions may require the application of the
factor theorem or some other theorems.

Example Solve:

(a) x 3 − 3 x + 2 = 0 (b) x 3 + 5 x 2 + 4 x = 0
(c) ( x − 1) ( x + 1) = 0 (d) x 3 + 3 x 2 + 3 x + 1 = 0
2 2

Solution:
(a ) Let p( x) = x − 3x + 2 . Then p(1) = 1 − 3(1) + 2 = 1 − 3 + 2 = 0 . Therefore ( x − 1) is a factor
3 3

of p according to factor theorem. Dividing x 3 − 3 x + 2 by ( x − 1) through long division gives

the other factor p as x 2 + x − 2 . Thus we can write the cubic polynomial of p as a product of a
linear polynomial and a quadratic polynomial namely x 3 − 3x + 2 = ( x − 1)(x 2 + x − 2) .
Furthermore, x 2 + x − 2 = ( x + 2)(x − 1) . Thus the equation x 3 − 3x + 2 = ( x − 1)2 ( x + 2) = 0 has
three roots namely x = 0, x = 0, x = −2 . Of course, the root x = 0 is repeated.
(b) For the polynomial x 3 + 5 x 2 + 4 x , it is obvious that x is a factor. Thus
x 3 + 5 x 2 + 4 x = x( x 2 + 5 x + 4) . Moreover, the quadratic expression x 2 + 5 x + 4 = ( x + 4)(x + 1) .
Thus x 3 + 5 x 2 + 4 x = x( x + 4)(x + 1) = 0 . The roots of the equation are x = 0, x = −1, x = −4 .
(c ) The equation ( x − 1) ( x + 1) = 0 is broken down into ( x − 1) = 0 and ( x + 1) = 0 .
2 2 2 2

The quadratic equation ( x − 1)2 = 0 has a repeated root, namely x = 1 . On the other hand the
equation ( x 2 + 1) = 0 has no real root. It however has complex roots namely x = + −1 and
x = − −1 .
(d ) . There seems to be no clear way of factoring the polynomial x 3 + 3 x 2 + 3 x + 1 .
However, by substituting a few numbers into the polynomial, through a judicious guess, we
observe that (−1)3 + 3(−1)2 + 3(−1) + 1 = −1 + 3 − 3 + 1 = 0. Thus, according to the factor theorem,
( x + 1) is a factor of x 3 + 3 x 2 + 3 x + 1 . We then use the process of long division to find the other
factor as x 2 + 2 x + 1 . This is obviously ( x + 1)2 . Thus x 3 + 3x 2 + 3x + 1 = ( x + 1)3 and the equation
x 3 + 3 x 2 + 3 x + 1 = 0 has a root x = −1 that is repeated thrice.

Example
Consider the polynomial equation x 3 − x 2 + 13x + 75 = 0 .
(a) What is the product of the roots?
(b) List all integers that are factors of your answer to part (a ) . Don’t forget the
negatives!
(c) Test all the possibilities from to see if any of the factors obtained is actually a root.

31
Solution:
(a ) Assume the three roots of the equation are  ,  , , then
( x −  )( x −  )( x −  ) = x − x + 13x + 75. Therefore  = 75. (b) The tripple integer factors
3 2

75 are 5  5  3 = −5  −5  3 = 5  −5  −3 = −25 −31 = −3 −151... . (c ) Substituting x = −3

into the polynomial gives (−3)3 − (−3)2 + 13(−3) + 75 = −27 − 9 − 39 + 75 = 0 . So x = −3 is a root.
The other factor of 75 are not roots since the other factor is x 2 − 4 x + 25 which has no real root.
The following theorem is sometimes very useful in finding roots of some polynomials.

Theorem (Rational Roots Theorem)


Let p( x) = an x n + ... + a1 x + a0 be a polynomial with integer coefficients. If
is a rational

number in lowest terms that is a root of p(x) , then a n is divisible by  and a0 is divisible by 
.

This theorem says that if there is any rational number solution at all, and there may not
be, the constant term of the polynomial must be a multiple of the numerator and the coefficient
of the highest degree term must be a multiple of the denominator. Otherwise, a fraction has no
hope of being a root of the polynomial.

Example
Given 3x 3 − 5 x 2 + 5 x − 2 = 0 .
(a) List all possible numerators of rational roots (there should be four).
(b) List all possible denominators of rational roots (there are four of these, too):
(c) List all possible rational roots of the polynomial (there should be eight):

Solution: (a ) The numerator must divide − 2 . Thus the possible numerators are − 2,2,1,−1 . (b)
The denominator must divide 3 . The possible denominator must divide − 3,3,−1,1 . (c ) the
 − 2 2 1 −1
possible rational root solutions are = , , , ,2,−2,−1,1. (c ) One of the roots of the
 3 3 3 3
2
equation is x − . This mean that the polynomial 3 x 3 − 5 x 2 + 5 x − 2 is divisible by 3x − 2 . The
3
other factor is x 2 − x + 1 which has no real root.

Practice Exercise 2.6

(1) Guess a root of each of the polynomials below:
(a) x 3 + 3 x 2 − 6 x − 8 = 0
(b) x 4 − x 3 + 2 x − 2 = 0
(c) x 3 − 8 x 2 + x + 42 = 0
(d) 2 x 3 + 5 x 2 − 2 x + 120 = 0
(e) x 3 + 3x 2 − 12x + 4 = 0
(f) 2 x 3 − 5 x 2 + x − 12 = 0
(2) Solve each of the polynomial equations below.
(a) x 4 − 6 x 3 + 3x 2 + 24x − 28 = 0
(b) x 4 − 3 x 3 − 12x − 16 = 0

32
 (c) x 4 − 13x 2 + 20x − 4 = 0

2.7 Inequalities
An Equation is a mathematical denotation of a statement of equality or equivalence of
two quantities or expressions. For example, 7 = 5 + 2 is an equation( 7 = 2 + 2 is also an equation
but it has no solution).
What then is an inequality ?
It is any mathematical expression that uses one of the symbols ``<"(less than), ``" (less
than or equal to), ``>" (greater than), or ``" (greater than or equal to), instead of ``=" to state the
relationship between two quantities.

Example
What is ``x < 7"?
It means that whatever value x has, it must be less than 7 .
Can you name ten numbers that are less than 7 ? Then each of those numbers satisfied the
inequality x < 7 . There are, of course, infinitely many real numbers that satisfy the inequality.
The bracket ( is used to represent the inequlity ``<" on the number line and ) is used to
represent ``>".On the other hand, [ to represent ``" and ] to represent ``" , respectively.
For example, numbers less than 7 are to the left of 7 on the number line. The inequlity
x < 7 is reprented as

You can name some of the numbers as − 20,−15,−1,−6 etc. There are also some other
5 17
numbers like 5 ,6 etc. However, the number 7 would not be a correct answer because 7 is
6 21
not less than 7 .

Example
What is x  2 ?
It means that whatever value x has, it is either 2 itself or else it must be greater than
2.Can you name ten numbers that are greater or equal to 2? The numbers greater than 2 are to the
right of 2 on the number line. The graph of the inequality is

If you mentioned 3,4,7,8 etc, you are right. But there are also other numbers like
5.23,7.44, 5 ,  , etc which satisfy this inequlity that are not neccesailry integers. In particular,
the number 2 is also a correct answer. Since 2 is equal to 2 , then it is greater or equal to 2 .

Example
What is − 3 < x  5 ?
33
 An inequality like this is called a double inequality because it involves two inequality
symbols. This means any number in between the two number − 3 and 5 excluding − 3 but
including 5. This says − 3 < x and x  5 . A number that must satisfy the given inequality must
satisfy the two inequalities − 3 < x and x  5 . This is graphed as

Example
What is x < −1 or x  2 ?
Unlike the last example, this is an ``or" problem. The inequality says x < −1 or x  2 . In
this case, both parts cannot be true at the same time since a number that is less than 1 cannot be
greater than 2 and vise versa. Once a number satisfies any of the two inequalities, then it is a
correct answer. For example, the numbers − 1.2,−3,2,7, 5 etc all belong to the solution set of
the inequality. This is graphed as

2.7.1 Linear Inequalities and absolute value function

A linear inequality is any inequlity involving a linear polynomial and any of the
inequality symbols. For example, x + 4 < 5,2 − 3x  6 etc.

Basic Principles.
To solve a linear inequality, we need to remember these properties of the order relation ``<" in
R.
• If x < y , then x + z < y + z for any z  R . (Translation invariant property).
• If x < y , then: xz < yz if z > 0 and xz > yz if z < 0 .

The second property is very crucial. Essentially, all of the properties that you learned to
solve linear equations would apply to solving linear inequalities with the exception that if you
multiply or divide by a negative you must reverse the inequality sign. Let it ring an alarm in your
brain any time you want to multiply or divide by a negative number.

Example
Solve 2x − 6 < 4x + 8 .
Solution:
Add − 4x to both sides: − 2x − 6 < 8 . Add 6 to both sides. − 2x < 14 . Divide both sides by − 2 :
We turned the sign to get x > −7 .

Example
Solve 4h + 8  −24 .

Solution
Add − 8 to both sides(we don’t need to change the sign of the inequality): 4h  −32 . Divide

34
both side by 4 : h  −8 (we don’t need to change the inequality).

Example
Solve 2x + 3  2 and 5x < 10
Solution:
The two inequalities are joined with the conjuntion ``and" . We must solve the two inequalities
and then the solution of the given ineqaulity is the intersection of the two component
inequalities. If they do not intersect, then there is no solution. The inequality 2x + 3  2 gives
−1
x while the inequality 5x < 10 gives x < 2 . Thus the solution to the inequality 2x + 3  2
2
−1
and 5x < 10 is  x < 2.
2

Example
Solve 5 < 3 − 2x < 17 .
Solution:
METHOD I Rewrite the compound inequality using the word ``and" , then solve each
inequality: 5 < 3 − 2x and 3 − 2x < 17 .
Add − 3 both sides: 2 < −2x and − 2x < 14 .
Divide both sides by − 2 and reverse the inequalities: −1 > x and x > −7 .
Join the two inequalities again by ``and" : − 7 < x < −1 .
METHOD II: To solve the inequality, subtract − 3 from all sides 2 < −2x < 14 . Divide
both sides by − 2 and reverse the signs: − 7 < x < −1 .

2.7.2 Absolute Value Inequality

The Absolute value of a number x is defined as follows:

 x, if x  0,

| x |= − x, if x < 0


The absolute value of a number x is simply the distance of the number from 0 . When an
absolute value is less than the variable you have a conjunction. For example, | x |< 4 is the same
as − 4 < x < 4 . On the other hand, when the absolute value of x is greater than a number, we
have a disjunction. For example, | x | 4 is the same as x  −4 or x  4 . In general, given a
positive number a , then | x |< a is solved by solving the equivalent inequality − a < x < a .

35
 Expressions of the form | 2(3 + 8 x) − 9 |> −2 is always true since the absolute value is
always positive. On the other hand, expressions of the form | ( x + 4) |< −2 is never true since
absolute values are never negative.

Example
Solve | 4 x + 3 |< 11 .

Solution:
The inequality is equivalent to −11 < 4x + 3 < 11. Subtract 3 from each side: −14 < 4x < 8 .
11
Divide both sides by 4 : − < x < 2 .
4

Practice Exercise 2.7A

(1) Write an inequality for each statement.
(a) A number m multiplied by 5 is less than 25 ..... .........
(b) The sum of a number y and 16 is no more than 100.............
(c) A number t increased by 7 is more than 11................................
(d) A number y plus 14 is greater than 21…………………………
(e) A number x is both less than 4 and greater than or equal to 2..........................
(f) A number t is either greater than 1 or less than or equal to 7.........................
(g) A number t is both greater than 9 and less than or equal to 18.5.................
(h) A number y is either greater than 5 or less than or equal to 1……........

(2)Solve and graph the following inequalities

(a) - 2 − 4x > 2x + 6 (b) − 4 < 2 − x < 8 (c) 2x < −4 or 3x  5
(d) - 2x < 4 or 4x  −8 (e) 3x + 4 < 1 and 2x > 8

(3) Find the set of solutions of (a) | 3 − 5 x |< 8 (b) | 4 x − 3 |< 13 (c) | 3x + 5 |< −1
(d) | 2 x − 1 | −3  6 (e) | 2 x − 4 | −6  1

2.7.3 Polynomial Inequalities

Basic Principle: When will the product ``ab" of two real numbers a and b be
positive?
The answer is when a and b are both positive or when a and b are both negative( + + = + and
36
 − − = + ).
Again: When will a product ``abcde of real numbers a, b, c, d , e be negative?
The answer in this case is when an odd number of the factors are negative. ( + − = − and
− + = − ). A quadratic inequality is any inequality of the form p ( x)  0, p ( x) > 0, p ( x) < 0 or
p ( x )  0 where p( x) = ax + bx + c, a > 0 . Given a quadratic inequality if b 2 − 4ac < 0 , then the
2

quadratic equation p (x ) = 0 has no real root. The solution to the quadratic inequality
ax 2 + bx + c > 0 is then the whole of R .
If b 2 − 4ac = 0 , then the polynomial has one repeated root, say  , and the inequality
ax 2 + bx + c > 0 which is a( x −  ) > 0 is true for all real numbers except  . On the other hand,
2

if b 2 − 4ac > 0 , then p has two distinct roots say  and  . Moreover, the quadratic equation
can be factorised as a ( x −  )( x −  ) . Assuming that  <  , then the real line is partitioned into
the following disjoint sets where the polynomial behave differently, namely:
(−,  ),{ }, ( ,  ),{ }, (  ,+ ). The numbers  ,  are called critical numbers.The inequality is
expected to change sign once it passes through a critical number. The inequality is resolved by
examing the sign chart as shown below.

− < x <  <x<  <x<

(x −  ) − ve + ve +ve
(x −  ) − ve − ve + ve
( x −  )( x −  ) +ve − ve + ve
a ( x −  )( x −  ) + − +

The sings are determined as follows: for example, if −  < x <  , then x <  and then
x −  < 0 . Thus the sign of x −  is − ve .On the other hand, x <  <  gives x <  and thus
x −  < 0 holds also.etc Thus ax 2 + bx + c > 0 when −  < x <  or  < x <  .

Example
Solve: x 2 − 3 x − 4 < 0 .
METHOD I
The related quadratic equation is x 2 − 3 x − 4 = 0 . This has solution x = −1 and x = 4 .
These solutions are the critical numbers. They divide the number line into three test intervals,
namely, (−,−1), (−1,4),(4,+) . Take a test point within each interval and check the sign.

Test Test x +1 x−4 ( x + 1)(x − 4)

Interval Point
(−,−1) −2 −1 −6 +6 > 0
(−1,4) 0 +1 −4 −4 < 0
(4, ) 5 +6 +1 +6 > 0

The correct option then is the test interval that gives x 2 − 3x − 4 = ( x + 1)(x − 4) < 0 that is
(−1,4) .

Example
37
 Solve: 6 x 2 + 13x − 5 < 0 .
Solution:
METHOD II
1 5
The quadratic equation x 2 + 13x − 5 = 0 has solutions x = and x = − .
3 2
1 5
Thus x 2 + 13x − 5 = 6( x − )( x + ) . The sign chart is shown below:
3 2

5 5 1 1
− < x < − − <x< <x<
2 2 3 3
5 − + +
(x + )
2
1 − − +
(x − )
3
1 5 − − = + + − = − + + = +
( x − )( x + )
3 2
5 1 + − +
6( x + )( x − )
2 3

5 1
Thus from the sign chart, 6 x 2 + 13x − 5 < 0 is true when − <x< .
2 3

Example
Solve ( x − 3)(x + 1)(x + 5)  0
Solution
The roots of the equation ( x − 3)(x + 1)(x + 5) = 0 are x = −5, x = −1, x = 3 .
The sign chart is presented below:

−  < x < −5 −1 < x < 3 3< x <

− 5 < x < −1

( x + 5) − + + +
( x + 1) − − + +
( x − 3) − − − +
( x − 3)(x + 1)(x + 5) − + − +

Thus the inequality is true when − 5 < x < −1 or 3 < x <  . But it is also true for
x = −5,−1,3 . Thus the solution is − 5  x  −1 or 3  x < 
For a general polynomial expression,there may not be any specific formula for finding
the roots. In that case, we resort to other means of finding the roots, for example factor theorem
and rational root theorem, etc in order to factor the polynomial. We then create a sign chart and
use it to deterime the solution.

Example
38
 Solve 3x 3 − 5 x 2 + 5 x − 2 < 0 .
Solution
2
We may resort to the rational root test to find the only real root, namely x = . We then
3
factor 3 x 3 − 5 x 2 + 5 x − 2 as (3x − 2)(x 2 − x + 1) where x 2 − x + 1 has no root. Thus the only
2
critical number is x = . The sign chart is presented below.
3

2 2
− < x < <x<
3 3
(3x − 2) − +
( x 2 − x + 1) + +
3x 3 − 5 x 2 + 5 x − 2 − +

2
From the sign chart, 3 x 3 − 5 x 2 + 5 x − 2 is negative when −  < x < .
3

Practice Exercise 2.7B

(1) Solve the followoing quadratic inequalities
(a) x 2 − 4 x + 3 < 0 .
(b) x 2 − 2 x − 3 > 0 .
(c) 3x 2 + 16x  −5
(2) By constructing the sign chart, solve the following
(a) ( x − 2)(2x − 3)(x + 5) < 0 .
(b) (3x + 2)(x − 4)(x − 1)2 > 0 .
(c) ( x + 2)(3 − x)( x − 4)  −5

2.7.4 Rational Inequalities

p( x)
A rational inequality is any inequality that can be expressed of the form > 0 or
q ( x)
p( x)
 0 where p, q are polynomials.
q ( x)
A rational inequality is solved by setting up a sign chart, just as in the case of polynomial
inequalities. However, the critical numbers are the roots of the two polynomials p and q .

x+4
Example 2.69 Solve > 0.
x−5

Solution: METHOD I: It is not correct to multiply both sides by x − 5 in order to cancel

the fraction because x − 5 being a variable contains both positive and negative values. To find
the critical numbers, we solve the equation x − 5 = 0 and x + 4 = 0 to obtain x = 5 and x = −4 .
These numbers dividie the number line into (−,−4) , (−4,5) and (5,) . We construct the sign
chart by taking one test point from each interval.

39
 Test Test x+4 x −5 x+4
Interval Point x−5
(−,−4) −5 −10 = −ve 1
+ = + ve
−1 = −ve 5
(−4,5) 0 +4 −5 4
− = −ve
5
(5,) 6 + 10 +1 +10 = +ve

x+4
Thus the solution to > 0 is (−,−4)  (5, ) .
x−5
METHOD II: Another technique is to observe that any rational inequality of the form
p( x)
> 0 has the same set of solutions as the corresponding polynomial ineqaulity p( x)q( x) > 0
q ( x)
. In this case ( x + 4)(x − 5) > 0 . This is then easily solve to obtain the solution set as s
(−,−4)  (5, ) .

Example
1
Solve  1.
x −5

1 6− x
Solution: Rewrite this in standard form as − 1  0 or  0 . Setting the
x−5 x −5
numerator and the denominator equal to zero, we get the critical numbers as x = 6 and x = +5.
These numbers give the test intervals (−,5), (5,6),(6, ) .

(−,5) (5,6) (6,)

(6 − x) + + −
( x − 5) − + +
6− x − + −
x−5

6− x 6− x
From the sign chart < 0 if x  (−,5)  (6, ) . On the other hand, = 0 if and
x −5 x −5
only if 6 − x = 0 or x = 6 . Thus the solution set is (−,5)  [6, ) .
Example
x
Solve 3
4− x
Solution:
x
Rewrite the inequality in standard form: − 3 < 0 . We find the LCM on both sides as
4− x
x − 3(4 − x) 4 x − 12
< 0 or < 0 . The critical numbers are obtained as x = 3 and x = 4 .
4− x 4− x
40
 Test Interval Test x+4 x −5 x+4
Point x−5
(−,3) 0 4 = +ve − 5 = −ve 4
− = −ve
5
(3,4) 3.5 + 2 = +ve 0.5 = +ve + 4 = +ve
(4,) 5 8 = +ve −1 = −ve − 8 = −ve

x
From the sign chart, < 3 is true if and only if x  (−,3)  (4, ) . On the other, if
4− x
3 x
x = 3 , then = 3 . Thus  3 if and only if x  (−,3]  (4, )
4−3 4− x

Real Life Example

20t
The function p(t ) = models the population in tens of thousands, of Abakaliki, t
t +1
years after 2015. The population, in tens of thousands, of nearby Afikpo is modeled by
240
q(t ) = . Determine the time period in years when the population of Abakaliki exceeds the
t +8
population of Afikpo.

Solution.
20t 240 20t 240
We need to solve the inequality > . This is the same as − > 0 or
t +1 t + 8 t +1 t + 8
20t 2 + 16t − 240t − 240 20(t − 6)(t + 2)
> 0 . The critical numbers for the inequality > 0 ought to
(t + 1)(t + 8) (t + 8)(t + 1)
be t = 6, t = −2, t = −8, t = −1 .However, since time cannot be negative, we are concerned with
t  0 . Therefore the only critical number here is t = 6 . We set up the sign chart as follows:

Test Test t −6 t +2 t +8 t +1 20(t − 6)(t + 2)

Interval Point (t + 8)(t + 1)
[0,6) | 0 −6 +2 +8 +1 −15

(6,) | 10 +4 + 12 + 18 + 11 960
+
198

20(t − 6)(t + 2)
Thus the inequality > 0 hold if t > 6 . Therefore the population of
(t + 8)(t + 1)
Abakaliki will overtake the population of Afikpo after 6 years.

Practice Exercise 2.7C

Solve the followoing rational inequalities

41
 x+5
(a) <0
x+2
x
(b) 2
x+2
7 + 2x
(c) <5.
x −5
2x x+3
(d) <
6 x + 5 3x − 1
5 3
(e) 
x −3 x−2
x 2 + 9 x + 14
(f) >0
x2 − 6x + 5
5 2
(g) 
x + 4 x +1
( x + 3)(x − 2)
(h) 0
( x − 4)(x − 1)(x + 4)
x 2 + 5x − 6
(i) 0
( x 3 − 3x 2 )

2.8 Algebraic and Partial Fractions

2.8.1 Rational functions
p( x)
A rational function is any function that has expression of the form f ( x) = where
q( x)
1
p and q are polynomials and 𝑞(𝑥) ≠ 0. For example, f ( x) = 2 , x  0,
x
x 2 − 5x + 6
g ( x) = , x  2,−2
x2 − 4 .
p
The domain of a rational function is all values of real numbers x such that q ( x)  0
q
x 2 − 5x + 6
For example, given g ( x) = , then
x2 − 4
Dom( g ) = {x  R : x 2 − 4  0}
= {x  R : x  2 or x  −2}
= R \ {2,−2}

p
A rational function is called proper if the degree of p is less than the degree of q .
q
Proper rational fractions may be further broken down into its partial fractions.

2.8.2 Partial Fractions

42
 We already know how to decompose numeric fractions.
2 3 (2  5) + (3  3) 19 5 3 (5  7) + (3  8) 59
For example, + = = , while + = = .
3 5 3 5 15 8 7 8 7 56
We can extend this idea to rational fractions.
3 5 3( x − 1) + 5( x + 2) 8x + 7
For example, + = = 2 .
x+2 x−2 ( x + 2)(x − 1) x + x−2
To find the partial fractions of a rational expression, we need to reverse the above process
8x + 7 3 5
of adding fractions: 2 = +
x + x−2 x+2 x−2

Example:
5x − 1
Find the partial fractions for .
( x − 2)(x + 1)
Solution:
5x − 1 A B
We need to write = + , for some real numbers
( x − 2)(x + 1) x − 2 x + 1
A and B . To find A and B , we follow these steps.
(a) Multiply both sides by the denominator of the left hand side:
5x −1 A B
( x − 2)(x + 1)  = ( x − 2)(x + 1)  + . So 5 x − 1 = A( x + 1) + B( x − 2).
( x − 2)(x + 1) x − 2 x +1

Remark:
If we understand the cancelling, we can in future go straight to the last line.

METHOD I:
Set x − 2 = 0, i.e x = 2 : Then 5(2) − 1 = A(2 + 1) + B(2 − 2) .
Thus 3A = 9 or A = 3 .
Set x + 1 = 0, i.e x = −1 . Then 5(−1) − 1 = A(−1 + 1) + B(−1 − 2) . Thus − 6 = −3B or B = 2
.
METHOD II:
Compare the coefficients. 5 = A+ B and − 1 = A − 2B . Solve the simultaneous
equation for the variables A and B to get A = 3 and B = 2 .

Substitute the values of A and B into the equation to get

5x −1 3 2
= + .
( x − 2)( x + 1) x − 2 x + 1

Example
x−2
Find the partial fractions for the expression .
( x − 3)(2x + 1)
x−2 A B
Solution: Let = + .
( x − 3)(2x + 1) x − 2 2 x + 1
Multiply both sides by the denominator of the left-hand-side:
x − 2 = A(2 x + 1) + B( x − 3) .
Substitute the value of x that makes the coefficient of B equal to zero and solve for A .
43
 Observe that x − 3 = 0 mean x = 3 .So 3 − 2 = A(2(3) + 1) + B(3 − 3) and solving gives
1
A= .
7
Substitute a value of x that makes the coeeficient of A equal to zero and solve for B .
−1 −1 −1 −1
Observe that 2x + 1 = 0 means x = . So − 2 = A(2( ) + 1) + B( − 3).
2 2 2 2
−5 −7 5
Thus = B or B = .
2 2 7
Alternatively, multiply the right hand side and compare the coefficients with the
right hand side. Then 1 = 2 A + B and − 2 = A − 3B . Solve the simultaneous equation
to obtain A and B .

Confirm your answer by reversing the method:

1 5 2 1 5 15
x+ + x−
7 + 7 =7 7 7 7 = x−2
.
x − 3 2x +1 ( x − 3)(2x + 1) ( x − 3)(2x + 1)

Remark:
If the denominator has 3 factors, we just extend the method natuarlly to accomodate it.

Example
x2 + 6x + 5
Find the partial fractions for .
( x − 1)(x + 3)(x + 2)
x2 + 6x + 5 A B C
Solution: Let = + + .
( x − 1)(x + 3)(x + 2) x − 1 x + 3 x + 2
Multiply both sides by ( x − 1)(x + 3)(x + 2) :
x 2 + 6 x + 5 = A( x + 3)(x + 2) + B( x − 1)(x + 2) + C ( x − 1)(x + 3)
x = 1 : 1 + 6 + 5 = A(4)(3),i.e 12 = 12 A or A = 1 .
x = −3 : 9 − 18 + 5 = 5(−4)(−1), i.e − 4 = 4 B or B = −1.
x = −2 : 4 − 12 + 5 = C (−3)(1), i.e − 3 = −3C or C = 1.
x2 + 6x + 5 A 1 1
So = − + .
( x − 1)(x + 3)(x + 2) x − 1 x + 3 x + 2

Remark:
If a linear factor is repeated in the denominator, the form in the examples above will not
work. We will use an alternative approach.

Example
4 x 2 − 3x + 5
Find the partial fractions for .
( x − 1)2 ( x + 2)
Solution: The partial fraction will be of the form
4 x 2 − 3x + 5 A B C
= + + .
( x − 1) ( x + 2) ( x − 1) ( x − 1) x + 2
2 2

44
 Multiply both sides by ( x − 1)2 ( x + 2) :
4 x 2 − 3x + 5 = A( x + 2) + B( x − 1)(x + 2) + C ( x − 1)2 .
x = −2 : 16 + 6 + 5 = C (−) 2 i.e C = 3.
x = 1 : 4 − 3 + 5 = 3 A i.e A = 2
There is no other obvious value of x to use. We can choose any value.
x = 0 : 5 = 2 A − 2B + C . Substitute for A and C to get B = 1 .
Thus
4 x 2 − 3x + 5 2 1 3
= + + .
( x − 1) ( x + 2) ( x − 1) ( x − 1) x + 2
2 2

Remark: If a quadratic factor cannot be factored into linear parts, then the partial fractions
are obtained as follows.

Example
x 2 + 10x + 8
Find the partial fractions of 2 .
( x + 4)(x + 3)
Solution: Let the partial fractions be of the form
x 2 + 10x + 8 Ax + B C
= 2 + .
( x + 4)(x + 3) x + 4 x + 3
2

Multiply both sides by ( x 2 + 4)(x + 3) :

x 2 + 10x + 8 = ( Ax + B)( x + 3) + C ( x 2 + 4)
x = −3 : 9 − 30 + 8 = C (9 + 4) or C = −1 .
x = 0 : 8 = 3B + 4C = 3B − 4 or B = 4 .
Comparing the coeeficients of x 2 , we have 1 = A + C = A −1 and then A = 2 .
Thus
x 2 + 10x + 8 2x + 4 1
= 2 − .
( x + 4)(x + 3) x + 4 x + 3
2

Remark:
If the degree of the numerator is greater than or equal to that of the denominator, then
it should first be reduced to a proper rational fraction before the partial fractions can be found.

Example
2x2 − 4x + 3
Find the partial fractions for .
x 2 − 5x + 6
2x2 − 4x + 3 6x − 8
Hint: The equation should first be written as = 2+ through long
x − 5x + 6
2
( x − 3)(x − 2)
6x − 8
division. The partial fractions for is then not difficult to obtain.
( x − 3)(x − 2)
Practice Exercise 2.8
Express each of these rational functions as a sum of their partial fractions.
5x − 5 2 7x + 3 1− 2x − x2
(a) (b) 2 (c) (d) 2
( x + 3)(x − 2) x −1 x( x + 1) x (2 x + 1)

45
 x2 x2 − 4x + 4 x2 − x − 5 x+4
(e) (f) (g) (h)
x2 + x − 2 4 − x2 ( x 2 + 4)(x − 1) ( x + 1)2
x 2 − 11x − 18 2 x 3 − 4 x 2 − 15x + 5 2 x3 − 4 x 2 − x − 3
(i) (j) (k)
x( x 2 + 3x + 3) x2 − 2x − 8 x2 − 2x − 3
x( x − 2) 4 − 2x et
(l) (m) 2 (n) 2t
(2x + 1)(x 3 − 1) ( x + 1)(x 2 − 1) e + 3et + 2
cos 1
(o) (p)
cos  + cos − 2 4tan  + tan − 4
2 2

2.9 Principles of Mathematical Induction

Theorem (Principle of Mathematical Induction).
Let P1 , P2 , ... , Pn ,... be propositions or statements, one for each positive integer, such that
1. P1 is true
2. For each positive integer n , Pn implies Pn +1 .
Then Pn is true for each positive integer n.

Example
Prove that Pn is the proposition that 1 + 2 + 3 + ... + n = n(n + 1) .
2
Solution
We start with when n = 1 , that is, for P1 . One observes that 1 = 1(1+ 1) = 1 , which shows that
2
P1 holds. Next assume that the statement is true for n = k , that is, Pk holds:
k ( k + 1)
1 + 2 + 3 + ... + k = (5.1)
2
and prove that the proposition holds for n = k +1 , that is for Pk +1 . To prove Pk +1 , we add k + 1 to
both sides of (5.1):
k (k + 1)
1 + 2 + 3 + ... + k + (k + 1) = + (k + 1)
2
k k +2
= (k + 1)[ + 1] = (k + 1)( )
2 2
(k + 1)[(k + 1) + 1]
= .
2
This implies that Pk +1 is holds and therefore Pn is true for all n  N .

Example
Prove by mathematical induction that 1 + 7 + 13 + ... + (6n − 5) = n(3n − 2).

Solution
We consider when n = 1 , we see that 1 = 1(31 − 2) = 3 − 2 = 1 which is true. Consider when
n = 2 , we have that 8 = 1 + 7 = 2(3 2 − 2) = 2(6 − 2) = 2  4 = 8 . Now assume true for n = k ,

46
that is,
1 + 7 + 13 + ... + (6k − 5) = k (3k − 2).
Add 6(k + 1) − 5 = 6k + 1 to both sides of equation above to obtain
1 + 7 + 13 + ... + (6k − 5) + (6k + 1) = k (3k − 2) + 6k + 1
= 3k 2 − 2k + 6k + 1
= 3k 2 + 4k + 1
= 3k 2 + 3k + k + 1
= 3k (k + 1) + 1(k + 1)
= ( k + 1)(3k + 1)
= (k + 1)(3(k + 1) − 2).
This is true for n = k +1 and therefore holds for all n.
Example
Prove that the sum of the squares of the first n odd integers is n (2n − 1)(2n + 1)
3
Solution
Let Pn be the proposition/statement 12 + 32 + 52 + ... + (2n − 1)2 = n (2n − 1)(2n + 1) . For P1 , we
3
1 3 Pn is true
consider when n = 1 and obtain 1 = (2 − 1)(2 + 1) = = 1 and P1 is true. Now assume
3 3
for n = k :
k
12 + 32 + 52 + ... + (2k − 1)2 = (2k − 1)(2k + 1)
3
and show that Pn is true for n = k +1 . Add (2k + 1) to both sides of above equation:
2

k
12 + 32 + 52 + ... + (2k − 1)2 + (2k + 1)2 = (2k − 1)(2k + 1) + (2k + 1)2
3
k
= (2k + 1)[ (2k − 1) + (2k + 1)]
3
1
= (2k + 1)[ (2k 2 − k + 6k + 3)]
3
1
= (2k + 1)[ (2k 2 + 5k + 3)]
3
1
= (2k + 1)(k + 1)(2k + 3)
3
(k + 1)
= [2(k + 1) − 1][2(k + 1) + 1].
3
Thus Pn is true for all n  N .

Example
Prove by mathematical induction that 1 + 5 + 52 + ... + 5n −1 = 1 (5n − 1).
4
Solution
Now consider when n = 1 , 1 = 1 (51 − 1) = 1  4 = 1 which holds. Consider again when n = 2 ,
4 4
1 2 1
we have that 6 = 1 + 5 = (5 − 1) =  24 = 6 . Now assume true for n = k , that is,
4 4

47
 1 k
1 + 5 + 52 + ... + 5k −1 = (5 − 1).
4
Add 5( k +1)−1 = 5k to both sides of equation above:
1 k
1 + 5 + 52 + ... + 5k −1 + 5k = (5 − 1) + 5k
4
1 k 1
= 5 − + 5k
4 4
5 k 1 1 k +1
= 5 − = [5 − 1],
4 4 4
which holds for n = k +1 and therefore valid for all n.

Practice Exercise 2.9

Prove by the principle of induction of the following
1. 1 + 4 + 7 + ... + (3n − 2) = 1 n(3n − 1).
2
1
2. 12 + 2 2 + 32 + ... + n 2 = n(n + 1)(2n + 1).
6
1
3. 13 + 23 + 33 + ... + n 3 = n 2 (n + 1)2 .
4
1
4. 14 + 2 4 + 34 + ... + n 4 = n(n + 1)(n + 2).
30
1
5. 1.2 + 2.3 + 3.4 + ... + n(n + 1) = n(n + 1)(n + 2).
3
1 1 1 1 n
6. + + + ... + = .
1.3 3.5 5.7 (2n − 1)(2n + 1) 2n + 1
7. 1.3 + 2.32 + 3.32 + ... + n.3n = 3 [(2n − 1)3n + 1].
4
3 4 5 (n + 2) 1
8. + + + ... + = 1 − .
1.2.2 2.3.22 3.4.23 n(n + 1)2n (n + 1)2n

48
 CHAPTER 3
Sequence and Series of real numbers
Introduction
The word ‘sequence’ is a constantly recurring one in everyday life, where it usually
implies the ordering of some set of events. Such sets of events can either be numerically
or non – numerically related. Examples:
1. The sequence of process required to produce drugs;
2. The sequence of numbers on recharge cards;
3. The sequence of months in a year
4. The sequence of page of a book
Our main pre- occupation in this book will be sequence of numbers, which is a
foundation to sequence of functions.

3.1 Understanding what a sequence is?

A sequence is a set of quantities that is enumerated in a definite order, with each successive
member following a fixed pattern or rule.
For example, the sequence; 4, 8, 12,16,…can be formed from the set of natural numbers
1, 2, 3, 4,… by multiplying each member by 4. The members of a sequence (Progression)
are called TERMS of the sequence. If a sequence is given as u 1 , u 2 , u 3 , u 4 ,…, the
general term of the sequence is written as 𝑈𝑛 where 𝑛 ∈ ℕ . Hence we can define a
sequence {U n } as a function having the ordered set of natural numbers {n} as its
domain.
Definition:
A finite sequence is a sequence that consists of numerable or finite number of terms while
infinite sequence has unending terms. Finite Sequence is of this form:
u 1 , u 2 , u 3 , u 4 ,…, u n , for 𝑛 a natural number. Infinite Sequence is of the form:
u 1 , u 2 , u 3 , u 4 ,…
Note: The three dots () is called Ellipsis and denotes continuation. If we know the terms of a
sequence, we can formulate a formula for the general terms of the sequence.

Example
Find the general formula for the following sequence
(a) 4, 7, 10, 13,…
(b) 3, 1, 1/3,…
(c) 4, − 8, 16, − 32,…
(d ) 11, 6, 1, − 4,…

Solution

49
 (a) u1 = 4
u2 = 4 + 3
u3 = 4 + 3 × 2
u4 = 4 + 3 × 3
 
u n = 4 + 3 × (n - 1) = 4 + 3n - 3 = 1 + 3n
 u n = 1 + 3n
(b) u1 = 3
1
u2 = 3 
3
2
1 1 1
u3 = 3   = 3   
3 3 3
3
1 1 1 1
u4 = 3    = 3   
3 3 3 3
 
n −1
1
 un = 3   
3
u1 = 4  (− 2)
0
(c )
u 2 = 4  (− 2)
1

u 3 = 4  (− 2)
2

u 4 = 4  (− 2)
3

 
 u n = 4  (− 2)
n −1

(d ) u1 = 11
u 2 = 11 − 5  1
u 3 = 11 − 5  2
u 4 = 11 − 5  3
 
u n = 11 − 5  (n − 1) = 11 − 5n + 5
 u n = 16 − 5n

Note: Generally, formula for obtaining the terms of a sequence is not uniquely obtained by
consideration of the few terms. For example 1, 3, 5, 7, … can be generated using
u n = 2n − 1 and u n = (2n − 1) + (n − 1)(n − 2)(n − 3)(n − 4) .
1
5
Sometimes a given rule can appropriately define a sequence but it cannot be easily
written as a formula.

50
Example
Find the first five terms of the sequence defined as u n+1 − u n − 2n = 2 if u n = 5 for all 𝑛
natural numbers.
Solution
u n+1 − u n − 2n = 2
For 𝑛 = 1, we have
u1+1 − u1 − 2(1) = 2
u 2 − u1 − 2 = 2
u 2 = 5 + 4 = 9
For 𝑛 = 2
u 2+1 - u 2 − 2(2) = 2
u3 − u 2 − 4 = 2
u3 = u 2 + 4 + 2
u 3 = 9 + 6 = 15
For n = 3
u 3+1 − u 3 − 2(3) = 2
u 4 −u 3 − 6 = 2
u 4 = 2 + 6 + u 3 = 8 + 15 = 23
for n = 4
u 4+1 − u 4 − 2(4) = 2
u 5 = 2 + 8 + u 5 = 10 + 23 = 33
∴ The sequence is 5,9,15,23,33,…

Example
If a sequence is such that u1 = −2, u 2 = 4 and u n +1 = u n − u n −1 for n > 2 . Find the first 6
terms.

Solution
If n = 2
u 3 = u 2 − u1 = 4 − (−2) = 6
if n = 3
u 4 = u3 − u 2 = 6 − 4 = 2
if n = 4
u 5 = u 4 − u 3 = 2 − 6 = −4
if n = 5
u 6 = u 5 − u 4 = −4 − 2 = −6

The first six terms of the sequence are − 2, 4, 6, 2, − 4, − 6,

Sequence of this form are said to have a recurrence relation.

51
Practice Exercise 3.1
1. Write down the first five terms of the sequences whose general formula 𝑈𝑛 is
given as the following for all positive integer
a) 2n + (− 1)
n

14
b)
2 −1
n

c) 4n − 3
d) 5 − (− 1)
n

e) n − (n + 1)
n

1
f) n −
n
g) 4n − (− 2 )
n −1

2. Find the general formula of the following sequences

a) 11, 32, 71, 134, 
b) −4 , 8, − 16, 32,
1 1 1
c) 1, − , , − ,
2 4 8
d) −1, 4 , − 27, 256,
e) 1, 3, 7, 15,
f) 1, 4 , 16, 64,
1
g) , 2 , 8, 16,
2
3. If a sequence is define by U n (4n + 2) = (4n − 3)U n +1 , for all positive integers 𝑛,
and if U1 = 2 and U 2 = 3 find the first 5 terms of the sequence.
4. Write down the first five terms of the sequence U n  in which U1 = 1,U 2 = 1, and
U n = un −1 + un −2 .
5. Given that u1 = −4, u 2 = −9 and U n = s + tn find s, t and u 6 .
6. A sequence U n  is defined by U n =  + n + n 2 . Given that u1 = 4, u 2 = 10 and
u 3 = 6 , find the values of 𝛼, 𝛽 and 𝜃.

3.2 Series
A series is the summation of the terms of a sequence. The sum of the terms of a finite
sequence gives us a finite series and the sum of the terms of an infinite sequence gives us
an infinite series. For finite series we write thus
S n = u1 + u 2 + u3 +  + u n (1)
For infinite series we write thus
S  = u1 + u 2 + u3 +  (2)

52
 S n and S  denotes the sum of 𝑛𝑡ℎ term of a series and sum of the terms of a series to
infinity respectively. From (1), the 𝑟𝑡ℎ term of a series is 𝑈𝑟 , hence (1) can be rewritten
as
n

u
r =1
r = u1 + u 2 + u 3 +  + u n (3)

Where ∑ is a Greek letter ‘Sigma’, which represents the summation of 𝑈𝑟 for all values
of 𝑟 from 1 to 𝑛. Example, if U r = r , we have
n

r = 1+ 2 + 3 ++ n
r =1

Observe that,
n

u
r= p
r = u p + u p +1 + u p + 2 +  + u n

Also
n −1
S n −1 =  u r = u1 + u 2 + u 3 +  + u n −1 (4)
r =1

Observe that,
n n −1
S n − S n −1 =  u r −  u r = U n
r =1 r =1

U n = S n − S n −1 (5)
Equation (5) relates series and sequence.

Example
5 8
Evaluate the following (a) r r =1
3
(b)  n(n − 1)
n=2

Solution
5
(a) r
r =1
3
= 13 + 2 3 + 33 + 4 3 + 5 3 = 1 + 8 + 27 + 64 + 125 = 225
8 8 8 8
(b)  n(n − 1) =  (n
n=2 n=2
2
− 1) =  n 2 −  n
n=2 n=2

= 2 2 + 32 + 4 2 + 5 2 + 6 2 + 7 2 + 8 2 − (2 + 3 + 4 + 5 + 6 + 7 + 8) = 203 − 35 = 168
Practice Exercise 3.2
4 7
1. Evaluate (a)  n(n
n =1
2
− 3n + 2) (b) n
r =3
2

5 6
(a) U n (b) U n.
2. If U n = 9 + 3n − 4n 2 , evaluate n =1 n=2

Generating of a sequence and series can be done in any defined pattern provided that the rule by
which the terms are generated is judiciously followed. Of all the patterns of generating a
sequence and series; there are two that are most widely used:
a) Sequence (or Series) obtained by adding or subtracting a constant number from a
preceding term to obtain its successor, it is called Arithmetic (linear) progression
or Sequence.

53
 b) Sequence (Series) generated by multiplying a preceding term with a constant
number to get the next term which is called geometric (Exponential) Progression
or Sequence.

3.3 Arithmetic Sequence

An arithmetic (or Linear) sequence is a sequence of numbers in which each new term is
obtained by adding a constant value to the preceding term. The constant value (+ or −) is called
the common difference; denoted by d . An arithmetic sequence (AS or AP) can be defined when
the first term a and the common difference is known. Hence terms of AP are
a, a + d , a + 2d , a + 3d ,, a + (n − 1)d ,
So the general formula for 𝑛𝑡ℎ term (U n ) of an AP is
U n = a + (n − 1)d (1)
Examples of AP are
(a) 1, 4, 7, 10,
(b) 2, 6, 10, 14, 
(c) − 5,−2,1,4,
(d) 3, 7, 11, 15,
1 1 5
(e) − , , 1, , 
3 3 3
The difference between successive or consecutive 𝑛𝑡ℎ terms is the common difference i.e
U 2 − U 1 = U 3 − U 2 = U 4 − U 3 =  = U n − U n −1 = d
The arithmetic mean (A) of the 𝑛 numbers u1 , u 2 , u3 ,, u n is defined as
u1 + u 2 + u3 +  + u n
A=
n
Arithmetic mean is commonly referred to as the average. Three consecutive terms of an AP
is more conveniently written as (a − d ), a, (a + d ); if there are four terms we write (a − 3d ),
(a − d ), (a + d ), (a + 3d ) . The sum of 𝑛𝑡ℎ terms of an arithmetic sequence, otherwise called
the arithmetic series is given as
S n = a + (a + d ) + (a + 2d ) + (a + 3d ) +  + (a + (n − 1)d )
= na + d 1 + 2 + 3 +  + (n − 1)
n 
= na + d  (n − 1)
2 
= na + dn(n − 1)
1
2
 S n = 2a + (n − 1)d 
n
(2)
2
Consequently, if 𝑎 is the first term and l = a + (n − 1)d is the last term of an arithmetic series,
then observe that
a + l = a + (a + (n − 1)d ) = 2a + (n − 1)d
Hence (2) can be written as
S n = (a + l )
n
(3)
2
54
Example
Find 3 numbers in arithmetic progression whose sum is 3 and whose products is −15.

Solution
Let the number be a − d , a, a + d
Then (a − d ) + a + (a + d ) = 3
3a = 3  a = 1
(a − d )  a  (a + d ) = −15
(a 2
)
− d 2 a = −15
a − d a = −15
3 2

but a = 1
 1 − d 2 = −15
d 2 = 16
d =4
 the numbers are − 3, 1, 5

Example
If x + 1,2 x − 1 and x + 5 are three consecutive terms of an AP. Find the value of 𝑥 and the terms.
Solution
Since the terms are consecutive, then
2 x − 1 − ( x + 1) = x + 5 − (2 x − 1) = d
 2x − 1 − x − 1 = d
x−2 = d (1)
x + 5 − ( 2 x − 1) = d
− x+6 = d (2)
Equating (1) and (2)
x − 2 = −x + 6
 2x = 8
x=4
 the terms are x + 1 = 4 + 1 = 5, 2 x − 1 = 2(4) − 1 = 7, x + 5 = 4 + 5 = 9

Example In an AP the 9𝑡ℎ term is −4 times the 4𝑡ℎ term and the sum of the 5𝑡ℎ and 7𝑡ℎ terms
is 9. Find the first and the common difference.

Solution
9th term, U 9 = a + 8d , 4𝑡ℎ term, U 4 = a + 3d
a + 8d = −4(a + 3d )
a + 8d = −4a − 12d
5a + 20d = a + 4d = 0 (1)
5th term, U 5 = a + 4d and 7𝑡ℎ term, U 7 = a + 6d
a + 4d + a + 6d = 9
2a + 10d = 9 (2)
Solving (1) and (2) simultaneously
55
 2  (2) : 4a + 20d = 18 (3)
(3) − 1 : 5a + 20d = 0
−a = 18
a = −18
Substituting a = −18 in (2)
2(−18) + 10d = 9
10d = 9 + 36
10d = 45
45 9 1
d= = =4
10 2 2
1
 first term, a = −18 and common difference d = 4
2
Example
What is the sum of all multiples of 7 or 11 less than1000?

Solution
The arithmetic series are
7 + 14 + 21 +  + 994
11 + 22 + 33 +  + 990
However, the multiples of 77 are included in both sequences, to avoid repetition; we
subtract it after adding the two sequences. The arithmetic series of multiples of 77 is
77 + 154 +  + 924.
994 990 924
To get the number of terms in each sequence, = 142, = 90, = 12
7 11 77
Using S n = (a + l )
n
2
The sum required is
142
(7 + 994) + 90 (11 + 990) − 12 (77 + 924)
2 2 2
= 1001(71 + 45 − 6)
= 1001 110
= 110110

Example
Given that the ratio of the 18𝑡ℎ term to the 6𝑡ℎ term in an AP is 3: 1, calculate the ratio of the
sum of the first 18 terms to the sum of the first 6 terms.

Solution
a + 17d
Given =3
a + 5d
a + 17d = 3a + 15d
a=d
Ration of the sum of the first 18 terms and that of the first 6 terms is given as

56
 S18
18
(2a + 17d ) 3(2a + 17d )
= 2 =
2a + 5d
S6 6
(2a + 5d )
2
But 𝑎 = 𝑑
S18 3(2a + 17a ) 3  19a 57
= = =
S6 2a + 5a 7a 7
Hence the ratio is 57: 7.

Practice Exercise 3.3

1. The sum of three numbers in an arithmetic progression is 18 and the sum of their squares
is 206. Find the numbers.
2. The first term of an arithmetic series is 7, the last is 70 and the sum is 385. Find the
number of terms in the series and common difference.
3. The first term of an AP is 35 and the last of 15 terms is − 109. Find the sum of these
terms.
4. How many terms are there in the AP − 12, − 7, , 133? Find the sum of the last 20
terms.
5. Find the first three terms of the AP, whose 7𝑡ℎ term is (3a + 5b) and 18𝑡ℎ terms is
19(2b − a ). .
6. What is the number of terms in the arithmetic sequence
− 1994, − 1992, − 1990,  , 1992, 1994?
7. The sum of 𝑛 terms of an arithmetic series is 5n 2 − 11n for all values of 𝑛. Determine the
common difference.
8. S n = 4n 2 + 1 represents the sum of the first 𝑛 terms of a particular series. Find the 10 term
of the series.
9. Find the fifth term of an AP, when:
i. The 4𝑡ℎ term is 5 and the sum of the first 10 terms is 70.
ii. The 3𝑟𝑑 term is (3x + 6) and the sum of the first 8 terms is 4(8x + 21)
10. The sum of 25 consecutive integers is 500. Determine the smallest of the 25 integers.
11. A drilling company had a contract of drilling a hole of 120𝑚. The cost of boring the first
metre is N 8,000.00 with an increase of N 3000.00 for more metre. Calculate the cost of
completing the contract.
12. A man invested $50 in a deposit account for his first child which compounded interest is
3% annually, and he continued to invest $50 in the same account on each of the child`s
subsequent birthdays. Find the amount in the account just before the 21𝑠𝑡 birthday, to the
nearest $1.
13. The sum of an arithmetic series is 100 times its first term, while the last term is 9 times
the first term. Calculate the number of terms in the series if the first term is not equal to
zero.
14. The sum of the first 𝑛 terms of a sequence is S n = 3n − 1 where n is a positive integer.

57
 i. If t n represents the 𝑛𝑡ℎ term of the sequence, determine t1 , t 2 , t 3
t n +1
ii. Prove that is constant for all values of 𝑛.
tn
3 f ( n) + 1
15. If 𝑓 is a function such that f (1) = 2 and f (n + 1) = for 𝑛 = 1,2,3, ⋯ What is the
3
value of 𝑓(100)?
16. If 𝑎 and 𝑟 are both positive, prove that the series log a + log ar + log ar 2 +  + log ar n −1 is
an arithmetic series and find the sum of the terms.
 n −3
17. Consider the sequence t1 = 1, t 2 = −1 and t n =  t n − 2 where 𝑛 ≥ 3. What is the
 n −1 
value of t1998 .
18. For the family of lines with equations of the form px + qy = r and which all pass through
the point (− 1, 2) , prove that 𝑝, 𝑞, and 𝑟 are consecutive terms of an arithmetic sequence.
19. Given that 𝑝, 𝑞 and 𝑟 are successive terms in an arithmetic sequence, calculate the value
of 𝑥 if (q − r ) x 2 + (r − p) x + ( p − q) = 0

3.4 Geometric Sequence And Series

Definition: A sequence is said to be geometric if the consecutive terms of the sequence
are all in the same ratio. For example, the sequence 27,9,3,1,  is geometric and the ratio
9 3 1
of any pair of the consecutive terms is = = . Examples of geometric sequence
27 9 3
(a) -32, 16, − 8, 4,
(b) 1, 3, 9, 27,
(c) 3, 12, 48, 192,
(d) 5, 5 / 4, 5 / 16, 5 / 64,
A geometric sequence can be generated if the first term (denoted by 𝑎) and the common
U
ratio i.e ratio of consecutive terms, denoted by 𝑟 are defined. In general, r = n . Hence
U n −1
𝑡ℎ
if 𝑈𝑛 is the 𝑛 term of a geometric sequence, then
U
U 1 = a, r = 2
U1
U 2 = U 1 r = ar
U 3 = U 2 r = ar.r = ar 2
U 4 = U 3 r = ar 2 .r = ar 3

U n = U n −1 r = ar n −1
Therefore the 𝑛𝑡ℎ term of a geometric sequence is given as
U n = ar n−1 (6)
58
 The geometric mean G of the 𝑛 numbers a1 , a 2 , a3 , , a n is the 𝑛𝑡ℎ root of the product
1

of the numbers, i.e. (a1 a 2 a3  a n ) . The geometric mean of 𝑝 and 𝑞 is given by

n

G = pq and a, ac, ac 2 is 3 a 3b 3 = ac . The sum of the terms of a geometric sequence

results to a geometric series, which in general we can write
S n = a + ar + ar 2 +  + ar n−1 (7)
If we multiply 𝑟 to (7), we have
rSn = ar + ar 2 +  + ar n−1 + ar n (8)
On subtracting (8) from (7), all term cancel except the first and the last
S n − rSn = a − ar n
(
S n (1 − r ) = a 1 − r n )
a(1 − r n )
Sn =
1− r
Subtracting (7) from (8), we have
rSn − S n = ar n − a
S n (r − 1) = a(r n − 1)
a(r n − 1)
Sn =
r −1
a(1 − r n ) a(r n − 1)
Sn = , is used when r n < 1 <>and S n = , is used when r n > 1
1− r r −1 .

Example
Three consecutive terms of a geometric sequence sums up to 37 and their products is 1728, find
the terms.

Solution
a
Let the terms be ,a,ar
r
a
  a  ar = 1728
r
 a 3 = 1728  a = 3 1728 = 12
a
Also + a + ar = 37
r
12
 + 12 + 12r = 37
r
12 + 12r + 12r 2 = 37r
12r 2 − 25r + 12 = 0
 ( 4r-3 )( 3r-4 )=0
3 4
r= or
4 3
12 4
Therefore, the required terms are , 12, 12  = 9, 12, 16 , the first 3 terms of the
4 3
3
59
 progression.

Example
1
The fifth term of an exponential sequence is greater than the fourth term by 13 2, and the fourth
term is greater than the third by 9. Find the common ratio and the first term.

Solution
Interpreting the statement mathematically,
1
First statement: ar 4 − ar 3 = 13
3
27
 ar 3 (r − 1) = (1)
2
Second statement: ar 3 − ar 2 = 9
 ar 2 (r − 1) = 9 (2)
Dividing (1) by (2)
ar 3 (r − 1) 27 1
= 
ar 2 (r − 1) 2 9
3
r =
2
3
Substituting  r = in (2)
2
9 9 8
a= 2 = = 9 = 8
r (r − 1)  3   3 
2
9
   − 1
2 2 
3
∴ the common ratio, 𝑟 = 2 and the first term, 𝑎 = 8.

Example
If N 1000 is invested on Dangote Sugar at compound interest of 8% per annum, determine (a)
the value after ten years, (b) the time, correct to the nearest year, it takes to reach more
than N 3000.

Solution
(a) Let the GP be a, ar, ar 2 ,, ar n−1

The first term 𝑎 = N 1000

8
The common ratio r = 1 + = 1.08
10
Hence the 2𝑛𝑑 term is
𝑎𝑟 = (1000)(1.08) = N 1008,
which is the value after one year,
the 3𝑟𝑑 term is
𝑎𝑟 2 = (1000)(1.08)2 = N 1166.4,
Thus the value after ten years
𝑎𝑟 10 = (1000)(1.08)10 = N 𝟐𝟏𝟓𝟖. 𝟗
(b) When N 3000 has been reached, 3000 = 𝑎𝑟 𝑛
60
 i.e. 3000 = 1000(1.08)𝑛
3 = (1.08)𝑛
Taking logarithms to base 10 of both sides gives:
log3 = log(1.08) = n log(1.08) by the laws of logarithms
n

log 3
n = = 14.3
log(1.08)
Hence it will take 15 years to reach more than N 3000.

Example
The first term of a geometric series is 3, the last term 768. If the sum of the terms is 1533, find
the common ratio and the number of terms

Solution
a = 3, U n = ar n −1
The last term will be, ar n −1 = 768
3r n −1 = 768  r n −1 = 256
r n = 256r (1)
a (r − 1)
n
Sn = = 1533
r −1
( )
 3 r n − 1 = 1533(r − 1)
r − 1 = 511(r − 1) = 511r − 511
n

r n = 511r − 510 (2)

Equating (1) and (2), we have
256r = 511r − 510
255r = 510
r = 2
Substituting 𝑟 = 2 in (1)
2 n = 256  2 = 512
 2 n = 29
Taking logarithm
n = 9
∴ Common ratio is 2 and there are 9 terms in the sequence.

Example
Three consecutive terms of a G.P are the third, sixth and tenth terms of an A.P. Find (𝑖) the
47
common ratio of the G.P, (𝑖𝑖) If the fourth term of the G.P is 1 81, find the sum of the
first 11 terms of the sequence.

Solution
(i) The consecutive terms of the G.P are 𝑎 + 2𝑑, 𝑎 + 5𝑑, 𝑎 + 9𝑑

61
 a + 5d a + 9 d
r = =
a + 2 d a + 5d
a 2 + 10ad + 25d 2 = a 2 + 11ad + 18d 2
7 d 2 − ad = 0  a = 7 d
7 d + 5d 12d 4
r = = =
7 d + 2d 9d 3
47
U 4 = ar 3 = 1
81
(ii) 3
4 128
a  =
3 81
128 27 2
a=  =
81 64 3

Therefore S n =
(
a 1− rn )
, since r < 1
1− r
2   4  
11

1−  
3   3  
S11 = = −45.4
2
1−
3

3.5 INFINITE GEOMETRIC SERIES

The general geometric series a + ar + ar 2 +  is given as

a(1 − r n ) a a n
Sn = = − r
1− r 1− r 1− r
The first term is independent of 𝑛 and the second term has a factor 𝑟 𝑛 , now if r < 1 , 𝑟 𝑛
becomes smaller and smaller as 𝑛 increases. Hence as 𝑛 increases, 𝑆𝑛 approaches the
a
limiting value since 𝑟 𝑛 approaches zero. Therefore the sum to infinity of a
1− r
geometric series if r < 1 is given as
a
lim S n = S  =
n → 1− r

Example
1 1 1 1
To what sum does the series , , , ,  converges?
5 25 125 625
Solution
1 1
First term, a = , r = < 1 . The series converges to
5 5

62
 1
1
S = 5 =
1 4
1−
5
Example
Express the following recurring fraction as a proper fraction (𝑎) 0.24 (𝑏) 0.666

Solution
(𝑎) The series can be written as
24 24 24
0.24 = + + +
100 1000 10000
24 1
Hence, a = ,r =
100 10
24
a 24 10 4
S = = 100 =  =
1− r 1 100 9 15
1−
10
6 6 6
(b) 0.666 = + + +
10 100 1000
6 1
a = ,r = <1
10 10
6
2
S  = 10 =
1 3
1−
10
Example
2
x  x 
For what values of 𝑥 do 1 + +  +  converges and to what?
1+ x 1+ x 
Solution
x x
a = 1, r = . The series will thus converge if < 1; x < 1 + x ; thus x <1 + x
1+ x 1+ x

which satisfies the condition x < 1 or x > −(1 + x ) i.e. x > − , therefore the series
1
2
1
converges within − 2 < 𝑥 < 1.
1
S = = 1+ x
x
1−
1+ x

Practice Exercise 3.4

1. The 3rd term of a geometric sequence is −1 , the 7𝑡ℎ term is − 81. Find the
term.
2. The second term of a geometric sequence is 24 , the 5th term is 81. Find the
seventh term.

63
3. Express 0.45 recurring as a fraction.
1
4. Insert four terms between 8 and 4 so that the six terms form a geometric sequence.
5. By how much does the sum of the first 59 even positive integers exceed the sum
of the first 59 odd integers?
1
6. Find (𝑖) the sum 𝑆𝑛 of the first 𝑛 terms of the geometric sequence 4, − 2, 1, −
2
(𝑖𝑖) the sum of the series as 𝑛 → ∞.
7. What is the common ratio of the following geometric sequence
( ) (
3 + 2 + 1+ 3 + )
8. In a geometric series, t 5 + t 7 = 1500 and t11 + t13 = 187500. Find all possible
values for the first three terms.
9. The sum of the first 𝑛 terms of a geometric series is 127 and the sum of their
127
reciprocals is . The first term is 1. Find 𝑛 and the common ratio.
64
10. The first and last terms of a geometric series are 2 and 2048 respectively. The
sum of the series is 2730. Find the number of terms and the common ratio.
3x 3x
11. For what values of 𝑥 does the series 3 x + + +  converges, and to
1 + x (1 + x )2
what does it converge?
12. Determine the range, or ranges of values of 𝑥 for which
2 3
7  7   7 
1+ +  +  +  converges and to what?
3x − 5  3x − 5   3x − 5 
13. The first term of a geometric series is 18 and the sum to infinity is 20. Find the
common ratio and the sum of the first six terms.
14. A loan N 15000 is made with interest at 5% p.a. it is repaid by annual payments
of N 2000, starting at the end of the first year. Find the amount outstanding, to the
nearest N 100, at the end of 8𝑡ℎ year.
15. A ball when dropped from any given height loses 20 per cent at each rebound. If
it is dropped from a height of 80 meters. Find how often it will rise to height of
over 16 metres. How far does the ball travel before coming to rest?
16. If the value of a machine, originally N 4000, depreciates each year by 15% of its
value at the beginning of the year, find its value after 3 years, to the nearest
naira.
17. A shrub of height 210𝑐𝑚 is planted. At the end of the first year, the shrub is
220𝑐𝑚 tall. Thereafter, the growth of the shrub each year is half of its growth in
the previous year. Show that the height of the shrub will never exceed 230𝑐𝑚.
18. A basketball is dropped vertically. It reaches a height of 2 m on the first bounce.
The height of each subsequent bounce is 90% of the previous bounce.
a. What height does it reach on the 8th bounce?
b. What is the total vertical distance travelled by the ball between the first and
sixth time the ball hits the ground?
64
19. A National Lottery is offering prizes in a new competition. The winner may
choose one of the following options.
Option one: $1000 each week for 10 weeks.
Option two: $250 in the first week, $450 in the second week, $650 in the third
week, increasing by $200 each week for a total of 10 weeks.
Option three: $10 in the first week, $20 in the second week, $40 in the third
week continuing to double for a total of 10 weeks.
a) Calculate the amount you receive in the tenth week, if you select
(i) option two;
(ii) option three.
(b) What is the total amount you receive if you select option two?
(c) Which option has the greatest total value? Justify your answer by showing
all appropriate calculations.

65
 CHAPTER 4
Permutation, Combination and Binomial theorem
4.1 Permutations

The arrangement of n objects in a definite order is called the permutation of the n objects
For instance, there are four books in a shelf, in how many ways could they be arranged in order?
If we label the books abcd . If we arrange with "a " being the first, we have
abcd abdc acbd
acbd adbc adcb
This gives us six different arrangements with book "a " coming first. Similarly, book with "b "
first, "c " first and, "d " first, giving us in all 24 different ways of the four books.
Also two letters ab may be written in two different ways in a line as ab and ba . Each of these
two arrangements is a permutation of two letters. In like manner, we can arrange three letters
abc in a number of different ways.
If we put "a " in the first place, the other letters may be written in the order bc or cb , so that we
have two permutations abc and acb in which "a " leads. Similarly, putting "b " in the first
place we have two orders to be ac or ca , so that we have two permutations bac and bca in
which "b " leads, by the same reasoning we can have cab and cba as two permutations in which
"c " leads.

Therefore, the required number of permutations of n different things taking r at a time is

denoted by n Pr and is defined as follows.
n!
n
Pr = ,r  n
(n − r )!
When n = r we have,
n! n!
n
Pr = n Pn = =
(n − n )! 0!
We take 0!= 1, so that
n!
n
Pn = = n!
1
The expression n! is called factorial n and is defined as
n!= n(n − 1)(n − 2)...
Here, it works only when it is required to find the number of permutations of unlike objects.
Example 4.1.1
How many different arrangements of the letters COME are possible?
Solution.
Hence, n = 4 and so the different permutations are 4!
4!= 4(4 − 1)(4 − 2)(4 − 3)
= 4.3.2.1
= 24 different ways

66
Permutations of Different Items
Let n be the number of objects of which p are all alike and the renaming n − p are different
from one another. It is required to find the number N of permutations of all the n objects. This
number is given by
n!
N=
p!
In general, the total number of permutations of n objects where there are p of one type and q
of another type and r of another type and so on is given by
n!
N=
pq!r!
Example 4.1.2
In how many ways can the letters of the word BETTER be arranged?
Solution.
Two letters E and T appear twice and so n = 6
6!
N=
2!2!
= 180 .

Example
How many different numbers of 5 digits can be made using the digits 2,2,3,3,3 .

Solution
If the 5 digits were all different, we call for 5! number but 2' s can be permuted in 2! ways and the
3' s in 3! ways. So that we have
5! 5.4.3.2.2
= = 10
2!3! 2.1.3.2,2

Conditional permutation
Example
Suppose the letters of the word SHALLOW are to be arranged. Find the number of ways in
which the letters are arranged if
(a ) The 2 L' s must not come together
(b ) The 2 L' s must come together
Solution.
Assume we missed out the 2 L' s , the letter SHAOW can be arranged in 5! ways. Hence, giving
us  S  H  A  O  W .The first L can be inserted in any of the 6 places. When this is done,
there are 5 possible places for the second L not next to the first L .So, we have 5!.6.5 if and
5!.6.5
only if the 2 L' s are not identical. If they are identical, we have
2!
(b ) Considering the 2 L' s one letter, we get SHALOW hence leading to
6!= 720

Example
There are 20 books on a shelf but the red covers of 2 clash, and they must not be put together.
In how many ways can the books be arranged?
67
Solution
Suppose the 2 red books are tied together. There are 19 books which can be arranged in 19!
ways. If the order of the 2 books is reversed, there will again 19! arrangements. Thus, there are
2.19! ways of arranging the books with red ones next to each order.
When there is no restriction, the 20 books can be arranged in 20! ways. Therefore, the number of
arrangements in which the red books are not together will be
20−! 2.19!
= 20.19!−2.19!
= 19!(20 − 2)
= 18.19!

Practice Exercise 4.1

(1) Find the number of possible permutation of the letters of the word “MARKETING”
(2) Find the number of arrangements using any three letters of the word “MARKETING”
(3) A man who works a five-day week, can travel to work on foot, by circle or by bus. In how
many ways can be arrange a week travelling to work
(4) Seven boys and two girls are to sit together on a bench. In how many ways can they arrange
themselves so that the girls do not sit next to each other?
(5) In how many ways can five different books be arranged in order?
(6) Show that, 90! = 90
8!

(7 ) Find the values of 12!

10!2!

4.2 Combinations
When a selection of objects is made without regard to order. It is referred to as a combination.
Thus, abc, acb and cba are different permutations but they are the same combination of letters.
In how many ways can r object be chosen from n unlike objects? This question can be solved
by recalling that the member of permutations of r objects can be arranged in r! ways.
Therefore, the number of combination of r objects chosen from n unlike objects, denoted by
n
C r is
n!
n
Cr =
(n − r )!r!
In permutation, order is important, abc, is different from cab, bac and cba , and
so on. In combination, order is irrelevant, abc, acb etc. are the same selection. Each set of three
can be permuted in 3! ways but these count as 1 combination. Hence,
n
P
n
Cr = r
r!
n(n − 1)(n − 2 )...(n − r + 1)
= .
r!
n
C1 = n , as this is the number of ways of solving 1 from n objects

68
 n(n − 1) n n(n − 1)(n − 2)
n
C2 = , C3 =
2! 3!
n
To find the value of Pr , i.e.the number of permutations of "n" objects taken "r" at a time ( n
and r must be positive integers with n  r ). It should be clear now that
n
Pr = n(n − 1)(n − 2)... to r factors
= n(n − 1)(n − 2)...(n − r + 1)
If r = n, then
n
Pr = n Pn
= n(n − 1)(n − 2)..... 2.1
= n!
Given that
n
Cr = x
n
Pr
x=
r!
xr!= n Pr
n
Pr
r! =
x
Corollary
n(n − 1)(n − 2 )...(n − r + 1)
n
Cr = .
r!
Let multiply both be r; numerator and denominator by (n − r )! and we get
n(n − 1)(n − 2)...(n − r + 1)(n − r )(n − r − 1).... 2.1
=
r!(n − r )(n − r − 1)..... 2.1
n!
=
r!(n − r )!
The result is very important and will be useful when we shall be discussing Binominal Theorem.

Corollary
n
C n is the number of ways of choosing n objects from n different things and since this can only
be done in one way by taking all the objects, it must follow that
n
Cn = 1 .

We have demonstrated that

n!
n
Cr =
r!(n − r )!
n! 1
n
Cn = =
r!(n − n )! 0!
Remember also that we have shown that
1
n
Cn =
n
Cn = 1 , and 0! .
1
Therefore, 0! = 1. That is, 0! = 1.

69
Theorem
n
C n−r = n C r
n!
n
C n−r =
(n − r )!n − (n − r )
To prove this theorem, we remove brackets to have
n!
(n − r )!r!
Hence,
n
C n−r = C r
n

We have established that

n
C n − r = C r and n C 0 = .1
n

Therefore,
n
C n −0 = C 0 = C n = 1 .
n n

Example
Nine people are going to travel in two taxis. The larger has five seats and the smaller has four. In
how many two ways can the party be split up?
Solution
The number of ways of choosing 5 people from 9 people is 9 C 5 .The remaining 4 people can be
chosen from 4 people is 4 C 4 way to find the second taxi
: The total number of ways of splitting the party is
9! 4!
9
C 5 .9 C 5 = . = 126 .
4!5! 0!4!

Example
Committee of 4 is to be chosen from 5 men and 3 women. In how many ways can this be done
so that the committee contains
(a) At least one man?
(b) At least one of each Sex?

Solution

(a ) In this case, the committee possibility are

1 man and 3 women
Or 2 men and 2 women
Or 3 men and 1 woman
Or 4 men and 0 woman
5
C1 . C 3 = 5.1 = 5
3

5.4 3.2
5
C 2 .3 C 2 = . = 30
2 2
5.4.3 3
5
C3 .3 C1 = . = 30
3.2 1
70
 5.4.3.2!
5
C 4 .3 C 0 = .1 = 5
4.3.2!
Therefore, total number of ways becomes
5 + 30 + 30 + 5
= 70

(b ) . The number of ways is 1st ,2 nd and 3 rd

, since at least one number of each sex is required
means that four men cannot be chosen, hence we have

= +6530 + 30 ways.
5

Example
A 5 member Senate of an Institution is to be form from 7 males and 6 females with a condition
that it must include at least 3 and at most 4 males. In how many ways can such a Senate be
selected?

Solution
The 5 member Senate must contain 3 or 4 males. Now, the 5 member Senate that contains
exactly 3 males out of 7 will contain exactly 2 females out of 6 , so it can be formed in
7
C 3 .6 C 2 ways
Again, the 5 member Senate with 4 males out of 7 contain 1 female out of 6 , and hence can be
selected in
7
C 4 .6 C1 ways
The required composition of the Senate becomes
7
C3 .6 C2 + 7 C4 .6 C1
= 35.15 + 35.6
= 525 + 210 = 735 ways.

Example
A dramatic group of 8 men and 4 women has to be constituted from a list of 10 men 7 women.
In how many ways can the group be formed.

Solution
10
C 8 .7 C 4
= 45.35
= 1525 ways.

Practice Exercise 4.2

(1) In how many ways can the 1st ,2 nd and 3 rd prizes be awarded in a race if there are 10
competitors
(2) How many numbers greater than 100 can be formed using the digits 1,2,3,4,5, if no digit
may be repeated
(3) There are 5 buses travelling between Onitsha and Asaba. In how many ways can a
businessman travel from Onitsha to Asaba, by one bus and return by a different bus

71
(4) Interview indicates that all the 4 Mathematics students, 5 Physics student and 7 Chemistry
students who applied for a scholarship in their respective disciplines qualified for an award. In
how many ways the award can be made if,
(a ) Only one scholarship is available in each of the disciplines
(b ) Only two scholarships are available in each of the disciplines
(5) Prove that,
C r +1 = n C r + n C r +1
n +1

(6) Prove that,

C r = ( n −1C r -1 )
n n
r
Evaluate the following
8
(7 ) 5C 4
C3

(8) 1 ( 7 C 4 )
20
(9) 5!( 7 C 5 )( 5C 3 )

4.3 The Binomial Theorem

In this section, our concern is to show how higher powers of (x + y ) are expanded with ease. This
is due to difficulty encountered by students. Now, consider the expansions below
(x + y )2 = x 2 + 2 xy + y 2
(x + y )3 = (x + y )2 (x + y )
= (x 2 + 2 xy + y 2 )(x + y )
= x 3 + 3x 2 y + 3xy 2 + y 3
(x + y )4 = x3 + 3x2 y + 3xy2 + y3 (x + y )
= x 4 + 4 x 3 y + 6 x 2 y 2 + 4 xy 3 + y 4

Suppose we write the coefficients alone, we have the following table of coefficients:
1 1
121
1331
14641

Following the table above, we may guess the next line and more importantly be able to see
how the table can be continued from the previous one
In a bid to construct the table of coefficients, the last three lines of the long multiplication are
Written, leaving out the letters to yield as fellows
(x + y )0 = 1
(x + y )1 = 11
( x + y )2 = 11
72
 = 121
(x + y )3 = 121
121
= 1331
(x + y )4 = 1331
1331
= 14641

(x + y )5 = 14641
14641
= 15101051
(x + y )6 = 15101051
15101051 = 1615201561
The table of coefficients may be written in triangle known as Pascal’s triangle named after the
man who propounded the formula as fellows

PASCAL’S TRIANGLE
1
11
121
1331
14641
15101051
1615201561

The Paschal triangle table is used to determine the coefficients, using (x + y ) as a standard
binomial
The expansion of (x + y )n , where n is a positive integer is of the form
(x + y )n = nC0 x n + nC1 x n−1 y+ nC2 x n−2 y 2 + ...+ nCr x n−r y r + ... + n Cn y n
n
=  n C r x n−r y r ,
r =0

where,
n!
n
Cr = .
(n − r )!r!
n(n − 1)x n − 2 y 2 n(n − 1)(n − 2)...(n − r + 1)x n − r y r
(x + y ) = x + nx y +
n n n −1
+ .... + ... + y n
2! r!
At this junction, it is necessary to remember the binomial theorem in the simple form as
( x + 1)n = nC0 x n + nC1 x n−1 + nC2 x n−2 + ...+ nCr x n−r + ... + n Cn
Or
(1 + x )n = nC0 + nC1 x + nC2 x 2 + ...+ nCr x r + ... + n Cn

Example
73
Expand (2x + 3 y )3 in descending power.

Solution
Here, a = 2x, b = 3 y
Thus, there will be four terms in the expansion involving,
(2x )3 , (2 x )2 (3 y ) , (2 x)(3 y )2 , (3 y )3
8x 3 , 12x 2 y , 18xy 2 , 27y 3 .

Their coefficients from Pascal’s triangle are

1331
Hence,
(2x + 3 y )3 = 8x 3 + 36x 2 y + 54xy 2 + 27y 3
Example
Evaluate (1.02 ) to three decimal places.
5

Solution
(1.02 )5 = (1+ 0.02)5
The required coefficients are
15101051
Therefore,
(1.02 )5 = (1 + 0.02)5 = 1 + 5(0.02) + 10(0.2)2 + 10(0.02)3 + 5(0.02)4 + (0.02)5
= 1 + 0.10 + 0.004 + 0.0008+ 0.0000016+ 0.0000000016
= 1.104816
 1.105

Example
Obtain the expansion of (1 + x − 2 x 2 ) as far as the term in x 3
8

Solution
(1 + x − 2 x )
2 8
= (1 + (x − 2 x 2 ))
8

Invoking the theorem, we have

(1 + x − 2 x ) = 1+ C (1) (x − 2x ) + C (1) (x − 2x ) + C (1) (x − 2x )
2 8 8
1
7 2 8
2
6 2 2 8
3
5 2 3
+ ....
= 1 + 8(x − 2 x ) + 28(x − 2 x ) + 56(x − 2 x ) + ....
2 2 2 2 3

= 1 + 8 x − 16x + 28(x − 2 x ) + 56(x − 2 x ) + ....

2 2 2 2 3

= 1 + 8 x − 16x + 28(x − 4 x + 4 x ) + 56(x − 6 x ) + ...

2 2 3 4 3 4

= 1 + 8 x − 16x 2 + 28x 2 − 112x 3 − 112x 4 + 56x 3 − 336x 4 + .

Hence, the required expansion of (1 + x − 2 x 2 ) becomes
8

1 + 8 x + 12x 2 − 56x 3 .

Example

74
 4
 1 
Expand  x − 2 
 x 

Solution
 1 
4
  1  
4
4 1 
4

 x − 2  =  x1 − 3   = x 1 − 3 
 x    x    x 
Employing the theorem,
(1 + x )n = nC0 + nC1 x + nC2 x 2 + ...+ nCr x r + ... + n Cn ,
Leads to
4 2 3 4
 1   1   1   1   1 
1 − 3  = 1 + 4 − 3  + 6 − 3  + 4 − 3  +  − 3 
 x   x   x   x   x 
4 6 4 1
= 1 − 3 + 6 − 9 + 12
x x x x
Hence,
4
 1  4 6 4 1 
x 4 1 − 3  = x 4 1 − 3 + 6 − 9 + 12  ,
 x   x x x x 
4 4 4 4
4x 6x 4x x
= x 4 − 3 + 6 − 9 + 12 , implying that
x x x x
4
 1 
 x − 2  = x − 4x + 6x − 4x + x
4 2 5 8

 x 

Example
Find the 5 th term in the expansion of (x + 2 y 3 )
17

Solution
The (r + 1) in the expansion of (x + 2 y 3 ) is n C r x n−r y r
th 17

Here, r = 4, since r + 1 = 5
Then on substitution we have
17
( ) 4
C4 x17 − 4 2 y 3 =17C4 x1316 y12

But,
17.16.15.14
17
C4 =
4!
= 2380
Therefore, the required coefficient on simplification is
2380x1316 y 12

4.4 Binomial theorem for negative and fractional exponent

If n is negative or fraction, then
(x + y )n = x n + nx n−1 y + n(n − 1)x y + .... + n(n − 1)(n − 2)(n − 3)...(n − r − 1)x y ... + y n
n−2 2 n−r r

2! r!
75
Example
Find the first four terms of (1 − x ) 5 . Hence, find the value of (1.2 ) 5 , correct to 5 places of decimal
1 1

Solution
(1 + x )n = 1 + nx + n(n − 1)x + n(n − 1)(n − 2)x
2 3

2! 3!
1
Here, x = − x, n = and substituting, we get
5
2
11  1  1  1 
 − 1  (− x )  − 1 − 2 (− x )
3

(1 − x )n = 1 + 1 (− x ) + 5  5  + 
5 5  5 
5 2! 3!
1 2 2 6 3
= 1− x − x − x
5 25 125
Now,
(1.2)5 = (1 + 0.2)5 16.656
1 1

Hence
(1.2 ) 5 1
(0.2) − 2 (0.2)2 − 6 (0.2)3
1
= 1−
5 25 125
= 1.03718.

Example
Evaluate 16.656 using Binomial theorem, correct to 3 decimal places.

Solution
16.656 = (16.656) 2
1

= (16 + 0.656) 2
1

1
  0.656  2
= 161 + 
  16 
1
1
 0.656  2
= 16 1 +
2

 16 
= 4(1 + 0.041) 2
1

Invoking the Binominal theorem yields

(1 + x )n = 1 + nx + n(n − 1)x + n(n − 1)(n − 2)x
2 3

2! 3!
1
Here, x = 0.041, n = and substituting, we get
2
2
11  1  1  1 
 − 1(0.041)  − 1 − 2 (0.041)
3

(1 + 0.041) 2 = 1 + 1 (0.041) + 2  2  + 
2 2  2 
1

2 2! 3!

76
= 1 + 0.041 − (0.041) + (0.041)
1 1 2 1 3

5 8 16
= 1 + 0.0205− 0.000210125+ 0.000004307
But,
4(1 + 0.041) 2 = 4 + 0.0830 − 0.0008405+ ...
1

= 4.0811595+ ...
= 4.081

Example
(a ) Find p and q if the coefficient of x and x 3 in the expansion of (1 + px + qx2 − 4x 3 )(1 + x)6 are
both zero
(b ) Show that if x is so small that x 3 and higher powers of x can be neglected,
1
 (1 + x )  2 1 1 181
  = 1 + x − x 2 . Hence, by putting x = .Show that 5= approximately
 (1 − x )  2 9 81

Solution
Expanding (1 + x ) leads to
6

(1 + x )6 = 6C0 x 0 + 6C1 x1 + 6C2 x 2 + 6C3 x 3 + C4 x 4 + 6C5 x 5 + C6 x 6

= 1 + 6 x1 + 15x 2 + 20x 3 + .......... ........
Then, (1 + px + qx2 − 4 x 3 )(1 + x ) becomes
(1 + px + qx 2 − 4 x 3 )(1 + 6 x1 + 15x 2 + 20x 3 + .......... ........ )
On expansion we have that
1 + px + qx 2 + 4 x 3 + 6 x + 6 px 2 + 60x 3 + 24x 4 + 15x 2 + 15 px3 + 15qx 4 + 20x 3 + 20 px 4
= 1 + px + 6 x + qx 2 + 6 px 2 + 15x 2 + 4 x 3 + 60x 3 + 24x 4 + 15 px 3 + 20x 3
= 1 + ( p + 6)x + (q + 6 p + 15)x 2 + (4 + 6q + 15 p + 20)x 3
p + 6 = 0, p = -6 and
4 + 6q + 15 p + 20 = 0, 24 + 6q + 15(− 6) = 0, 6q = 66, q = 11,
1

(b )  (1 + x )  = (1 + x ) 2 (1 − x )− 2
2 1 1

 (1 − x ) 
Recall that
(1 + x )−n = 1 + nx + n(n − 1)x + n(n − 1)(n − 2)x .
2 3

2! 3!
Then.
2
11  1  1  1  3
 − 1(x )  − 1 − 2 (x )
(1 + x ) 2 = 1 + 1 x + 2  2  + 2  2  2 
1

2 2! 3!
1 1 1 3
= 1− x − x2 − x + .......... ....
2 8 64
Similarly,
(1 − x )− 2 = 1 − 1 x − 1 x 2 − 1 x 3 + .......... ...
1

2 8 64
77
Then,
1
 (1 + x )  2  1 1 1 3  1 1 1 3 
  = 1 − x − x 2 − x + .......... ...  1 − x − x 2 − x + ........ 
 (1 − x )   2 8 64  2 8 64 
(1 + x ) 2 (1 − x )− 2 = 1 − 1 x − 3 x 2 − 1 x + 3 x 3 − 1 x 2 − 1 x 3 − 3 x 4
1 1

2 8 2 16 8 16 64
1 2
= 1− x + x
2
1
Hence, putting, x = , we get
9
1
 1   2
 1 + 9  
  = 4 = 2
 1   5 5
 1 − 9  
 
2 1
= 1− x + x2
5 2
2 1 1
= 1− +
5 5 162
2 145
=
5 162
5 162
=
2 145
Hence,
324 181
5= =
145 81

Example
When (1 + ax ) is expanded in ascending powers of x , the first three terms are 1 − 10x + 40x 2 ,
n

find n and a .

Solution
(1 + ax )n = n C0 (1)n (ax)0 + nC1 (1)n−1 (ax)1 + nC2 (1)n−2 (ax)2 + .......... ....
=
n!
(1) + n! (1)(ax) + n! (1)(ax)2 + .......... ....
0!n! (n − 1)!1! (n − 2)!2!
n(n − 1)! n(n − 1(n − 2))!
= 1+ ax + (ax)2 + .......... ....
(n − 1)! (n − 2)!2!
= 1 + nax +
(n 2
− n)
(ax)2 + .......... ....
2
Hence,
1 + nax +
(n
− n) 2
(ax)2 = 1 − 10x + 40x 2
2
Equating the coefficient yields
78
na = 10
and
(n 2 − n) a 2 = 40
2
Simplifying, we get
2
 10  10
  −
n = , and  
10 a a 2
a = 40
a 2
Further simplification gives
100a − 10a = 80 , a = 2
But,
10
n= .
a
Substituting for a gives
n=5

Example

Use the binominal theorem to find the values of

(a ) (9 − 18x ) 2
1

(b ) 2x
,
2+ x

and state the range of x for which the value is valid.

Solution
(9 − 18x ) 2 = 9(1 − 2 x )2
1 1

(a ) = 3(1 − 2 x ) 2
1

But,
n(n − 1)x 2 n(n − 1)(n − 2)x 3
(1 + x ) = 1 + nx +
n
+ , n<1
2! 3!
11  1  1  1 
1
(− 2 x ) 2  2 − 1(− 2 x ) 2  2 − 1 2 − 2 (− 2 x )
2 3

(1 − 2 x ) 2 = 1 + 2 +   +   
1

1! 2! 3!
2 3
x x
=1− x − −
2 3
So that
3 3
3(1 − 2 x ) 2 = 3 − 3x − x 2 − x 3
1

2 2
The expansion (1 − 2 x )2 is valid if
1

− 2x  1 ,
1
2 x  1 , implies that x 
2
79
The range of values of x for which the expansion is valid is
1 1
− x
2 2
(b ) 2 x = 2x (2 + x )−1
2+ x
−1
 x
= x1 + 
 2
n(n − 1)x 2 n(n − 1)(n − 2)x 3
(1 + x )n = 1 + nx + + , n<1
2! 3!
x
Here n = −1, x =
2
2 3

(− 1) x  (− 1)(− 2) x  (− 1)(− 2)(− 3) x 

= 1+ 2 + 2 + 2
1! 2 6
2 3
x x x
=1− + −
2 4 8
Hence,
2x  x x2 x3 
= x1 − + − 
2+ x  2 4 8 
x2 x2 x4
= x− + −
2 4 8
The range of values of x for which the expansion is valid is
x
 1.
2

80
 CHAPTER 5
Circular measure and Trigonometry

5.1 Measurement of angle

An angle measures the magnitude of rotation between two straight lines. One of the ancient units
of measurement of angle is the degree. It originated amongst the Greeks and Indians who divided
the angle in a circle into 360 equal parts with each part corresponding to angle of magnitude 1 .
Each degree was further divided into 60 equal parts called minutes and each minute divided into
60 seconds ( 60'' ), so that the relationship between these units becomes
60'' = 1'
60' or 3600'' = 1
A second and very useful unit for measurement of angles is the radian. This unit for measuring
angle relates the magnitude of an angle to the length of an arc. Consider a circle with centre O
and two points X and Y, on its circumference as shown in Fig 1

Y X

r r

Figure 1
As the line OX rotates into position OY, point X covers a distance equal to the length of the arc
l
XY. Let this length be l, the ratio of this length to the circumference of the circle is . Note
2 r
that the ratio of the angle (in radian) subtended by the arc at the centre to the angle in one
revolution equals the ratio of the arc length lto the circumference. [1] That is,
 l
= (1)
2 2 r
l
Thus, the measure of the angle (in radian) subtended by the arc XY is  = ; that is, the measure
r
of the angle subtended at the centre of a circle by a given arc is the ratio of the length of the arc
to the radius of the circle. In particular, when l = r , then  = 1 radian, so that one radian is seen as
the angle subtended at the centre of a circle by an arc whose length is equal to that of the radius
of the circle. It is easy to see from the relation above that for a unit circle (circle whose radius is
1), the measure of the angle XOY in radians is equal to the arc length l and the circumference is
2 units. With this results for a unit circle, we have the conversion formula as
360 = 2 radians
Or equivalently,
180 =  radians .
 180  
Hence, one radian is equal to   or approximately 57.3 while 1 = 180 radians. Therefore, any
  
angle measurement can be converted from degrees to radians and vice versa by multiplying the
81
  180  
given angle by   and 180 rad. respectively.
  
Example 1
Express the following angles in radian (a.) 75 , (b.) 103.15
Solution
a. 180 =  rad

 75 = ( 75 )
180
5
=  or 0.417
12
b. 180 =  rad

 103.15 = (103.15 )
180
= 0.573
Example 2
3
Express the following angles in degrees (a.) (b.) 0.425 rad.
5
Solution
a.  rad = 180
3 180  3 
=
5   5 
= 108
b.  rad = 180
180
0.425 rad. = ( 0.425 )

= 24.36

5.2 Arc lengths, sectors and segments

We now use the notion of angle measurement in radians to derive simple formulas for the arc
length and area of a sector of a circle. Consider a circle of radius r with an arc length l
subtending an angle  rad at the centre of the circle (see figure2..).
l

The formula for arc length is obtained from equation (1) as

l = r (2)

82
 Similarly, using the notion of ratio, we can obtain the formula for the area of a sector of circle
of radius r whose arc subtends an angle  rad. at the centre of the circle (see figure 3). The ratio
of the area A of the sector to the total area of the circle is equal to the ratio of the angle in the
sector  (in radians) to one revolution (i.e 2 rad ). That is;

A 
=
r 2
2
1 2
 A= r (3)
2
The area of a segment of a circle is easily obtained by subtracting the area of the triangle from
the area of the sector.

Example 3
A cylindrical storage drum 100cm long, lies with its axis parallel with the horizontal as shown
in the accompanying figure, and is partially filled with liquid to the level of thechord AB as
13
indicated in the figure. If the radius of the cross section is cm, the area of OAB is
4
2
3.7575 cm , and the central angle AOB is 2.35 rad. , find the volume of the liquid content,
correct to three significant figure. (Question 21 Exercise 5.1 in [2]).

o
A B

C
Solution
The required volume V is given by V = 100S cm3 , where S is the area of the cross section of
the liquid. To find S, we note that S is the area of the circular segment ABC, thus,
S = A− ,
whereA is the area of the sector OACB, and  = the area of OAB = 3.7575cm3.We apply
formula (3) to get
2
1 1  13 
A = r 2 =   ( 2.35) = 12.4109
2 2 4 
Therefore,
S = A −  = 8.6534 and V = 100S cm3 = 865.34cm3 865cm3

5.3 Orientation and Magnitude of angles

Consider a circle of radius say r whose centre is at the origin O, of the coordinate plane and a
radial line OB from the centre to a point B on the circumference of the circle intersecting the
positive x-axis. If the radial line rotates in an anti-clockwise direction to a position OA, it
generates a positive angle AOB (see figure 4). If the sense of rotation is clockwise, the angle
generated is taken to be negative. The orientation of an angle is usually specified by attaching a
negative (-) or positive (+) sign to its magnitude indicative of a clockwise or anti-clockwise
rotation respectively. The positive sign is usually omitted in the case of anti-clockwise
orientation, thus when an angle is specified without a sign attached, it is generated by anti-
clockwise rotation. See illustrations in figure 5…..

83
 y
y
y y

A
P

x x
O B O O
x O x

Figure 4 Figure 5.

We now define quadrants of a circle and magnitudes of angles in each quadrant. A quadrant is
one quarter of a circle, and quadrants are conventionally labelled going anti-clockwise as shown
in figure 6. Any angle  , such that 0    360 can be located in one of the quadrants
y

Quadrant Quadrant

Quadrant I: 0    90
II I

x
Quadrant II: 90    180
Quadrant
Quadrant III: 180    270
Quadrant

III IV

Quadrant IV: 270    360

Angles can also have magnitude greater than 360 (one complete revolution) depending on the
number of complete revolutions. For example, the angles 405 make one complete revolution
while 750 make two complete revolutions as illustrated below;

Practice Exercise 5.1

1. Express the following angles in terms of 𝜋 radians
a. 390𝑜 b. 75𝑜 c. 105𝑜 d. 173𝑜 e.
2. Convert each of the following angles from degrees to radians

5.4 Trigonometric Functions

For any acute angle  we can construct a right triangle with one of the angles being  . In this
triangle (as with every triangle) there are three sides. We will call these the adjacent (denoted by
adj which is the leg of the right triangle that forms one side of the angle  ), the opposite
(denoted by oppwhich is the leg of the right triangle that is opposite the angle  , i.e. does not
form a part of the angle) and the hypotenuse (denoted by hyp which is the longest side of the
right triangle). Let us consider the right-angled triangle ABC with an acute angle  .

84
 B

A C
Figure 1
The side opposite the right angle (that is AB ) is called the hypotenuse. The side opposite angle 
(that is BC ) is called the opposite while the remaining side (that is AC ) is known as the
adjacent. The standard trigonometric functions of  are defined thus:
opposite adjacent opposite
sin  = , cos  = , tan  = .
hypotenuse hypotenuse adjacent
Associated with these are the reciprocal ratios, cosecant, secant and cotangent which are given
as:
1 hypotenuse 1 hypotenuse 1 adjacent
cosec = = , sec = = , cot = = .
sin  opposite cos  adjacent tan  opposite
sin  = x   = sin −1 x
Note that: cos  = x   = cos −1 x
tan  = x   = tan −1 x

5.5 Solving right-angled triangles

Trigonometric ratios can be used to obtain the lengths and angles in right-angled triangles. We
consider three cases that can occur:
Case 1: Given the hypotenuse and an angle
We use sin or cos as appropriate
B

h
y

A C
x Figure 2
Assume h and  in figure 2 are given, then
y
sin  = which gives y = h sin 
h
x
Similarly, we can calculate the value of x from cos  = which gives x = h cos  .
h
Case 2: Given a side other than the hypotenuse and an angle
We use tan:

85
 y
(i) If y and  are given then, in figure 2 tan  = so the value of x is obtained as
x
y
x=
tan 
y
(ii) If x and  are known then, tan  = so that y = x tan 
x
The hypotenuse can then be obtained using Pythagoras’ theorem: h = x 2 + y 2
Case 3: Given two of the sides
We use tan −1 or sin −1 or cos −1 to obtain the angle 
B

A y  y
(i) x C
Figure 3 tan  = so  = tan −1  
x x
B

h
y

y  y
(ii) A C Figure 4 sin  = so  = sin −1  
h h
B

A x x
(iii) x C
Figure 5 cos  = so  = cos −1  
h h
Since two sides are given in this case, we can obtain the length of the third angle using
Pythagoras’ theorem.
Example 1
Obtain all the angles and the remaining side for the triangle shown:
A

B C
5
Solution
4 4
tan B =  B = tan −1   = 38.66
5 5
Sum of angles in a triangle is 180 , therefore angle A is obtained by
180 − (90 + 38.66 ) = 51.34
By Pythagoras’ theorem the side c = 42 + 52 = 41 = 6.4
86
 Example 2
Obtain the remaining sides and angle for the triangle shown:
A

15
b

30040/
B
C
a
Solution
 40 
31 40' =  31 +  = 31.67
 60 
b
sin 31.67 =  b = 15  sin 31.67 = 7.87
15
a
cos 31.67 =  a = 15  cos 31.67 = 12.77
15
Angle at A = 90 − 31.67 = 58.33 (sum of angles in a triangle is 180 )

5.6 Some special angles

Angles of magnitude 30 , 45 and 60 are special angles because their trigonometric ratios
can easily be expressed as fractions and surds using the sides of triangles as shown below:

450 0
60
1 2 2
1

450 30
0

1 3

1 2 3
sin 45 = = sin 60 = = cos 30
2 2 2
1 2 1
cos 45 = = sin 30 = = cos 60
2 2 2
1
tan 45 = 1 tan 30 = ; tan 60 = 3
3

5.7 Trigonometric functions of angles of different magnitude

A. First quadrant (0    90 )

Sine, cosine and tangent of angles in the first quadrant are all positive. It also follows that
their reciprocal ratios cosecant, secant and cotangentin this quadrant are equally positive.
B. Second Quadrant (90    180 )

Only sine and cosecant of angles in the second quadrant are positive. All other trigonometric
functions are negative. For any obtuse angle  that is (90    180 ) , its supplement  say
defined by  = 180 −  is an acute angle. The trigonometric functions of  and  are related
as follows:

87
(i) sin  = sin(180 −  ) = sin 
(ii) cos  = − cos(180 −  ) = − cos 
(iii) tan  = − tan(180 −  ) = − tan 

C. Third quadrant (180    270 )

In the third quadrant, all trigonometric functions are negative except tangent and cotangent.
If  is in the third quadrant, then the angle  =  − 180 is an acute angle and their
trigonometric functions are related as follows:
(i) sin  = − sin( − 180 ) = − sin 
(ii) cos  = − cos( − 180 ) = − cos 
(iii) tan  = tan( − 180 ) = tan 

D. Fourth quadrant (270    360 )

Cosine and its reciprocal ratio secantare the only trigonometric functions whose values are
positive for angles in the fourth quadrant. The angle  defined by  = 360 −  such that
270    360 is acute and their trigonometric ratios are related as follows:
(i) sin  = − sin(360 −  ) = − sin 
(ii) cos  = cos(360 −  ) = cos
(iii) tan  = − tan(360 −  ) = − tan 

E. Rotation beyond the fourth quadrant (  360 )

For angles greater than 360 , we employ the periodicity of the given trigonometric function
followed by the appropriate case (A-D) above. Recall that angle 360 represents one complete
rotation, thus any angle greater than 360 would represent a certain number of complete
rotations say, n plus an angle of magnitude less than 360 .For example an angle say, 405
represents one complete revolution plus 45 . In general, if n is any integer, we have that
sin( + 360n)  sin  and cos( + 360n)  cos  (  is measured in degrees)
or sin( + 2 n)  sin  and cos( + 2 n)  cos  since 360  2 radians
We say that the trigonometric functions sin  and cos alongside their reciprocal ratios
cosecant and secant are periodic with period of 360 . However, the periodicity of tangent and
cotangent is 180 thus
tan( + 180n)  tan  and cot( + 180n)  cot  (  is measured in degrees)
or tan( +  n)  tan  and cot( +  n)  cot 

NOTE: It is important to note the following identities of trigonometric functions of negative

angles:
sin( − ) = − sin  , cos( − ) = cos  , tan(− ) = − tan( )
cos ec( − ) = − cos ec , sec( − ) = sec  , cot( − ) = − cot( )

88
 Example 3
Without using calculators or figure tables evaluate (a) sin 840 (b) sec930
Solution
(a) sin 840 = sin[(2  360) + 120] = sin120 (by periodicity)
120 lies in the second quadrant and sine is positive in second quadrant. From B above,
we have
3
sin120 = sin(180 − 120 ) = sin 60 =
2
1
(b) sec930 = (by definition)
cos 930
1 1
= = (by periodicity)
cos[(2  360) + 210] cos 210
210 lies in the 3rd quadrant and we know that cosine is negative in the 3rd quadrant. From C
above, we have
3
cos 210 = − cos(210 − 180 ) = − cos 30 = −
2
Example 4
Obtain all the values of  between 0 and 2 such that
1 1 3
(a) cos  = (b) tan  = − (c) sin  = −
2 3 2

Solution
1
(a) cos  =
2
1 
The acute angle whose cosine is is .
2 4
Since the given cosine is positive and we know that cosine is positive in 1st and 4th
quadrant, we obtain the corresponding angles in both quadrants as follows:
  7
In 1st quadrant;  = and in 4th quadrant;  = 2 − =
4 4 4
 3
Therefore, the values of  between 0 and 2 are and
4 4
1
(b) tan  = −
3
1 
The acute angle whose tangent is is .
3 6
Since the given cosine is negative and we know that tangent is negative in 2nd and 4th quadrants,
we obtain the corresponding angles in both quadrants as follows:
 5  11
In 1st quadrant;  =  − = and in 4th quadrant;  = 2 − =
6 6 6 6
5 11
Therefore, the values of  between 0 and 2 are and
6 6
89
 3
(c) sin  = −
2

3 
The acute angle whose sine is is .
2 3
Since the given sine is negative and we know that sine is negative in 3rd and 4th
quadrants, we obtain the corresponding angles in both quadrants as follows:
 4  5
In 3rd quadrant;  =  + = and in 4th quadrant;  = 2 − =
3 3 3 3
4 5
Therefore, the values of  between 0 and 2 are and .
3 3
Example 5
Find (a) the basic solutions of sin  = 0.9649
(b) all possible solutions.
Solution
The given sine is positive, thus the values of  must lie in the 1st and 2nd quadrants.
The acute angle whose sine is 0.9649 is 74.8 .
The corresponding values of  in 1st and 2nd quadrants are 74.8 and 180 − 74.8 = 105.2
respectively. Thus, the basic solutions are  = 74.8 and 105.2
(b) The rest of the infinitely many solutions are obtained from periodicity as
 = 74.8  (360k ) and 105.2  (360k ) for k = 1, 2,...

5.8 Trigonometric identities

A trigonometric identity is a relation between trigonometric expressions which is always true for
all magnitudeof angles. Many of the relationships between the trigonometric functions are very
useful especially in calculus.The most fundamental of these identities is the Pythagorean
identityis proven below:
Let us consider the right-angled triangle ABC with hypotenuse of unit length.
B

C A
With ACB =  , we see that BC = cos and AB = sin  . Hence from Pythagora’s theorem, we
have that
sin 2  + cos 2  = 1 (1)
Dividing (1) respectively by sin 2  and by cos 2  , we obtain,
1 + cot 2  = cos ec 2 (2)
tan 2  + 1 = sec 2  (3)
Further identities are:
sin( A + B) = sin A cos B + cos A sin B (4)
cos( A + B ) = cos A cos B − sin A sin B (5)
Dividing (4) by (5) we obtain
90
 sin( A + B) sin A cos B + cos A sin B
tan( A + B) = =
cos( A + B) cos A cos B − sin A sin B
Dividing every term by cos A cos B we obtain
tan A + tan B
tan( A + B) = (6)
1 − tan A tan B
Replacing B by − B in (4) and (5) and recalling that cos(− B) = cos B, sin(− B) = − sin B we obtain
sin( A − B) = sin A cos B − cos A sin B (7)
cos( A − B) = cos A cos B + sin A sin B (8)
Using the identities (7) and (8) in same manner for obtaining identity (6) above, we have

tan A − tan B
tan( A − B) = (9)
1 + tan A tan B

5.9 Double and Half angle formulae

If we replace B with A in the identities given in (4) and (5), we obtain the following double -
angle formulae:
sin 2 A = sin Acos A + cos Asin A = 2sin Acos A (10)
cos 2 A = cos A cos A − sin A sin A = cos 2 A − sin 2 A (11)
Substituting cos 2 A with 1 − sin 2 A and sin 2 A with 1 − cos 2 A in (11), we obtain the additional
formulae for cos 2A
cos 2 A = 1 − 2sin 2 A (12)
cos 2 A = 2 cos 2 A − 1 (13)
Double-angle formula for tangent is given by:
2 tan A
tan 2 A = , for tan   1 (14)
1 − tan 2 A
A
Replacing A by , and consequently 2 A by A , in (10), (13) and (14) above, we obtain half-angle
2
formulae for sine, cosine and tangent as follows
 A  A
sin A = 2 sin   cos  
2 2
 A
cos A = 2 cos 2   − 1
2
 A
2 tan  
tan A = 2
 A
1 − tan 2  
2

5.10 Product formula

For any pair of angles A, B we have
1
cos A cos B = cos( A + B) + cos( A − B)
2
1
sin A sin B = cos( A − B) − cos( A + B )
2

91
 1
sin A cos B = sin( A + B) + sin( A − B)
2
5.11 Sum and Difference formulae
For any pair of angles A, B we have
A+ B A− B
cos A + cos B = 2 cos cos
2 2
A+ B A− B
cos A − cos B = −2sin sin
2 2
A+ B A− B
sin A + sin B = 2sin cos
2 2
A+ B A− B
sin A − sin B = 2 cos sin
2 2
Example 6
Using the addition formulae, evaluate the following ratios without calculators.
(1) tan105 (2) cos ec120
3 12
(3) sin( +  ) , if sin  = and sin  = with (a)  ,  acute (b)  acute,  obtuse
5 13
Solution
(1) 105 = 60 + 45
tan 60 + tan 45
tan105 = tan(60 + 45) =
1 − tan 60 tan 45
3 +1
= = −(2 + 3)
( )
1 − 3 (1)
(2) cos ec120 = cos ec 2(60)
1
=
sin 2(60)
1 1 2 3
= = =
2sin 60 cos 60  3  1  3
2  
 2  2 
(3) sin( +  ) = sin  cos  + cos  sin 
3 4 12 5
If sin  = , then cos  = .Similarly by Pythagoras’ theorem, since sin  = then cos  = .
5 5 13 13
(a)Both  ,  are acute, which implies that all trigonometric ratios of both angles are positive. So,
 3   5   4   12 
sin( +  ) = sin  cos  + cos  sin  =     +    
 5   13   5   13 
 3   48   63 
= +  = 
 13   65   65 
(b) Since  is acute and  is obtuse, cos  would be negative (Recall that while sine is positive in
2nd quadrant, cosine is negative. So,
 3   5   4   12 
sin( +  ) = sin  cos  + cos  sin  =    −  +    
 5   13   5   13 

92
  3   48   33 
= − +  =  
 13   65   65 

Practice Exercise 5.2

1. Solve the right-angled triangle shown:

b
8

B C
15
2. Express each of the following ratios as trigonometric ratios of an angle in −90    90
(a) cos ec190 (b) cos 635 (c) tan 95
3. Write the following ratios in terms of one of the special angles, and hence evaluate without
5 7
tables (a) sin (b) cot
3 4

4. Obtain all the values of between 0 and 360 (correct to 2 decimal places)such that
1 1
(a) tan  = − (b) sin  = 0.4305 (c) cos  = − (d) 2 cos 2  + 7 cos  + 3 = 0 (e)
2 2
cos ec 2 = 5.7208
4
5. Find tan 2 without tables or calculator, if sin  = and  is acute
5
6. Calculate the value of d1 + d 2 in the accompanying figure

d1 200

150 280
d2
7. Calculate h and hence the length AB , in the triangle shown, if the sum of the other two sides is
10.21cm
C

210 120
A B
8.Obtain the remaining sides and angles for the triangles shown:

First triangle second triangle

9. Prove the following identities:
2 cos3  − cos 
(a) (1 − sin  )(1 + sin  ) = cos 2  (b) = cot 
sin  cos 2  − sin 3 

93
 CHAPTER 6
Introduction to complex numbers
6.1 Introduction
This chapter introduces readers to some basic properties of complex numbers and their
applications in science and engineering. It covers the elementary aspects of complex numbers.

6.2 The theory of complex numbers

The solution of m 2 + 1 = 0 gave rise to the theory of complex numbers. This is true since
m 2 = −1 has no real solution. In order to evaluate m 2 = −1 , we denote (−1) by an imaginary
number i so that i 2 = −1. With this denotation, we solve m 2 + 1 = 0 as follows.
m2 +1 = 0
m 2 = −1
m2 = i2
m =  i2
m = i
6.2.1 Representation of complex numbers
Every complex number z could be represented in terms of two real numbers x and y
as follows.
z = x + iy ,
z = x + yi
or
z = ( x, y )
The real part of z is denoted by Re z and the imaginary part of z is denoted by Im z . It follows
that
Re z = x
Im z = y
In particular, z1 = 3 + 4i, z 2 = −1 + i and z 3 = (5, 7) are complex numbers.
Re z1 = 3, Im z1 = 4
Re z 2 = −1, Im z 2 = 1
Re z 3 = 5, Im z 3 = 7
The absolute value of z is denoted by | z | and the modulus of z is denoted by mod z where
mod z = | z |= x 2 + y 2
The conjugate of z is denoted by z where
z = x − yi

Remark 6.2.1
In electricity, the letter i is used to denote current. In order to avoid ambiguity, most
engineers denote − 1 by j . In this sense, they use z1 = 3 + 4 j and z 2 = −1 + j as two complex
94
numbers instead of z1 = 3 + 4i and z 2 = −1 + i . Also, they denote the complex number
z = x + yj by z = m 0 where
x = m cos ,
y = m sin 
and
z = m 0 represents a vector of magnitude m at angle of  0 with the real axis.

6.2.2 Arithmetic operations on complex numbers

Let z1 = a + bi and z 2 = c + di where a, b, c and d are real numbers. We define the
following rules.
z1  z 2 = (a + bi)  (c + di)
= (a  c) + (b  d )i
z1 z 2 = (a + bi)(c + di)
= ac + adi + cbi + bdi 2
= (ac − bd ) +(cb + ad )i
cz1 = c(a + bi)
= ca + cbi
z1 = z 2 iff a = c and b = d
z1 a + bi
=
z 2 c + di
a + bi c − di
= , c − di is the conjugate of c + di .
c + di c − di
ac − adi + cbi − bdi 2
=
c 2 − (di) 2
(ac + bd ) + (cb − ad )i
=
c2 + d 2
z1 z1
=
z2 z2
z1 z 2 = z1 z 2
z1 z 2 = z1 z 2
 z1  z1
  =
 z2  z2
z1 z1 = ( a + bi )( a − bi )
= a 2 − abi + abi − (bi) 2
= a2 + b2
= z
2

Therefore,
z1 z 1 = z .
2

95
Example
If z1 = 2 + 3i and z 2 = 4 − 5i, compute
− z1
(a) z1 + z 2 (b) z1 − z 2 (c) z1 z 2 (d ) | z1 | ( e) z 2 ( f )
z2
Solution
(a) z1 + z 2 = (2 + 3i) + (4 − 5i)
= (2 + 4) + (3i − 5i )
= 6 − 2i
(b) z1 z 2 = (2 + 3i )(4 − 5i )
= 8 − 10i + 12i − 15i 2
= 8 + 2i − 15(−1)
= 8 + 15 + 2i
= 23 + 2i
(c ) | z1 |=| (2 + 3i) |
= 2 2 + 32
= 13
−
(d ) z 2 = 4 − (−5)i
= 4 + 5i
z1 2 + 3i
( e) =
z 2 4 − 5i
2 + 3i 4 + 5i
= .
4 − 5i 4 + 5i
(2 + 3i )(4 + 5i )
=
4 2 − (5i ) 2
− 7 + 22i
=
16 + 25
− 7 + 22i
=
41
− 7 22
= + i
41 41

6.2.3 Argand diagram

An Argand diagram is a plot of complex numbers as points z = x + yi, x, y  R, in the
complex plane. The complex plane consists of the x − axis as its real axis and y − axis as its
imaginary axis.

Example
Sketch the Argand diagram of
(a) z = x + yi (b) z1 + z 2 if z1 = x1 + y1i and z 2 = x2 + y2 .
Solution
(a ) z = x + yi
96
Figure (Argand diagram of z = x + yi )

Imaginary
axis ((

y z = x + yi

|z


| y

0 x Real
axis

Figure 6.2.2 (Argand diagram of z1 + z 2 ):

(b) z1 + z 2 if z1 = x1 + y1i and z 2 = x2 + y2 .

z1 + z 2 = ( x1 + x2 ) + ( y1 + y2 )i

Imaginary y2
axis |z2| | z1 + z 2 |
Z y1 + y 2
y1 2 z1 = x1 + y1i

| z1 |
y1
2 |
1
0 x1 Real
axis
x1 + x2

97
6.2.4 Argument of z
The argument or amplitude of z = x + yi is the angle,  , made by a line which passes
through the points (0, 0) and ( x, y ) with the positive real axis. It is given by
 y
 = tan −1  
x
as shown in figure (5.2.1). The argument of z is denoted by arg z . Thus,
The argument of z in the interval −      , denoted by Argz , is called the principal value of
arg z . In operational form we have
 y
Argz = tan −1   ; if z = x + yi , Re z  0 and Im z  0.
x
 y
Argz = − tan −1   ; if z = x − yi , Re z  0 and Im z  0.
x
 y
Argz = − tan −1   +  ; if z = − x + yi , Re z  0 and Im z  0.
x
 y
Argz = tan −1   −  ; if z = − x − yi , Re z  0 and Im z  0.
x
The following are some of the properties of arg z .
arg z = Argz + 2n, n = 0,  1,  2, ...
arg( z1 z 2 ) = arg z1 + arg z 2
z1
arg = arg z1 − arg z 2
z2
arg z r = r arg z

Example
Find the principal argument of
(a) z = 1 + i (b) z = 1 − i (c) z = −1 + i (d ) − 1 − i
1+ i
( e) z = ( f ) z = (1 + i )i
i
Solution
(a) z = 1 + i
1 
Argz = tan −1   = tan −1 (1) =
1 4
(b) z = 1 − i
 −1 1 
Argz = tan −1   = − tan −1   = − tan −1 (1) = −
 1  1 4
( c ) z = −1 + i
 1  1  3
Argz = tan −1   = − tan −1   +  = − tan −1 (1) +  = − +  =
 −1 1 4 4
(d ) z = −1 − i

98
  −1 1  − 3
Argz = tan −1   = tan −1   −  = tan −1 (1) −  = −  =
 −1 1 4 4
1+ i
( e) z =
i
1+ i
Argz = Arg = Arg (1 + i ) − Arg (0 + i )
0+i
1 1
= tan −1   − tan −1  
1 0
= tan (1) − tan (0)
−1 −1

   − 2 
= − = =−
4 2 4 4
( f ) z = (1 + i )i
Argz = Arg ((1 + i )i ) = Arg (1 + i ) + Arg (0 + i)
1 1
= tan −1   + tan −1  
1 0
= tan −1 (1) + tan −1 (0)
   + 2 3
= + = =
4 2 4 4
Example
Find arg(1 + i ) .
Solution
Let z = 1 + i .
1 
Argz = tan −1   = tan −1 (1) =
1 4
arg z = Argz + 2n; n = 0,  1,  2, ...

arg z = + 2n; n = 0,  1,  2, ...
4
    
arg z = , + 2 , − 2 , + 4 , − 4 , ...
4 4 4 4 4
  + 8  − 8  + 16  − 16
arg z = , , , , , ...
4 44 4 4
 9 7 17 15
arg z = , , − , , − , ...
4 4 4 4 4

99
6.2.5 Polar representation of a complex number
Consider the complex number z = x + yi sketched in the diagram below.
Figure 5.2.3(Argand diagram of z = x + yi )

Imaginary
axis

y z = x + yi

r =| z |
y = r sin 
z = x + yi
z = r cos + ir sin 
0 x = r cos Real
axis

Therefore,
z = r (cos + i sin  )
is the principal polar representation of z = x + yi. The general polar representation of z = x + yi
is
z = r (cos( + 2n) + i sin( + 2n))
n = 0,  1,  2, ...

Example 6.2.5
Find the principal and general polar representation of z = 1 + ( 3 ) i.
Solution
( )
z = 1 + 3 i.

r = z = 12 + ( 3) 2
= 1 + 3 = 4 = 2.
 3 
 = tan −1  =
 3
 1 
From equation (5.2.5), the principal polar representation of z = 1+ ( 3 ) i is
  
z = 2 cos + i sin  .
 3 3
From equation (5.2.6), the general polar representation of z = 1+ ( 3 ) i is
    
z = 2 cos + 2n  + i sin + 2n  , n = 0,  1,  2, ... .
 3  3 

100
Theorem 6.2.1 (De Moivre’s theorem).
De Moivre’s theorem states that
(cos + i sin )m = cos m + i sin m
holds for any integer m and   R.

Proof:
By Euler’s formula,
e i = cos + i sin 
(e i )m = e im = cos m + i sin m
Therefore,
(cos + i sin )m = cos m + i sin m
Example 6.2.6
(a) Expand (cos  + i sin  )4 by De Moivre’s theorem.
(b) Expand (cos  + i sin  )4 by binomial expansion method.
(c ) Use your results in (a ) and (b) to express cos 4 and sin 4 in terms of powers of cos
and sin  .

Solution
(a ) By De Moivre’s theorem,
(cos + i sin )4 = cos4 + i sin 4
(b) By binomial expansion,

(cos + i sin  )4 = (cos )4 + 4(i sin  )(cos )3 + 4(3) (i sin  )2 (cos )2

2!

+
4(3)(2)
(i sin  )3 (cos )1 + 4(3)(2)(1) (i sin  )4
3! 4!
= cos4  + i 4 sin  cos3  + (i 2 )6 sin 2  cos2  +
+ (i 2 )i 4 sin 3  cos + sin 4 
= cos4  + i 4 sin  cos3  − 6 sin 2  cos2 
− i 4 sin 3  cos + sin 4 
= (cos4  + sin 4  − 6 sin 2  cos2  )
+ 4(sin cos3  − sin 3  cos ) i
(c ) From the results in (a ) and (b) , we see that
cos 4 + i sin 4 = (cos4  + sin 4  − 6 sin 2  cos2  ) + 4(sin cos3  − sin 3  cos ) i
Equate the real parts and the imaginary parts in the equation above to get the expression for
cos 4 and sin 4 in terms of powers of cos and sin  :
cos 4 = (cos4  + sin 4  − 6 sin 2  cos2  )
sin 4 = 4(sin cos3  − sin 3  cos )

101
6.2.6 The roots of a complex number
Every nonzero complex number has exactly n distinct roots. Suppose we seek the n
distinct roots of a complex number
w = r (cos + i sin  ).
Let
z = p(cos + i sin )
be the sought solution such that
zn = w
must to solved. Thus, we have to solve
 p(cos + i sin )n = r(cos + i sin )
By De Moivre’s theorem, we have to solve
p n (cos n + i sin n ) = r (cos + i sin  )
The above equation holds if
1

p = r or
n
p=r n

and

n =  or  = .
n
From the above equations, substitute for  and p in
z = p(cos + i sin )
to have
 
1

z = r n  cos + i sin 
 n n
Equation (5.2.29) is not the only solution of equation (5.2.28). Equation (5.2.28) is also satisfied
for all angles
 + 2k
= , k = 0, 1, 2, ..., n − 1
n
The above values of k are used for distinct roots. The complete n roots of w are
 + 2k  + 2k 
1

z k = r n  cos + i sin , k = 0, 1, 2, ..., n − 1
 n n 

Example 6.2.4
Find all the four distinct roots of z = 1 + ( 3 )i
Solution
z = 1 + ( 3 )i
r = 12 + ( 3 ) 2 = 4 = 2
 3 
 = tan −1  =
 3
 1 
The n roots of a complex number are given by
 + 2k  + 2k 
1

z k = r n  cos + i sin , k = 0, 1, 2, ..., n − 1
 n n 
The n = 4 roots of z = 1 + ( 3 )i are given by
102
  / 3 + 2k  / 3 + 2k 
1

z k = 2 4  cos + i sin , k = 0, 1, 2, 3.
 4 4 
1   2k  1  + 6k  + 6k
. + = . =
43 1  4 3 12
 + 6k  + 6k 
1

z k = 2  cos
4
+ i sin , k = 0, 1, 2, 3.
 12 12 
For k = 0, we have
 + 6 (0)  + 6 (0) 
1

z 0 = 2  cos
4
+ i sin 
 12 12 
  
1

= 2  cos + i sin 
4

 12 12 
For k = 1, we have
 + 6 (1)  + 6 (1) 
1

z1 = 2  cos
4
+ i sin 
 12 12 

 + 6  + 6 
1

= 2 4  cos + i sin 
 12 12 
7 7 
1

= 2 4  cos + i sin 
 12 12 
For k = 2, we have
 + 6 (2)  + 6 (2) 
1

z 2 = 2 4  cos + i sin 
 12 12 
 + 12  + 12 
1

= 2 4  cos + i sin 
 12 12 
13 13 
1

= 2 4  cos + i sin 
 12 12 
For k = 3, we have
 + 6 (3)  + 6 (3) 
1

z 3 = 2 4  cos + i sin 
 12 12 
 + 18  + 18 
1

= 2 4  cos + i sin 
 12 12 
19 19 
1

= 2 4  cos + i sin 
 12 12 

6.2.7 Geometric view of n roots of a complex number

Geometrically, the n roots of a complex number are the vertices of a regular polygon
1
n
with n sides. To see this, we plot a circle of radius r and trace the vertices with the angles

103
  + 2k
= , k = 0, 1, 2, ..., n − 1 made with the real axis.
n
Example 6.2.6
Solve the equation z 3 = 4 + 3i and show the roots geometrically.

Solution
1
Let w = 4 + 3i and z = w3 .
r = 4 2 + 32 = 16 + 9 = 25 = 5
3
 = tan −1   = tan −1 (0.75)
4
= 0.6435radian or 36.87 0
The n roots of a complex number are given by
 + 2k  + 2k 
1

z k = r n  cos + i sin , k = 0, 1, 2, ..., n − 1
 n n 
The n = 3 roots of w = 4 + 3i , solution of z = (4 + 3i ) 3 , are given by
1

0.6435 + 2k 0.6435 + 2k 

1

z k = 5  cos
3
+ i sin , k = 0, 1, 2.
 3 3 
0.6435 + 6.2832k
.(0.6435 + 6.2832k ) =
1
= 0.2145 + 2.0944k
3 3
1

z k = 5 (cos(0.2145 + 2.0944k ) + i sin(0.2145 + 2.0944k )), k = 0, 1, 2

3

For k = 0, we have
1

z 0 = 5 (cos(0.2145 + 2.0944(0) ) + i sin(0.2145 + 2.0944(0) ))

3

= 5 (cos(0.2145) + i sin(0.2145))
3

= 5 3 (0.9771+ 0.2129i )
= 1.71(0.9771+ 0.2129i )
For k = 1, we have
1

z1 = 5 3 (cos(0.2145 + 2.0944(1)) + i sin(0.2145 + 2.0944(1)))

1

= 5 3 (cos(2.3089) + i sin(2.3089))
= 1.71(− 0.6729 + 0.7397i )
For k = 2, we have
1

z 2 = 5 3 (cos(0.2145 + 2.0944(2) ) + i sin(0.2145 + 2.0944(2)))

1

= 5 3 (cos(4.4033) + i sin(4.4033))
= 1.71(− 0.3042 − 0.9526i )
Geometrically, the roots z 0 , z1 and z 2 are the vertices of a triangle (Z0Z1Z2) while

104
 1 1

r 3 = 5 3 = 1.71 is the radius of the circle shown below. The angles of the roots are
z0 : 0.2145 radian (12.290 ) ; z1 : 2.3089 radians (132.290 ) ; z 2 : 4.4033 radians (252.290 )

Figure 6.2.4 (Geometric view of z 0 , z1 and z 2 ).

Imaginary axis
1.71i

Z1 252.290
Z0
132.29
0 12.29
0
0 1.71
-1.71 Real axis

Z2 -1.71i

Example 6.2.7
Solve the equation
(2 + 4i) 2 − 2(n + iy) = n + iy
Solution
(2 + 4i) 2 − 2(n + iy) = n + iy
4 + 16i 2 + 16i − 2n − 2 yi = n + yi
4 − 16 − 2n + 16i − 2 yi = n + yi
(− 12 − 2n ) + (16 − 2 y )i = n + yi
− 12 − 2n = n
− 12 = 3n
n = −4
16 − 2 y = y
16 = 3 y
1
y=5
3

Example 6.2.8
A circuit has a resistance of 5 in series with a resistance of 7 as shown in figure (5.2.5)
below.

105
Figure 6.2.5 (Series resistors)

R = 5 R = 7

Represent the impedance by a complex number.

Solution
The impedance, in rectangular form, is
z = 5 + 7 j; j = − 1
The magnitude of the impedance is 5 2 + 7 2 = 8.6 . The phase angle is
7
tan −1   = 0.95055 or 54.5 0.
5
The impedance expressed in polar form is
z = 8.6(cos54.5 0 + j sin 54.5 0 ) or z = 8.654.5 0

Example 6.2.8
Find the combined impedance of the circuit in figure (5.2.6) below.
Figure 6.2.6 (Parallel resistors)

R = 40 R = 30

Z1
R = 40 R = 30

Z2
R = 40 R = 30

Z3
R = 40 R = 30

Z4

Solution
Let
106
 Z 1 = 40 + 30 j
Z 2 = Z 3 = Z 4 = Z1 .
4
Z1 Z 2 Z 3 Z 4 Z 1 3
Total impedance = = 1 = Z1
Z1 + Z 2 + Z 3 + Z 4 4Z1 4
Z1 = 402 + 302 = 50

 30 
Phase angle = tan −1   = 0.6435radian or 36.9 0
 40 
In polar form,
Z1 = 50(cos 0.6435 + j sin 0.6435)
Z1 = 503 (cos0.6435 + j sin 0.6435)
3 3

By De Moivre’s theorem,
Z1 = 125000(cos0.6435(3) + j sin 0.6435(3))
3

= 125000(cos1.9305 + j sin1.9305)
1 3 125000
Z1 = (cos1.9305 + j sin1.9305)
4 4
= 31250(cos1.9305 + j sin1.9305)
= −11000 + j 29250
1.9305radians = 110.6 0 .
The combined impedance is
Z = −11000 + j 29250 or Z = 31250110.6 0

Practice Exercise 6.1

1. Given that z1 = 2 + 3i, z 2 = −3 + i and z 3 = 3 + i, evaluate
− −
(a ) z1 + z 2 (b) z1 − z 2 (c) z1 z 2 (d ) z1 (e) z1 z 2 ( f ) z1 ( g ) z 2
− − 1 1 z1
3 5
(h) z1 z 2 (i ) z1 z 2 ( j ) z1 (k ) z 2 (l ) z 3 4 (m) z1 3 (n)
z2
 
z1 z1 + z 2 z 
(o ) ( p) (q ) Argz1 (r ) Argz3 ( s ) Argz3 z 3 (t ) Arg  −3 
z1 − z 2 z1 − z 2 z 
 3
2. Express z1 = 5 − 8i and z 2 = −3 − 4i in polar forms.
3. Draw the Argand diagrams for z1 = 5 − 8i and z 2 = −3 − 4i .
4. Solve for x and y if
(x 2 − 2 x − y ) + (2 x − y − 3) i = 0
5.
(a) Expand (cos  + i sin  )5 by De Moivre’s theorem.
(b) Expand (cos  + i sin  )5 by binomial expansion method.
107
 (c ) Use your results in (a ) and (b) to express cos 5 and sin 5 in terms of powers of cos
and sin  .
6. If z = 40 − 50 j , find arg z.
7. Express the combined impedance in the diagram below as a complex number.

Figure 6.2.7(Parallel resistors)

Figure 6.2.7 (Parallel resistors)

108
 CHAPTER 7
Matrices and Determinant

7.1 Introduction
The problems of solving large systems of equations which often arise for example in
finding the forces on members of a large structure in electrical engineering, or shortest-distance
problems in operations research or even cost management analysis in Economics can much
easily be handled by isolating the coefficients of the variables of the equations as a block or array
of numbers called matrix in mathematics. Specifically, in this chapter, we shall concern
ourselves with
• how to develop the terminology and basic properties of matrices,
• how to conveniently represent large systems of such equations, and
• how we might find solution of such equations.

7.2 Basic Definitions and Concepts

Matrix is rectangular array of objects arranged in horizontal rows and vertical columns.
The plural form of ``matrix" is ``matrices". We usually denote a matrix by an upper case letter,
for example A and represent this in box bracket or parentheses
 a11 a12  a1n   a11 a12  a1n 
a  a 
a  a  a  a 
A= 21 22 2 n =

21 22 2 n
. (2.1)
          
   

am1 am 2  amn   am1 am 2  amn 
The matrix A can as well be simply represented as
A = (aij ) where i = 1,2,, m and j = 1,2,, n. (2.2)
The objects a11 , a12 ,, amn are called the elements or entries of the matrix A. These elements
can be real numbers, complex numbers or any other objects. The matrix A has m rows and n
columns and it is said to be a matrix of order m  n. It can as well be said that the dimension or
size of A is m  n. We will keep to use of parentheses in this book and may use any of the other
notations when the need arises.

Example
• Given
 a11 a12 a13   1 2 3 
   
A =  a21 a22 a23  =  4 5 6 .
a   
 31 a32 a33   7 8 9 
We observe that a11 = 1, a12 = 2, a13 = 3, a21 = 4, a22 = 5, a23 = 6, a31 = 7, a32 = 8 and a33 = 9 are
the elements of the matrix A. Also, A is a 3 3 matrix.
• If

109
  0 −1 − 3 − 4 5 
1 3   
  1 2 3 4   0 6 10 0 1 
B =  2 − 2 , C =  , and D = 
2 4   5 6 7 10  5 7 9 3 9 
   
 − 1 7 11 6 − 7 
 
then B is a 3 2 matrix, C is a 2 4 matrix and D is a 4 5 matrix.

8.3 Types of matrices

There are a number of forms matrices appear and each are given special names. Some of
these basic types are presented as follows.

Zero or null matrix

The zero or a null matrix is the matrix all of whose elements are zero. There are zero
matrices for all dimensions.

Example
0 0 0 0 0
  and  
0 0 0 0 0
are 2 2 and 2 3 null matrices respectively. Null matrices of any dimension are denoted by
0.

Square matrix
A matrix whose number of rows and number of columns are equal is called a square
matrix (i.e m = n ) of order n  n or simply n. The square matrix A can therefore be
represented by
 a11 a12  a1n 
 
 a21 a22  a2 n 
A= . (3.1)
    
 
a 
 n1 am 2  ann 
In the square matrix A, the leading diagonal (or principal diagonal) is the north-west to
south-east collection of elements a11 , a22 , a33 ,, ann . The sum of the elements in the leading
diagonal of A is called the trace of the matrix A and this is denoted by Tr ( A). That is,
n
Tr ( A) = a11 + a22 + a33 +  + ann = a .
i , j =1
ij (3.2)

A square matrix in which all the elements below the leading diagonal are zero is called an
upper triangular matrix. Typical upper triangular matrices are

110
  u11 u12  u1n 
 
 0 u 22  u2n  1 2 3
 
U = 0 0 
 u3n , uij = 0 when i > j and B =  0 5 6 .
 
      0 0 9
 
 0 
 u nn 
 0

A square matrix in which all the elements above the leading diagonal are zero is called a
lower triangular matrix. Typical lower triangular matrices are
 l11 0  0 
 
 l21 l22  0  1 0 0
   
L = l31 l32  0 , lij = 0 when i < j and C =  2 3 0 .
 
      4 5 6
 
l 
 n1 ln 2  lnn 

A square matrix where the only non-zero elements are along the principal diagonal is
called a diagonal matrix. A typical diagonal matrix is
 d11 0 0 0 
 
 0 d 22 0 0 
D= , where at least one of d 22  0 and d ij = 0 when i/neqj.
0 0 d 33 0 
 
 0 
 0 0 d 44 

A square matrix, usually denoted by I n , is a diagonal matrix in which all non-zero

element are 1 (unity). Typical identity matrices are
1 0 0
1 0  
I 2 =  , and I 3 =  0 1 0 
0 1 0 0 1
 
For any order n, a positive integer,
1 0  0
 
0 1  0
In =  0 0  0 .
 
   
0 0  1 


Row and Column matrices

An m 1 matrix with m  1 is called a row matrix or row vector. That is, a row matrix
has only one column and many rows. On the other hand, an 1 n matrix with n  1 is called a
column matrix or column vector. So, a column matrix has only one row and many columns.
The matrices

111
 1
 
R =  − 1 and C = (1 0 9 3 4)
5
 
are row and column vectors respectively.

Equal matrices
Two or more matrices are said to be equal only if they have the same number of rows
and same number of columns and if each corresponding elements are equal. The matrices
1  7 1 2 7 
   
A =   3 0  and B =  2 3 0  are equal if only  = 2 and  = −8.
4 5   4 5 − 8
   

Transpose Matrix
The transpose of a matrix A, denoted by Ac or AT , is the matrix whose row elements
are the column elements of A and vice versa. The following matrices are transposes of each
3 2 1 
3 1 4 2  
  1 0 −2 
other A =  2 0 − 1 1  and A = c  .
1 − 2 3 0  4 −1 3 
   
2 1 0 
Note, a matrix is called symmetric if it the transpose of itself. That is, a matrix A is
 1 −1 2
 
symmetric only if A = A . Quickly check that  − 1 0 3  is symmetric.
c

 2 3 7
 
Conjugate Matrix
The conjugate of a matrix A, denoted by A is the matrix of the conjugate of the
3 1 4 2 3 1 4 2
   
elements of A . By this definition, if A =  2 0 − 1 1  then A =  2 0 − 1 1  since the
1 − 2 3 0 1 − 2 3 0 
  
 1 − 3i 2i 
conjugate of a real number is again the same real number. Similarly, if B =  
 3 + 4i − 1 − 5i 
 1 + 3i − 2i 
then B =   since i = −i, for i  C.
 3 − 4i − 1 + 5i 
If a matrix A is such that ( A )T = A, we call A a Hermitian matrix.

Practice Exercise 7.1

1). Indicate whether or not the following are square, upper triangular or lower triangular
matrices and identify the order in each case.
112
 1 2 3 8
 
 1 2 3  1 1 0 − 2 − 4 0 
(a) A =  , (b) B =  , (c) C = 
 2 5 4  2 3 0 1 0 − 1
 
 
 

1 0 0 
 
 0 0 − 4
(d ) D =  , and (e) E = 0.
0 1 0 
 
 
 
2). Classify the following matrices and, where possible, find the trace
1 2 3 4 
a e o  
  1 + i 2i  5 6 7 8 
(a) V =  u i o , (b) C =  , (c) S = 
e a u  − i − 3i  −1 − 3 − 2 − 4 
   
 
 
 groupA groupB groupC groupD
 
 groupB groupC groupD groupA
(d ) M =  .
groupC groupD groupA groupB
 
 groupD groupA groupB groupC
 

3a). Let groupA, groupB, groupC, and groupD be sets of students in each of the groups offering
MTH101 this semester in FUNAI. Suppose the cardinalities of the sets are 100, 120, 98 and 208
respectively. Form a matrix F whose each element is the element of the matrix M in (2d) being
replaced by its cardinality. Compute Tr ( F ).
3b). Find F c . Is the matrix F symmetric?
a e o
 
4). Suppose V =  i a o  is equal to the 3 3 identity matrix, find the value of a, e, i, o and u .
 e u a
 

 2 3 − 1
 
5). If A =  0 1 2  then show that AAT is symmetric.
4 5 6 
 

 i 1 + i 2 − 3i 
 
6). Verify that  − 1 + i − 2i 1  is Hermitian.
 − 2 − 3i 1 0 


 0 1
11 0  0 1 2  
7). If B =   and C =   verify that ( BC ) =  11 3  = C T B T .
T

 2 1 1 1 3  22 7 
 

113
 7.4 Addition and Subtraction of Matrices
Two matrices A and B can be added or subtracted together to obtain another matrix
(say) C only if they are of the same order. When they have the same order, we say they are
conformable for addition/subtraction. Conformable matrices for addition/subtraction are
added or subtracted by adding or subtracting their corresponding elements. So in general, given
 a11 a12  a1n   b11 b12  b1n 
   
 a21 a22  a2 n   b21 b22  b2 n 
A= and B =  with both dimensions m n, we have
         
   
a   b  
 m1 a m2 a mn   m1 m 2
b bmn 

A  B = C with C given by
 a11  b11 a12  b12  a1n  b1n   c11 c12  c1n 
   
 a b a22  b22  a2 n  b2 n   c21 c22  c2 n 
C =  21 21 =  . (4.1)
       
   
a  b  amn  bmn   cm1 cm 2  cmn 
 m1 m1 am 2  bm 2

Example
2 3 1
1 2  5 6  
Let A =  , B =  ,
C =  0 − 2 4 ,
3 4 7 8  3 1 2
 
 1 −1 5 
2 4 1 0 − 3 1   
D =  , E =   and F =  0 3 2 .
 3 0 2  5 1 − 2  4 2 − 3
 
We have
1 2  5 6 1+ 5 2 + 6  6 8 
• A + B =   + B =   =   =  .
3 4  7 8   3 + 7 4 + 8  10 12 
1− 5 2 − 6  − 4 − 4
• A − B =   =  .
3 − 7 4 − 8  − 4 − 4
 4 4
• B − A =  .
 4 4
• B + C do not exist because they are not conformable for addition.
• C − D do not exist because they are not conformable for subtraction.
1 4 − 4
 
• C − F =  0 − 5 2 .
 −1 −1 5 
 

114
 Properties For any matrices A, B and C , it can be shown that
• A+ B = B + A
• A− B  B − A
• ( A + B) + C = A + ( B + C )
• ( A − B ) − C  A − ( B − C ).

7.5 Multiplication of a Matrix by a number

Matrices can also be multiplied with a number (scalar) naturally. To see this, observe
1 2  1 2   1 2  2 4   2(1) 2(2) 1 2
that for A =  , we have A + A =   +    =   = 2   . In
3 4  3 4   3 4  6 8   2(3) 2(4) 3 4
 a11 a12  a1n 
 
 a21 a22  a2 n 
general, given a number  and a matrix A =  , their product A is given
    
 
a 
 m1 am 2  amn 
by
 a11 a12  a1n     a11   a12    a1n 
   
a a22  a2 n     a21   a22    a2 n 
A =   21 = . (5.1)
          
   
a  
 amn     am1   am 2    amn 
 m1 am 2

3 1 5 
 1 
 3 1 5 2  2 2 2 
   − 2 .
1 1
Example If A =  − 2 0 1 − 4  then  A = − 1 0
 4 −1 3 − 2  2  2 
   1 3 
2 − −1 
 2 2 
The following properties also hold for the matrix algebras presented so far
• Let A and B be any two matrices. Multiplication of the matrix A by a scalar 
satisfies the law of distribution
 ( A + B ) = A + B

• ( A + B)T = AT + BT
• ( A − B)T = AT − BT
• ( AT )T = A.

115
 Practice Exercise 7.2
2 3 1
1 2  5 6  
1). Given A =  , B =  , C =  0 − 2 4 ,
3 4 7 8  3 1 2
 
2 4 1 0 − 3 1 
D =   and E =   and
 3 0 2  5 1 − 2
 1 −1 5 
 
F = 0 3 2 .
 4 2 − 3
 
Find (a) A + B − G (b) Show that ( A + B) + C = A + ( B + C )
(c) Show that ( A − B ) − C  A − ( B − C ).
2). Find (a) − 2 A (b) A + B given  = 2 and
1
 = −3 (c) 2C − F + 3I 3 .
3
3 − 2 + i 4i 
3). Given H =  . Suppose  = 1 − 3i, find (a) H (b) H − D.
 i 3 − 5i 6 − 2i 
 1 2 3
 
4). If G =  4 5 6 , determine (3G T − F )T .
0 0 1
 

 −1 4 
 1 2 3  
5). If S =   and Q =  0 1 , verify that 3(S T − Q) = (3S − 3QT )T .
 4 5 6  2 7
 

7.6 Matrix Multiplication

Two or more matrices may be multiplied together. However, unlike arithmetic and
algebra of real numbers, even when the product of two matrices exists, the order of
multiplication matters. We will now go through how matrices may be multiplied. Let A be a
c
1 2 row matrix and B a 21 column matrix: A = (a b ) and B =  . The product of A and
d 
c
B written as AB is the 11 matrix AB = (a b )   = (ac + bd.) so that corresponding
d 
elements are multiplied and the results added together. For example,
(2 − 3)   = (12 − 15)(− 3).
6
 6

116
  3 
 
 3 
Also (2 − 4 3 2)   = (6 − 12 − 6 + 10)(− 2). Thus, AB = C if the i th row and j th
−2
 
 5 
 
column of C is obtained by multiplying the i th row of A with j th column of B and the result
added up.
In the same way, suppose A is a 2 2 matrix and B also a 2 2 matrix:
 a b w x
A =   and B =  . The product of A and B written as AB is the 2 2 matrix
c d   y z
  w  x 
 a b  a b  
a b   w x 
AB =      =   y  z   =  aw + by ax + bz . (6.1)
 c d   y z   c d  w x   
  y c d     cw + dy cx + dz 
   z 
For example
  2  4 
 2 − 1  2 − 1  
 2 −1   2 4   6  1  =  − 2 7 .
     = 
 3 − 2   6 1   3 − 22 4   
 6  3 − 2    − 6 10
   1  

Continuing in this manner, we see that for 3 3 matrices A and C

a b c  r s t
   
A =  d e f  and B =  u v w
g h i  x y z 
  
their product C = AB is given by
 ar + bu + cx as + bv + cy at + bw + cz 
 
C =  dr + eu + fx ds + ev + fy dt + ew + fz 
 gr + hu + ix gs + hv + iy gt + hw + iz 
 
For example
1 2 −1   2 −1 3   4 − 8 7 
     
 3 4 0    1 − 2 1  = 10 − 11 13.
 1 5 − 2   0 3 − 2   7 − 17 12 
     
The general rule for multiplication of matrices A and B with order m  n and i  j respectively
are as follows
• A and B are conformal for multiplication only if the number of columns n of the
first matrix A equals the number of rows i of the second matrix B.
• The element in the i th row and j th column of the product matrix C (say) is obtained
by multiplying the i th row of A with the j th column of B and adding the result together.
• More compactly, we write this as
n
AB = (ab) = aij b jk = ai 2b2 k + ai1b1k +  + ainbik = cik . (6.2)
j =1

117
 Example
1 4
 1 3 5  
Given A =   and  2 5 , A and B have dimensions 2 3 and 2 2 and hence are
 2 4 6  3 6
 
conformable for multiplication. Their product AB is the 2 2 C given by
1 4
1 3 5    1 + 6 + 15 4 + 15 + 30   22 49 
C =     2 5  =   =  .
 2 4 6   3 6   2 + 8 + 18 8 + 20 + 36   28 64 
 
The following table is a list of the basic properties of matrix multiplication algebra for any
matrices A, B and C compared with the corresponding algebra for numbers a, b and c.

Matrix algebra Number algebra

A( B + C ) = AB + AC a (b + c) = ab + ac
AB  BA ab = ba
A( BC ) = ( AB )C (ab)c = a(bc)
AI n = I n A = A 1 a = a 1 = a
A0 = 0A = 0 0 a = a  0 = 0

Practice Exercise 7.3

 3 4
1 2  −1 5   2 4  2 1 5  
Given A =  , B =  , C =  , D =   and E =  − 1 2 .
 2 5  3 4 1 2 1 3 2  2 1
 
Find (1 a) AB (b) BC (c) DC
(2) Show that A( B + E ) = AB + AE
(3) 2 A2 + 3B − BC − ABC .
 − 3 0
(4) Let Q =   and f ( x) = 2 x 2 + 3x − 5. Find f (Q).
 1 2 
 1 2
(5) Given P =  . Show that P 2 − 4 P − I 2 = 0.
 2 3 

7.7 Determinant
Determinant as a mathematical concept has various uses among whom is to determine
which system of linear equations have a unique solution. The evaluation of determinant is a key
skill in econometrics, engineering, statistics and mathematics. We devote this section to
evaluation of determinants of small size matrices. For large size matrices, one can often use

118
 some fundamental properties of determinant to simplify the computation.

Definition (Minor of a matrix). Minor M ij of an element a ij of a matrix A is smaller

sized matrix obtained by removing the column and row in which the element a ij lies.

Example
 1 2 3
   2 3
Given the matrix A =  4 5 6 . The minor of a31 = 7 is the matrix M 13 =   and the minor of
7 8 9  5 6
 
 1 3
a22 = 5 is the matrix M 22 =  .
7 6

Definition: (Determinant). Given a matrix

 a11 a12  a1n 
 
 a21 a22  a2 n 
A= .
    
 
a  
 m1 a m2 a mn 

The determinant of A denoted by det ( A) or | A | is the number (or scalar) given by

n
det ( A) =  (−1)
i , j =1
i+ j
aij | M ij | (7.1)

where M ij is the minor of aij .

The formula (7.1) is known as the Laplace expansion formula. The Laplace expansion
technique can be used to evaluate determinant of a matrix of any order. Of course, one can fix
either i or j in expanding the sum (7.1).
In particular
• If A = (a11 ) a 11 matrix, then, fixing j = 1, we have
1
det ( A) =| A |= (−1)1+1 a11 | M 11 |= (a11 ). (7.2)
i =1

a a12 
• If A =  11  a 2 2 matrix, then, fixing i = 1, we have
 a21 a22 
2
det ( A) =| A |= (−1)1+ j a1 j | M 1 j |
j =1

= (−1) (a11 ) | M 11 | +(−1)1+ 2 (a12 ) | M 12 |

1+1

= a11a22 − a12a21.
So,
det( A) =| A |= a11a22 − a12a21 (7.3)

119
 a a12 
for any 2 2 matrix  11 . Continuing this way we have for a 3 3 matrix
 a21 a22 
 a11 a12 a13 
 
• A =  a21 a22 a23  on fixing j = 1 that
a 
 31 a32 a33 
3
det ( A) =| A |= (−1)1+ j a1 j | M 1 j | = (−1)1+1 (a11 ) | M11 | +(−1)1+2 (a12 ) | M12 | +(−1)1+3 (a13 ) | M13 |
j =1

a22 a23 a21 a23 a21 a22

= a11 − a12 − a13 .
a32 a33 a31 a33 a31 a32
So,
det( A) =| A |= a11 (a22 a33 − a23a32 ) − a12 (a21a33 − a23a31 ) − a13 (a21a32 − a22 a31 ) (7.4)
 a11 a12 a13 
 
for any 3 3 matrix  a21 a22 a23 .
a a33 
 31 a32

Example
 1 3
• If A =   then,
 − 1 6 
det ( A) =| A |= 6 − (−1)(3) = 9.

 5 −1 2
 
• If B =  1 1 3  then,
 −1 0 6 
 
1 3 1 3 1 1
det ( B) = 5 − (−1) +2
0 6 −1 6 −1 0
= 5(6 − 1) + 1(6 − −3) + 2(0 − (−1)) = 30 + 9 + 2 = 41.

Properties
Some of the basic properties of determinants are quoted below. These properties can be used to
simplify computation of determinant of large matrices.
• If two rows or two columns of a matrix are interchanged, then the determinant of the
resulting matrix is multiple of − 1.
4 2 1 3
For example, = 12 − 2 = 10 while = 2 − 12 = −10.
1 3 4 2

120
 • The determinant of a matrix and that of its transpose are equal.
1 2 1 3
For example, = 4 − 6 = −2 and = 4 − 6 = −2 too.
3 4 2 4

• If two rows or two columns of a matrix are equal then its determinant is zero.
For example,

1 2 3
1 2 3 2 3 1 3 1 2
= −2 +3 = (12 − 15) − 2(6 − 12) + 3(5 − 8) = −3 + 12 − 9 = 0.
4 5 6 5 6 4 6 4 5

• If the elements of one row or one column of a matrix is multiplied by a constant k

then its determinant is multiplied by k .
For example,
1 2 3 1 2 3
4 8 6 2 4 3
=2 .
7 8 9 7 8 9

So if one row or one column of a matrix is a multiple of another row or another column, then its
determinant is zero.
For example
2 4 −1 2 4 −1
4 2 1 4 2 1
=2 = 0.
− 4 −8 2 −2 −4 1

• If we add or subtract a multiple of one row or one column of a matrix to another, its
determinant remains unchanged.
1 2
For example, . Add 2 row1 to row 2. We have
4 5
1 2 1 2
= = 9 − 12 = −3.
4 5 4+ 2 5+4

• The determinant of a lower triangular, an upper triangular matrix or a diagonal matrix

is the product of the elements on the leading diagonal.
For example, check and convince yourself that the following determinants are each 24.
1 2 3 1 0 0 1 0 0
0 4 5 , 2 4 0 and 0 4 0 .
0 0 6 3 5 6 0 0 6

121
 Practice Exercise 7.4
1 2 3
1 2 −1 2
1). Determine the following determinants (a) (b) (c) 7 4 5
0 4 4 5
1 3 6
3 −1 4
2). (a) Use the Laplace’s expansion along the first row to determine 6 4 5
−1 2 1
2 3 4
(b) Use the Laplace’s expansion along the first column to determine − 1 2 3
3 2 5

3). Evaluate the following determinants using any of the properties of determinants
2 4 6 4 1 4 8 2
12 27 12
0 4 6 9 2 −1 1 − 3
(a). 28 18 24 (b) (c)
2 1 4 0 0 2 4 2
70 15 40
1 2 3 2 0 3 6 3
1 4  1 − 1
5). Given A =   and B =  . Evaluate 3 A − 2B + A3 .
 2 3   0 2 

7.8 Inverse of a Matrix

1
We know that every non-zero real number c has an inverse usually written as or
c
1
simply c-1 for instance, is the inverse of 4. In matrix algebra however,not all non zero matrix
4
have inverse.
The following points are important for matrices.
• Non-square matrices do not have an inverse.
• There is no division in matrix algebra. If a square matrix A has an inverse, it is
1
usually written an A −1 not .
A
• Not all square matrices have an inverse.
• If a matrix has an inverse, it is unique. That is to say no matrix have more that one
inverse.
• If A −1 is the inverse of A then AA −1 = A −1 A = I n .

We now define some few concepts that lead to give a general formula for the inverse of a
matrix if it exists.

Definition (Cofactor). The cofactor Aij of the element a ij of a matrix A is given by

122
 Aij = (−1)i + j det ( M ij ) (8.1)
where M ij is the minor matrix of aij .

 1 2
For example, for A =  . The cofactors of A are
 −1 4
A11 = (−1)1+1 (4) = 4, A12 = (−1)1+ 2 (−1) = 1,
A21 = (−1)2+1 (2) = −2 and A22 = (−1)2+ 2 (1) = 1.
 3 4 5
 
Also for B =  2 − 1 8 . The cofactors of B are
5 − 2 7
 
−1 8 2 8
B11 = (−1)1+1 = 9, B12 = (−1)1+ 2 = 26,
−2 7 5 7
2 −1 4 5
B13 = (−1)1+3 = 1, B21 = (−1)2+1 = −38,
5 −2 −2 7
3 5 3 4
B22 = (−1)2+ 2 = −4, B23 = (−1)2+3 = 26,
5 7 5 −2
4 5 3 5
B31 = (−1)3+1 = 37, B32 = (−1)3+ 2 = −14
−1 8 2 8
3 4
and B33 = (−1)3+3 = −11.
2 −1

Definition (Adjoint). The adjoint of a matrix C (say), denoted by Adj (C ) is the transpose matrix
of the matrix formed by replacing the elements of C by its corresponding cofactors. So if
C = (cij ), then
T
 C11 C12  C1n 
 
 C21 C22  C2 n 
Adj (C ) =  (8.2)
    
 
C 
 n1 Cn 2  Cnn 
where Cij is the cofactor of cij .
 1 2
For instance, the adjoint of A =   is
 −1 4
 4 − 2
T
 4 1
Adj( A) =   =  
 − 2 1 1 1 
and that of B is

123
  9 − 38 37 
T
 9 26 1 
   
Adj( B) =  − 38 − 4 26  =  26 − 4 − 14 .
 37 − 14 − 11 1 26 − 11
  

Finally, we can now give a definition of the inverse of a matrix.

Definition (Inverse of a matrix A ). The inverse A −1 of a matrix A is given by

1
A−1 =  Adj( A). (8.3)
det ( A)

For example, the inverse of the specified matrices A and B above are
1  4 − 2
 
6  1 1 
and
 9 − 38 37 
1  
 26 − 4 − 14 .
136 
1 26 − 11
a a12 
It can be easily shown that the inverse of any 2 2 matrix G =  11  is
 a21 a22 
1  a22 − a12  1  a22 − a12 
G −1 =   =   (8.4)
det (G )  − a21 a11  a11a22 − a12 a21  − a21 a11 
provided det(G )  0.

7.9 Systems of Linear Equations

Consider a system of linear n equations in m variables
a11 x1 + a12 x2 +  + a1n xn = b1
a21 x1 + a22 x2 +  + a2 n xn = b2
 
an1 x1 + an 2 x2 +  + ann xn = bn .
We can separate three facets of this system, namely, the coefficients of a11 , a12 ,, ann , the
variables x1 , x2 ,, xn and the numbers on the right-hand-side b1 , b2 , , bn . The facets can as well
be isolated and put into matrix form as follows.
 a11 a12  a1n  x1   b1 
    
 a21 a22  a2 n  x2   b2 
  = . (9.1)
        
    
a     b 
 n1 a n2 a nn  xn   n

124
  a11 a12  a1n   x1   b1 
     
a a22  a2 n   x2  b 
Suppose we name the matrices thus A =  21  X =   and B =  2  then we
     
     
a  ann     
 n1 an 2  xn   bn 
can simply write (9.1) as
AX = B (9.2)
where again A is the matrix of the coefficients of the system (??), X is column vectors of its
variables (unknowns) and B is the column vector of the numbers on its right-hand-side.
For example, the system
3x + 2 y − z = 3
x− y+z = 4
2x + 3y + 4z = 5
can simply be rewritten
 3 2 − 1 x   3 
    
AX = B   1 − 1 1  y  =  4 .
 2 3 4  z   5 
    
We append to A the column of the right-hand-side,we obtain and augmented matrix for the
system:
 3 2 −1 3 
 
 1 −1 1 4 .
2 3 4 5 
 
Similarly for the two system
2x + 3y = 1
4x − 2 y = 4
we have
 2 3  x   1 
AX = B     =  
 4 − 2  y   4 
and the augmented matrix becomes
2 3 1 
 .
4 − 2 4 
To solve the system of equations, one can apply the basic property of the inverse
A−1 A = AA −1 = I n to have
A−1 AX = A−1B
→ X = A−1 B. (9.3)
For example to solve the system
2x + 3y = 1
4 x − 2 y = 4,
1  − 2 − 3
We pre-multiply both sides of AX = B with the inverse matrix A−1 = −   of A.
16  − 4 2 
That is,

125
 1  − 2 − 3  2 3  x  1  − 2 − 3  1 
−     = −   
16  − 4 2  4 − 2  y  16  − 4 2  4 
 x 1  − 2 − 3  1 
 X =   = −   
 y 16  − 4 2  4 
 7 
 
=  8 .
 − 1 
 4
7 1
Therefore, x = and y = − .
8 4

Similarly, to solve the system

2x + 3y + z = 9
x + 2 y + 3z = 6
3x + y + 2 z = 8 we have
 2 3 1  x   9 
    
AX = B   1 2 3  y  =  6 
 3 1 2  z   8 
    
 x  1 − 5 7  9 
  1   
 X =  y =  7 1 − 5  6 
 z  118  − 5 7 1  8 
  
35 29 5
. Therefore, x = , y = and z = .
18 18 18

7.10 Cramer’s Rule

System of equations can also be solved by merely computing determinants. Cramer’s
rule involves the calculation of determinants and their ratios.
Suppose a system of equations has been resolved into the following matrix form
AX = B
 a11 a12  a1n   x1   b1 
     
 a21 a22  a2 n   x2   b2 
where A =  , X = and B =   .
      
     
a  x  b 
 n1 an 2  ann   n  n
Cramer’s rule says

126
  det ( Ax1 ) 
 
 x1   det ( A) 
   det ( Ax ) 
x   2 
X =  2  =  det ( A)  (10.1)

    
x  
 n det ( Ax ) 
 n

 det ( A) 
where
a11 a12  a1n
a21 a22  a2 n
det ( A) = ,
   
an1 an 2  ann

b11 a12  a1n

b21 a22  a2 n
det ( Ax ) = ,
1    
bn1 an 2  ann

a11 b12  a1n

a21 b22  a2 n
det ( Ax ) =
2    
an1 bn 2  ann
and

a11 a12  b1n

a21 a22  b2 n
det ( Ax ) = .
n    
an1 an 2  bnn

Example
• To solve 2 x + y = 7, 3x − 4 y = 5 simultaneously using Cramer’s rule, let
2 1   x  7
A =   =   =  .
 3 − 4  y   5
2 1 7 1 2 7
So, det ( A) = = 11, det ( Ax ) = = −33, det ( Ay ) = = −11.
3 −4 5 −4 3 5
− 33 − 11
Therefore, ( x, y ) = ( , ) = (3,1).
− 11 − 11
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 • Given the system
x1 − 2 x2 + x3 = 3
2 x1 + x2 − x3 = 5 ,
3 x1 − x2 + 2 x3 = 12
we have

 1 − 2 1   x1   3 
     
B =  2 1 − 1 =  x2  =  5 .
 3 − 1 2   x  12 
   3  
So,
1 −2 1 3 −2 1
1
det ( B ) = 2 1 − 1 = 10  x1 = 5 1 − 1 = 3,
10
3 −1 2 12 − 1 2
1 3 1 1 −2 3
1 1
x2 = 2 5 − 1 = 1 and x2 = 2 1 5 = 2.
10 10
3 12 2 3 − 1 12
Therefore, ( x1 , x2 , x3 ) = (3,1,2).

7.11 Gaussian Elimination Method

Gaussian elimination (also called elementary row elimination technique) is a
technique of reducing augmented matrix associated with the system of equations to a triangular
matrix and then matching variables with constants.
To reduce the associated augmented matrix to a triangular matrix, the following three
operations may be possible.
• Interchange two rows. For example, you may interchange rows one and three, which
one can write in shorthand as R1  R3.
• Multiple or divide a row by a non-zero constant factor. For example R1 3, and
2  R2.
• Add to, or subtract from, one row a multiple of another row. For example R1+ 3R2
1
and 2 R3 − R 2.
2

Example Solve the system

x1 + 3x2 + 5x1 = 14
2 x1 − x2 − 3x3 = 4
4 x1 + 5 x2 − x3 = 5
can simply be rewritten

128
 1 3 5  x1  14 
    
AX = B   2 − 1 − 3  x2  =  3 .
 4 5 − 1  x   7 
  3   
The associated augmented matrix is
1 3 5 14 
 
 2 −1 − 3 3 .
 4 5 −1 7 
 
So now proceed to reduce the rows.

1 3 5 14  1 3 5 14 
  R 2 − 2  R1  
 2 −1 − 3 3   0 − 7 − 13 − 25 
 4 5 −1 7  R3 − 4  R1  0 − 7 − 21 − 49 
   
1 3 5 14   1 3 5 14 
R 2  (−1)    
 0 − 7 13 25  R3 − R 2  0 − 7 13 25 .
R3  (−1)    0 0 8 24 
 0 7 21 49   
The augmented matrix is now triangular form and we can easily observe that
8 x3 = 24  x3 = 3,
7 x2 + 39 = 25  x2 = −2 and
x1 − 6 + 15 = 14  x1 = 5.
Therefore ( x1 , x2 , x3 ) = (5,−2,3).

Practice Exercise 7.5

1 2  −1 0 1 
1). Find the inverses of (a)   (b)   (c)  3 cos− 7 sin  
.
3 4  0 5  
 sin  4 cos 
 1 2 3  2 0 1
   
2). Determine the inverses of the following matrices (a)  4 5 6  (b) 1 0 0
7 8 9  4 1 3
   
1 8 2 1 0 0  2 0 0
     
(c)  0 1 5  (d)  3 1 0  (e)  0 3 0 
0 0 1 3 0 6 0 0 1
     

3). Solve
x + 2 y = 1 and 2 x − 3 y = 10 simultaneously using the (a) inverse matrix method
(b) Gaussian elimination method and (c) Cramer’s rule.

4). Solve
3x + 2 y − z = 3
x− y+z = 4

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 2x + 3y + 4z = 5
using the (a) inverse matrix method (b) Gaussian elimination method and (c) Cramer’s rule.

130
 CHAPTER 8
Introduction to coordinate Geometry
Geometry is an aspect of mathematics that studies shapes, sizes, and relative position of figures

and general properties of shapes. The key concepts in coordinate geometry are point, line, plane,

surface, angle, curve and their generalisations, namely manifolds, metrics and geodesics.

In coordinate geometry, those concepts are studied using coordinate system. The method of

describing the location of points using coordinate system was proposed by a French

mathematician René Descartes, (Pronounced "day CART"), who lived between 1596 - 1650.

He proposed further that curves and lines could be described by equations using coordinate

system, thus being the first to link algebra and geometry. In honour of his work, the coordinates

of a point are often referred to as its Cartesian coordinates, and the coordinate plane as the

Cartesian Coordinate Plane.

We shall briefly discuss distance between points, midpoint of line segments, ratios, gradient,

parallel and perpendicular lines, and some equations of geometric objects in this chapter.

8.1 Distance

Consider the right-angled triangle of Figure 8.1 below given in coordinate system. By

Pythagoras rule,

𝐴𝐵 2 = 𝐵𝐶 2 + 𝐴𝐶 2 = (8 − 1)2 +

(5 − 1)2 = 65.

∴ 𝐴𝐵 = √65.

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In general, the distance between any two coordinate points 𝑝(𝑥1 , 𝑦1 ) and 𝑞(𝑥1 , 𝑦1 ) is given by

𝑑(𝑝, 𝑞) = √(𝑥1 − 𝑦1 )2 + (𝑥2 − 𝑦2 )2 .

Similarly, the distance, 𝑑, between any 𝑛-coordinate points 𝑥 = (𝑥1 , 𝑥2 , … , 𝑥𝑛 ) and 𝑦 =

(𝑦1 , 𝑦2 , … , 𝑦𝑛 ) is given by

𝑑(𝑥, 𝑦) = √(𝑥1 − 𝑦1 )2 + (𝑥2 − 𝑦2 )2 + … + (𝑥𝑛 − 𝑦𝑛 )2 = √∑𝑛𝑖=1(𝑥𝑖 − 𝑦𝑖 )2 . (8.1)

So in our previous example, the distance between 𝐴(1,1) and 𝐵(5,8) is simply

𝑑(𝐴, 𝐵) = √(1 − 5)2 + (1 − 8)2 = √65 . The distance formula given by equation (8.1)

above is known as the Euclidean distance.

Example

The distance between the points 𝑀(5,5) and 𝑁(1,2) is

𝑑(𝑀, 𝑁) = √(5 − 1)2 + (5 − 2)2 = 5.

Practice Exercise 8.1

Find the distance between the following pair of points. (i.) (3,6) and (−1,2). (ii.) (2, −3) and

(5, −4). (iii.) (−4, −4) and (−1,5). (iv.) (4, −4) and (−1,5). (v.) (−3, −6) and (−1, −2).

8.2 Midpoint and Ratios

Given a pair of points 𝑃(𝑥1 , 𝑦1 ) and 𝑄(𝑥2 , 𝑦2 ) on a line segment. If 𝑀(𝑥, 𝑦) is a midpoint

between 𝑃 and 𝑄, see Figure 8.2.

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Then, 𝑥1 − 𝑥 =
𝑥1 + 𝑥2 𝑦1 + 𝑦2
𝑥 − 𝑥2 ⇒ 𝑥 = and 𝑦1 − 𝑦 = 𝑦 − 𝑦2 ⇒ 𝑦 = . These imply that
2 2

𝑥1 + 𝑥2 𝑦1 + 𝑦2
𝑀(𝑥, 𝑦) = 𝑀 ( , ). (8.3)
2 2

Example

The midpoint 𝑀 between 𝑝(1,3) and 𝑞(1,3) from equation (8.3) is

1+ 7 3+ 7
𝑀(𝑥, 𝑦) = 𝑀 ( , ) = 𝑀(4,5) .
2 2

Line segment can be divided into given ratios. To divide 𝐴(𝑥1 , 𝑦1 ) and 𝐵(𝑥2 , 𝑦2 ) into the
ratio 𝑚: 𝑛, consider Figure 8.3: So,
𝐴𝑃 𝑚 𝑥1 −𝑥 𝑚𝑥2 +𝑛𝑥1
= 𝑛 = 𝑥−𝑥 ⇒ 𝑥 = 𝑚+𝑛 . Similarly
𝑃𝐵 2
𝐴𝑃 𝑚 𝑦1 −𝑦 𝑚𝑦2 +𝑛𝑦1
= = 𝑦−𝑦 ⇒𝑦= . Thus the
𝑃𝐵 𝑛 2 𝑚+𝑛
point 𝑃(𝑥, 𝑦) has coordinates
𝑚𝑥2 +𝑛𝑥1 𝑚𝑦2 +𝑛𝑦1
𝑃(𝑥, 𝑦) = 𝑃 ( 𝑚+𝑛 , 𝑚+𝑛 ).

Q.E:
Let 𝐴(11,1), 𝐵(2,6) and 𝑃(𝑥, 𝑦) be points on a
line segment such that 2𝐴𝑃 = 𝑃𝐵. Find the
coordinates of 𝑃.
𝐴𝑃 1 𝑚
To find the coordinates (𝑥, 𝑦) of 𝑃 we use that 2𝐴𝑃 = 𝑃𝐵 ⇒ =2= . Therefore,
𝑃𝐵 𝑛
1×2+2×11 1×6+2×1 24 8
𝑃(𝑥, 𝑦) = 𝑃 ( , ) = 𝑃 ( 3 , 3).
3 3

Practice Exercise 8.2

(i). Suppose 𝐴(9,2), 𝐵(2,4) are points on a line segment divided at 𝑃(𝑥, 𝑦) such that 2𝐴𝑃 =
3𝑃𝐵. Find the coordinates of 𝑃.
(ii). At what ratio is the line segment connecting 𝐴(9,2) and 𝐵(2,4) divided at 𝑃(3,3)?

8.3 Gradient, Parallel and Perpendicular lines

The gradient (also known as the slope) of a line segment connecting two points 𝐴(𝑥1 , 𝑦1 ) and
𝐵(𝑥2 , 𝑦2 ), which denote here by grad(𝐴𝐵) is defined by the ratio of change in the 𝑦-coordinates
and the change in the 𝑥-coordinates, that is,
𝑦 −𝑦
grad(𝐴𝐵) = tan 𝜃 = 𝑥2 − 𝑥1 (8.4)
2 1
Geometrically, the slope, grad(𝐴𝐵), is shown in Figure 8.4. From Figure 8.4, grad(𝐴𝐵) is the

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 tangent of angle the line makes with the 𝑥-axis. That is,
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑦 −𝑦
grad(𝐴𝐵) = tan 𝜃 = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 = 𝑥2 − 𝑥1 .
2 1

Example

The gradient of the straight line joining

𝑝(−2, 4) and 𝑞(7,9) is (using the formula 8.4) is
9− 4 5
grad(𝑝𝑞) = 7−−2 = 9.

Practice Exercise 8.3A

Find the gradient of the lines (i). (3,6) and (−1,2). (ii). (2, −3) and (5, −4).
Definition (Parallel lines): Two lines are said to be parallel to each other if the angle between
them is zero (0).

Remarks:
1. Suppose a line segment passing through the points (𝑥1 , 𝑦1 ) and (𝑥, 𝑦) has
gradient 𝑚. Then the equation of the line is 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ). We can pick
any point for (𝑥1 , 𝑦1 ). For instance let (𝑥1 , 𝑦1 ) = (0, 𝑏) then 𝑦 = 𝑚𝑥 + 𝑏 is
the equation of the line.
2. Consider the two line segments 𝑙1 = 𝑚1 𝑥 + 𝑏1 and 𝑙2 = 𝑚2 𝑥 + 𝑏2 , see
Figure 5 below. So gradients 𝑚1 = 𝑡𝑎𝑛 𝜃1 and 𝑚2 = 𝑡𝑎𝑛 𝜃2 respectively. The
angle between the two
lines is therefore

134

Figure 14a Figure 5

 𝑡𝑎𝑛 𝜃2 − 𝑡𝑎𝑛 𝜃1 𝑚2 − 𝑚1
tan ∅ = 𝑡𝑎𝑛(𝜃2 − 𝜃1 ) = = .
1 + 𝑡𝑎𝑛 𝜃2 𝑡𝑎𝑛 𝜃1 1 + 𝑚2 𝑚1
Let the angle between the two lines be ∅ = 0. We know that tan 0𝑜 = 0. Then,
𝑚 −𝑚
tan ∅ = 1+2𝑚 𝑚1 = 0 ⇒ 𝑚1 = 𝑚2 .
2 1
Therefore, two lines are parallel to each if they have the same gradient.
Definition (Perpendicular lines): Two lines are said to be perpendicular to each other if the angle
between them if 90𝑜 .
Remark: Recall that tan 90𝑜 is undefined which is equivalent to the angle between the two lines
being ∅ = 0. Since tan 0𝑜 = 0. Then,
𝑚 −𝑚
tan 90𝑜 = 1+2𝑚 𝑚1 is undefined if 1 + 𝑚2 𝑚1 = 0 where 𝑚1 and 𝑚2 are the
2 1
gradients some two lines as defined before. So, 1 + 𝑚2 𝑚1 = 0 implies that
𝑚2 𝑚1 = −1. Therefore, it follows that two lines are perpendicular to each other if the product
of their gradients equals −1.

Practice Exercise 8.3B

1. Find the gradients of the following line segments: (i). 𝐴(1,1); 𝐵(4,5). (ii). 𝐴(1,1); 𝑃(4,1).
(iii). 𝑋(2,2); 𝑌(2,6).
2. Find the equation of a straight line whose gradient is 3 and passes through the point (4,1).
3. Show that 𝐴(1,1); 𝑃(4,1) is perpendicular to 𝐵(4,5); 𝑃(4,1).
4. A straight line has gradient 2 and cuts an intercept 5 on the 𝑦-axis. Find its equation.
5. Show that 𝑋(2,2); 𝑌(2,6) and 𝐴(1,1); 𝐵(4,5) are parallel to each other.

8.4 The Circle

Definition (Locus): Locus of a point is the path traced out by a point moving in accordance to a
given law.
For example, the locus of a point 𝑝 lying on a straight line segment joining (3,4) 𝑎𝑛𝑑 (2,1) is
4−1
the path 𝑦 = 𝑚(𝑥 − 𝑥1 ) + 𝑦1 where here 𝑚 = 3−2 = 3. Therefore 𝑦 = 3(𝑥 − 3) + 4 = 3𝑥 − 5
is the locus (path/line) of the point 𝑝.
Definition (Circle): A circle is the locus of a point 𝑝 which is of a constant distance 𝑟 (radius)
from a fixed point (centre).
Let 𝑝(𝑥, 𝑦) be a given point on a circle of centre 𝑐(ℎ, 𝑘) and radius 𝑟. Then the locus of
𝑝(𝑥, 𝑦) on the circumference of the circle is called the equation of the circle and it is given by
(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 . (8.5)
When the coordinate of the centre 𝑐 is (0,0) then ℎ = 𝑘 = 0 and equation (8.5) reduces to
𝑥2 + 𝑦2 = 𝑟2 . (8.6)
Now consider a circle with centre 𝑐(−𝑔, −𝑓) and radius 𝑟 with a point 𝑝(𝑥, 𝑦) on the
circumference of the circle. The equation (8.5) becomes
(𝑥 + 𝑔)2 + (𝑦 + 𝑓)2 = 𝑟 2 ⇒ 𝑥 2 + 𝑦 2 + 2𝑥𝑔 + 2𝑦𝑓 + 𝑘 = 0 (8.7)
where
𝑘 = 𝑓 2 + 𝑔2 − 𝑟 2 (8.8)
is some constant. Equation (8.7) is the general equation of a circle with centre 𝑐(−𝑔, −𝑓)
and of radius 𝑟 over a point 𝑝(𝑥, 𝑦). Observe from (8.8) that
𝑟 = √𝑓 2 + 𝑔2 − 𝑘. (8.9)

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Example
1. Find the equation of a circle with centre (2, −5) and radius 3.
To this we use equation (8.5) to write (𝑥 − 2)2 + (𝑦 + 5)2 = 32 ⇒ 𝑥 2 + 𝑦 2 − 4𝑥𝑦 +
10𝑦 + 20 = 0 is the equation of the circle.
2. Find the coordinates of the centre of a circle given by the equation 3𝑥 2 + 3𝑦 2 − 12𝑥𝑦 +
24𝑦 + 17 = 0.
17
Re-arrange the equation to have 𝑥 2 + 𝑦 2 − 4𝑥𝑦 + 8𝑦 + 3 = 0. Completing squares:
17
(𝑥 − 2)2 + (𝑦 + 4)2 + + (−2)2 + 42 = 0. Therefore the coordinates of the centre is
3
(2, −4).
3. Find 𝑘 if the equation of a circle is 𝑥 2 + 𝑦 2 + 10𝑥𝑦 − 8𝑦 + 𝑘 = 0. Doing this, we have
𝑥 2 + 𝑦 2 + 10𝑥𝑦 − 8𝑦 = −𝑘. Completing the squares:
(𝑥 + 5)2 + (𝑦 − 4)2 = −𝑘 + 52 + (−4)2 = 32 .
Therefore, −𝑘 = 9 − 25 − 16 ⇒ 𝑘 = 32.

4. Find the radius of a circle whose equation is 2𝑥 2 + 2𝑦 2 − 3𝑥 + 4𝑦 + 2 = 0.

3
Re-arrange and complete the squares: 𝑥 2 + 𝑦 2 − 2 𝑥 + 2𝑦 + 1 = 0
3 2 3 2 3
⇒ (𝑥 − 2) + (𝑦 + 1)2 = −1 + 12 + (− ) = 𝑟 2 . Therefore, 𝑟 = 2.
2

Practice Exercise 8.4

1. Find the equation of a circle whose centre and radius is (i). (4, −3) and √6.
(ii). (4, −5) and 3.
2. Find the coordinates of the centre of a circle whose equation is (i). 4𝑥 2 + 4𝑦 2 − 16𝑥 +
24𝑦 + 14𝑦 + 5 = 0. (ii). 5𝑥 2 + 5𝑦 2 − 20𝑥 + 30𝑦 + 10 = 0.
3. Find 𝑘 if 𝑥 2 + 𝑦 2 + 6𝑥 − 7𝑦 + 𝑘 = 0 is equation of a circle of radius 4.

136