UNDERSTANDING

SENIOR THREE MATHEMATICS

BASED ON THE ABRIDGED CURRICULUM

by

KAZIBA Stephen and Ronald DDUNGU

Key Topics covered in the book

• MAPPINGS AND RELATIONS
• VECTORS AND TRANSLATION
• NUMERICAL CONCEPT (Indices ,logarithms and Surds)
• SIMILARITIES AND ENLARGEMENT
• GEOMETRY
.

TERM ONE 2022

The only way to learn mathematics is to do mathematics.

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To the students

This book continues to help you to learn ,enjoy ,understand and progress through mathemat-
ics in the abridged curriculum.As well as a clear and concise text the book offers a wide range
of learning activities and practical questions that are relevant to the mathematics you are learning.

Exercises have been included at the end of each subsection that will help you to understand
the work, further more an assessment is included at the end of each chapter for extra practice

Remember if you dont understand something ,ask some one/friend who can explain it to you,if
you still dont understand ,ask again ,again and again.

The last three topics will be accessed online .

NOTE:This book is free ,and available on the internet for easy access

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 Chapter 0

TABLE OF CONTENTS

Table of Contents iii

1 MAPPINGS AND RELATIONS 1

1.1 Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.1.1 Representing relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Domain and Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Mapping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.3.1 Types of Mappings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.4 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.4.1 Function Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2 VECTORS AND TRANSLATION 21

2.1 Translations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.1.1 Translation in the Cartesian plane . . . . . . . . . . . . . . . . . . . . . . . 22
2.2 Vectors and Scalars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
2.2.1 Notation and representation of vectors . . . . . . . . . . . . . . . . . . . . 31
2.2.2 Position Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
2.2.3 Column Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
2.2.4 Displacement Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
2.2.5 Operations on vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3 NUMERICAL CONCEPTS (Indices , logarithms and Surds) 46

3.1 Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
3.1.1 Rounding off . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
3.1.2 Decimal Places . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
3.1.3 Significant figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
3.2 Indices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
3.2.1 Index Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
3.2.2 Laws of Indices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
3.2.3 The Zero index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
3.2.4 Negative Indices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
3.2.5 Same power Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
3.2.6 Fractional Power (or Root) Law . . . . . . . . . . . . . . . . . . . . . . . . 59
3.2.7 Solving equations involving indices . . . . . . . . . . . . . . . . . . . . . . 62

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3.3 Standard Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

3.4 Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
3.4.1 Rules of Logarithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
3.4.2 Using Logarithms for calculation . . . . . . . . . . . . . . . . . . . . . . . 70
3.5 Surds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

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 Chapter 1

MAPPINGS AND RELATIONS

Learning objectives

By the end of this topic,the learners should be able to

• Use arrow diagrams /mappings to represent relations and functions
• Identify domain and range of a mapping ♣
• Describe and distinguish between function and non function mapping
• Evaluate functions using the given domains

Introduction
In mathematics, we study relations between two sets of numbers, where members of one set are
related to the other set by a rule.A relation shows a connection(relationship) between sets of
values(numbers),items,events or people
Activity
• Write down as many as possible the biological relations you have with your family
members.
• Write down any four relations you have among your classmates.
• Are there any students in your school that you are biologically related to? If any, write
down their names and the relations you have with them.
• Write down any four mathematical relations you know.

From the activity ,you have realized that you have many relations. You have used statements
such as Sylvia is my aunt and so on. In a statement such as ‘Stephen is the uncle of Martin’;
the phrase “is the uncle of” indicate that there is a biological connection between Stephen and
Martin. This tells us that relation is a connection between two or more things.
Examples of mathematical relations
For a set A = {1, 2, 3, 4, 5, 6, 9}

• 2 is less than 3. • 2 is a factor of 6.

• 4 is greater than 2 • 2 is half of 4.
• 6 is a multiple of 3. • 9 is a square of 3

1
1.1. RELATION

1.1 Relation

A relation is a pairing of input values with output values. Relations are also described as mappings.
When we map a set of numbers onto another set of numbers, we often express the rule for the
mapping using mathematical relationships instead of words.

1.1.1 Representing relations

In this example, shown below, we define a relation between the set A={−2, 1, 3} and the set B=
{−4, 2, 6} as ‘multiply by 2’. Notice that −2 is mapped onto −4, 1 onto 2,and 3 onto 6.Therefore
we can represent the relation between sets A and B in the following ways.
• Ordered pairs
Relations can be shown as a set of ordered pairs (x,y), where x is an input and y is an
output(Input , output).The ordered pair preserves the directional property of the relation.
.For example Daniel is a brother to Timothy.The relation is a brother and the ordered pair is
(Daniel,Timothy).It is also consistent with the order of points plotted on a Cartesian Plane
represented by (x, y): (−2, −4), (1, 2) and (3, 6)
• Table
We can also use a table to show the connection between the input (x−values) and output
(y−values)

x −2 1 3
y −4 2 6

• Gaph
When representing our relation on the cartesian plane ,we plot (x, y).

Range

0
Domain
-5 -4 -3 -2 -1 0 1 2 3 4 5 6
-1

-2

-3

-4

-5

• Mapping diagram(Arrow diagram)

This is shown by drawing arrows to connect members of the set A to the members of set
B.We refer to the members of the set A as the input and members of the set B as the output.
The direction of the arrows is always from the input to the output.

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 1.2. DOMAIN AND RANGE

-2 -4

1 2

3 6

Figure 1.1: A mapping diagram(Arrow diagram)

In a mapping notation ,we describe the relationship that exists between the input and output
values.i.e What has been done to the input elements to produce the output elements.For
the example above ,our mapping notation is x −→ 2x for x = −2, 1, 3
NOTE:A relation exists between two sets of numbers if we can find a rule that maps members
of the first set (domain) onto members of the second set (codomain). The rule must hold for all
possible pairs that are connected.

1.2 Domain and Range

• Domain(D):Is the set of input values for a relation.i.e The set of all values that the first
elements in the ordered pair can take.
• Range(R):Is the set of output values for a relation.i.e The set of all values that the second
elements in the ordered pair can take
Example 1.2.1
State the domain and range of the relation {(−2, −4), (1, 2), (3, 6)}
SOLUTION
• Domain={−2, 1, 3}
• Range={−4, 2, 6}

Example 1.2.2
Consider the relation on the arrow diagram

1 4

2 7

3 10

4 13

• Write down the ordered pairs of the mapping.

• Determine the domain and range

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1.2. DOMAIN AND RANGE

SOLUTION
• Ordered pairs={(1, 4), (2, 7), (3, 10), (4, 13)}
• Domain={1, 2, 3, 4}
• Range={4, 7, 10, 13}

Example 1.2.3
For the mapping x −→ 2x + 1,x ∈ {−1, 0, 1, 2}.
(a) Construct a table of values
(b) State the range of the relation
(c) Represent the relation on an arrow diagram
SOLUTION
(a) Table of values
x 2x + 1
−1 −1
0 1
1 3
2 5
(b) Range={−1, 1, 3, 5}
(c) Arrow diagram

−1 −1

0 1

1 3

2 5
Domain Range

1.1 Exercise Set

1. State the domain and range for each of the relation shown

(a) {(−3, −2), (5, 8), (6, 9), (7, 5)} (e) {(Syson,84),(Stacy,94),(Pauline,74)}
(b) {(−1, 1), (4, 11), (5, 13), (1, 5)} (f) {(Museveni,1),(Robert,2),(Patrick,3)}
(c) {(−2, 4), (−1, 4), (0, 4), (2, 4)} (g) {(−1, 1), (0, 0), (2, 4), (3, 9)}
(d) {(1, 1), (2, 4), (3, 9)} (h) {(A,a),(B,b),(D,d),(E,e)}

2. Given set A = {0, 1, 2, 3, 4, 5} and the relation “multiply by 2 ”, Find set B, the range.

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 1.2. DOMAIN AND RANGE

3. Copy and complete this arrow digram.State the range for the relation

3x + 5
−1 2

2
Domain Range

4. For the mapping x −→ 4x ,State the missing value in each ordered pair

(a) (−3, −) (c) (−2, −) (e) (−, −4) (g) (−, 12)
(b) (−1, −) (d) (1, −) (f) (−, 8) (h) (−, 16)

5. For the mapping x −→ 4x ,

(a) State the missing value in each ordered pair

(i) {(−2, −), (0, −), (1, −), (3, −)} (ii) {(−, −4), (−, 4), (−, 8)}

(b) Represent the relations on an arrow diagram

(c) State the domain and range for the relations.

6. For the mapping x −→ 2x − 4, x ∈ {−3, 0, 2, 4}

(a) Construct the table of values for the mapping

(b) Represent the relation on an arrow diagram

(c) State the ordered pairs of the relation

(d) State the range of the relation

7. Write the ordered pairs for the relation shown on the mapping diagram

2 4

4 10

5 12

6 14

8. Consider the domain A = {0, 1, 3, 4} set A is mapped onto set B by the relation x −→ 2x.
Find the range, the ordered pair and hence draw the graph of the relation

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1.3. MAPPING

1.3 Mapping

Activity
• Write down set A of the first six whole numbers.
• Form another set B by squaring each number in set A above.
• Write the numbers in set A in an ellipse/rectangle and their squares in set B in a
different ellipse/rectangle on the same level side by side.
• Use arrows to match the numbers in set A to their squares in set B. What would you
call such matching?
• We say that squaring elements in set A maps onto set B. Your diagram showed the
mapping. Set A is called the domain(Input) and set B is called the range(Output).

1.3.1 Types of Mappings

There are four different types of mappings:

• One to One mapping:

This is a mapping in which each element in the domain is mapped onto a unique element
in the range.For example x −→ 2x + 1, x ∈ {−1, 0, 1, 2}

x −1 0 1 2
2x + 1 −1 1 3 5

Range={−1, 1, 3, 5}

−1 −1

0 1

1 3

2 5
Domain Range

• One to many mapping:

This is a mapping where an element of the domain is mapped onto more than one element
√ √ √
of the range.For example x −→ x ,x ∈ {4, 9}. 4 = ±2, 9 = ±3.Therefore the range is
{2, −2, 3, −3}

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 1.3. MAPPING

4 2

−2

−3

9 3
Domain Range

Activity
In your groups, work as pairs ,and carry out the following activity.
1. Give three names of your siblings. List the names down on a piece of paper.
2. On the domain set A, write the names of each one of you in the group.
3. On the range set B, write the six names of the siblings listed in step 1 above.
4. Match the name on the domain set A to the names of the siblings in the range
set B with arrows.
5. What name would you call such a mapping?

• Many to one mapping:

Activity
Some S2 students of Taibah international school obtained the following grades:
Name Keziah Akasha Priscilla Mimmi Philemon Stacy
Grade D1 C3 D2 D1 C3 D1
1. Use the names as the domain and the grades as the range to show the mapping
on a mapping diagram.
2. What name would you call such a mapping?
3. From the activity, you have observed that several elements in the domain set can
be mapped onto one element in the range set

Many to one is a mapping where more than one element of the domain is mapped onto the
same element of the range.i.e Two or more elements of the domain have the same range.For
example x −→ x2 ,x ∈ {0, 1, −2, 2}.

x 0 1 2 −2
x2 0 1 4 4

Range={0, 1, 4}

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1.3. MAPPING

0 0

1 1

2 4

−2
Domain Range

• Many to many mapping:

This is a mapping where more than one element of the domain is mapped onto more than
one element of the range.For example in the mapping Is a factor of ,Is a multiple of.

2 4

4 10

5 12

6 14
Domain Range

Example 1.3.1
For each of the examples below, state the type of mapping.

16 4 16 4 0 0

25 5 −4 1 1

36 6 −5 3 9

49 7 25 5 −3
A B A B A B
(i) (ii) (iii)

SOLUTION
(i) One to one mapping.This is because each member of B is related to only one member
of A
(ii) One to many mapping.This is because members of A are mapped onto more than one
element of B.
(iii) Many to one mapping.This is because more than one member of A is mapped onto the
same element of B.

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 1.3. MAPPING

Example 1.3.2
A relation, A is defined by the set of ordered pairs; (−3, 2), (−2, 4), (0, 5), (−2, 6)
(i) List the members of the domain.
(ii) List the members of the range.
(iii) What type of mapping is A?
SOLUTION
(i) {−3, −2, 0}
(ii) {2, 4, 5, 6}
(iii) First draw a mapping digram

−3 2

−2 4

0 5

6
Domain Range

One to many mapping.This is because members of the domain are mapped onto more
than one element of range.

1.2 Exercise Set

1. Draw a mapping diagram for x −→ x2 for x ∈ {−2, −1, 0, 1, 2}.

(a) Which type of mapping is it?
(b) State the range of the relation.
2. Copy and complete this arrow digram.

1 2
2 4
3 5
4 6
Domain Range

(a) Show the relation is a factor of

(b) State the type of mapping
3. Set A={1, 2, 3, 4, 5} is mapped onto set B by the relation “multiply by 4”.

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1.3. MAPPING

(a) List the elements of set B

(b) Map A onto B.

(c) State the type of mapping

4. A set A maps onto set B by the operation “multiply by 3 and add 1”. The elements of
set A are {5, 6, 7, 8, 9}

(a) List the element of set B.

(b) Map set A onto B.

(c) What type of mapping is this?

5. Copy and complete this arrow digram.

−4 16

−3 9

0 −9

3 0
D R

(a) Show the relation x −→ x2

(b) State the type of mapping

6. Some S2 students of Taibah international school obtained the following marks in a math
test:

Name Daniel Tendo Christopher Hannah Nicole Joshua

Grade 80 78 84 80 92 84

(a) Illustrate the mapping of marks to students.

(b) What type of mapping is this?

7. Show the mapping of the relation “is a factor of” for the sets A= {2, 3, 4, 5, 6} and B =
{30, 32, 33, 34, 35, 36}.

8. Consider the domain A = {0, 1, 3, 4} set A is mapped onto set B by the relation x −→ 3x.
Find the range, the ordered pair and hence illustrate the relation on an arrow diagram
BE WARE!!
Do you know that one to many or many to many sexual relationships between members
of the opposite sex can spread HIV/AIDS? You are advised to stay away from sex before
marriage.

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 1.4. FUNCTIONS

1.4 Functions

A function is a relation in which for each member of the Domain there is a single corresponding
member of the Range. OR A function is a type of mapping in which every object on the domain has
one and only one image in the range. Mappings, which are functions, are one to one mappings
and many to one .A function involves two sets(Ordered pairs) and a rule of correspondence
between them.The rule specifies how to pair the elements of one set with those in the other set.

For any mapping/relation to be a function

• Every member(element) of the domain is mapped onto one and only one member of
the range
• An input cannot have more than one output.
• Two or more members of the domain can be mapped onto the same member of the
range

DID YOU KNOW?

In the real world, we may think of a function as a mapping onto the set of sons to a
corresponding set of biological mothers . Each son will be associated with one and only one
mother, and two or more sons can be associated with the same mother but one son cannot
be associated with two or more mothers.

Example 1.5.1
For each of the mapping diagrams below, state whether it represents a function and if not
why.

16 4 16 4 0 0

25 5 −4 1 1

36 6 −5 3 9

49 7 25 5 −3
Domain Range Domain Range Domain Range
(i) (ii) (iii)

SOLUTION
(i) This is a function because each object in the domain is mapped onto only one image
in the range.This is a one to one mapping,and it meets the conditions for a mapping
to be a function.
(ii) This is not a function because each object in the domain is mapped onto more than
one image in the range.i.e 16 is mapped onto 4,−4 ,25 is mapped onto 5,−5.This is
a one to many mapping, and so does not meet the conditions for a mapping to be a
function.

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1.4. FUNCTIONS

(iii) This is a function because different objects in the domain are mapped onto only one
image in the range.This is a many to one mapping,and it meets the conditions for a
mapping to be a function.

Example 1.5.2
A mapping is defined by {(−1, 2), (4, 6), (5, 0), (4, 3)}.State whether it represents a function
and if not why.
SOLUTION

−1 2

4 6

5 0

3
Domain Range

This is not a function because an object in the domain is mapped onto more than one
image in the range.i.e 4 is mapped onto 6,and 3 .This is a one to many mapping.

Example 1.5.3
State whether the mapping diagram below represents a function and if not why.

−1 a

4 b

5 c

−3
Domain Range

SOLUTION
This is not a function because at least one object in the domain is not associated to
any image in the range.i.e 5 is not assigned to an element in the range.

1.3 Exercise Set

1. A mapping is defined by {(−1, 1), (4, 11), (5, 13), (1, 5)}.

(a) Represent the mapping on an arrow diagram

(b) Which type of mapping is it?

(c) Is the mapping a function?

2. For the mapping x −→ 2x − 4, x ∈ {−3, 0, 2, 4}

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 1.4. FUNCTIONS

(a) Represent the relation on an arrow diagram

(b) State the type of mapping
(c) Is the mapping a function?
3. For each of the mapping diagrams below, state whether it represents a function and if not
why.

1 W 2 2 0 A
4
2 X 3 1 B
6
3 Y 4 9 3 C
15
4 Z 16 −3 D
Domain Range Domain Range Domain Range
(i) (ii) (iii)

4. Do these mapping diagrams represent functions ? Give reasons for your answers.

a d a p

b e q
b
c f c r
(a)
d s
(c)

-1 2

a 4 6
d

e 5 0
b

c f 3
(b) (d)

1.4.1 Function Notation

We can describe a function using mathematical notation, written as f(x).We read f(x) as "f
of x" or "f at x" since f(x) gives the value of f at x.The notation f(x) represents the second
element in the ordered pair that has x as its first element.The ordered pair can be represented
as (x, f (x)).Often we replace f(x) by y and write y = f (x).For example f (x) = 2x + 3 may be
written as y = 2x + 3

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1.4. FUNCTIONS

Mappings that represent functions can be written as follows;

f
• x→
− 2x + 3:Read as the function f maps x onto 2x + 3
• f : x −→ 2x + 3 : Read as a function f such that x is mapped onto 2x + 3
• f (x) = 2x + 3 :An algebraic formula that gives the values of the Range given particular
values of x in the domain.

Example 1.5.1.1
Given the function f (x) = 2x + 3, find
(a) f (0)
(b) f (1)
(c) f (−1)
 
1
(d) f 2

SOLUTION
(a) f (0)
f (0) means we Substitute 0 for x in the function
f (x) = 2x + 3
f (0) = 2 × 0 + 3 Substituting 0 for x
f (0) = 0 + 3
f (0) = 3
(b) f (1)
f (0) means we Substitute 1 for x in the function
f (x) = 2x + 3
f (1) = 2 × 1 + 3 Substituting 1 for x
f (1) = 2 + 3
f (1) = 5
(c) f (−1)
f (0) means we Substitute 0 for x in the function
f (x) = 2x + 3
f (−1) = 2 × −1 + 3 Substituting −1 for x
f (−1) = −2 + 3
f (−1) = 1

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 1.4. FUNCTIONS
 
1
(d) f 2
f (x) = 2x + 3
1 1 1
 
f =2× +3 Substituting for x
2 2 2
1 1
= 2

 × 1 + 3
2


=1+3
=4

Example 1.5.1.2
Given that f : x −→ 6x − 2 .Find

 
(a) f (0) (b) f (2) (c) f (−4) (d) f 1
2

SOLUTION
f : x −→ 6x − 2 can be written as f (x) = 6x − 2
(a) f (0)
f (x) = 6x − 2
f (0) = 6 × 0 − 2
f (0) = 0 − 2
f (0) = −2
(b) f (2)
f (x) = 6x − 2
f (2) = 6 × 2 − 2
f (2) = 12 − 2
f (2) = 10
(c) f (−4)
f (x) = 6x − 2
f (−4) = 6 × −4 − 2
f (−4) = −24 − 2
f (−4) = −26
 
1
(d) f 2
f (x) = 6x − 2
1 1
 
f =6× −2
2 2
1
  3 1
f = 6

 × 1 − 2
2
2


1
 
f =3−2
2
1
 
f =1
2

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1.4. FUNCTIONS

Example 1.5.1.3
Given that g(x) = 2x + 3, find the range g(x),given the domain {−1, 1, 4, 5}

SOLUTION
The range={g(−1), g(1), g(4), g(5)}.

(a) g(−1) (b) g(1) (c) g(4) (d) g(5)

g(x) = 2x + 3 g(x) = 2x + 3 g(x) = 2x + 3 g(x) = 2x + 3
g(−1) = 2 × −1 + 3 g(1) = 2 × 1 + 3 g(4) = 2 × 4 + 3 g(5) = 2 × 5 + 3
g(−1) = −2 + 3 g(1) = 2 + 3 g(4) = 8 + 3 g(5) = 10 + 3
g(−1) = 1 g(1) = 5 g(4) = 11 g(5) = 13

The Range ={1, 5, 11, 13}

Example 1.5.1.4
Given that h(y) = y 2 + 3y, find h(2)

SOLUTION
h(y) = y 2 + 3y
h(2) = 22 + 3y
h(2) = 2 × 2 + 3 × 2
h(2) = 4 + 6
h(2) = 10

Example 1.5.1.5
The function f (x) = bx2 − 2 and f (2) = 18. Find:
(a) the value of b
(b) f (2)
(c) f (0)

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SOLUTION
(a) (b) f (2)
2
f (x) = bx − 2 f (x) = 5x2 − 2
f (2) = 18 f (2) = 5 × 22 − 2
f (2) = b × 22 − 2 f (2) = 5 × 4 − 2
f (2) = 4b − 2 f (2) = 20 − 2
4b − 2 = 18 f (2) = 18
4b − 2 + 2 = 18 + 2
4b = 20 (c) f (0)
4b 20 f (x) = 5x2 − 2
=
4 4 f (0) = 5 × 02 − 2
5
4
b 20
 
> f (2) = 5 × 0 − 2
= 1
4
f (2) = 0 − 2
4


b=5 f (2) = −2

Example 1.5.1.6
1
If f (x) = 2x−10
,find the value of :
(i) f (6)
(ii) x for which f (x) is undefined

SOLUTION
(i) f (6)
1
f (x) =
2x − 10
1
f (6) =
(2 × 6) − 10
1
f (6) =
12 − 10
1
f (6) =
2
(ii) x for which f (x) is undefined
A function is undefined or meaningless if its denominator part is equal to zero.
2x − 10 = 0
2x − 10 + 10 = 0 + 10
2x = 10
2x 10
=
2 2
5
2
x 10
 
>
= 1
2

2


x=5

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1.4 Exercise Set

1. Given that h(x) = 6x + 2, Find

(a) h(−4) (b) h(0) (c) h(1) (d) h(3)

2. The function f (y) = 4y 2 + 3y.Find

(a) f (0) (c) f (1) (e) f ( 31 ) (g) f (7)

(b) f (−2) (d) f (0.5) (f) f (−1) (h) f (4)

3. Given that g : x −→ 31 x − 4 , find

 
1
(a) f (0) (b) f (9) (c) f (−18) (d) f 2

4. Determine the range corresponding to the domain {−1, 1, 4, 5} for the function g(x) = 2x+3.
5. Given that g(x) = 5(x2 + 4).Find

(a) g(−4) (b) g(0) (c) g(1) (d) g(3)

6. Given that h(x) = 6x + 4, find the range h(x),given the domain {−2, 0, 3, 7}
7. The function f (x) = 4x − 2 and f (x) = 6.Find the value of x
8. Given h(x) = 5x + 4, If f (a) = 14, find the value of a.
x−4
9. Given that f (x) = 5
.Find

(a) f (24) (b) f (−6) (c) f (5) (d) f (−16)

10. If f(x) = 2 − 21 x has the domain {−2, 0, 2, 4, 6}, find the range.
y 2 −1
11. If h(y)= y+1
, find:

(a) h(0) (b) h(2) (c) h(−4) (d) h(−5)

12. The function h(x) = 21 x2 + p and h(−2) = 6. Find the value of p.

Summary

1. Relation−It is a connection between two or more numbers or things.

2. Domain−It is the set of input values for a mapping
3. Range−It is the output set in mapping.
4. Papygram−It is a circular representation of relationships that exist between a given set ♣
of things.
5. Function−It is a relation in which for each member of the Domain there is a single
corresponding member of the Range.

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ASSESSMENT
1. Given that B={(2, 1), (4, 5), (6, 7), (8, 9)}.Determine
(a) the domain and range
(b) Whether B is a function and if not why.
(c) the type of mapping
2. Ethan,Joel,Daniel and Martin like the following types of foods:matooke,rice,meat and
matooke respectively.
(a) List the elements of the domain and range of the relation "likes"
(b) Draw an arrow diagram to illustrate the relation
(c) Identify the type of mapping
3. Given that f (x) = 4x − 3,find

 
(i) f (2) (ii) f (−1) (iii) f 1
2

4. Consider the function f (x) = 2x. If the domain is {0, 1, 2, 3, 4} .Find the range of
values. Hence, draw the arrow diagram.
5. Given that h(y) = 4y − 2,find the value of
(i) h(−2)
(ii) y when h(y) = 6
6. The function is defined as f : x −→ 3 − 2x.Determine the range if the domain is
{0, 1, 2, 3}.
7. If g(x) = qx + 3 and g(5) = 23,Find the value of

(i) q (ii) g(0) (iii) g(−5)

8. (a) For the mapping x −→ 4x + 5,find the domain when the range is {1, 13}
(b) The function f (x) = 17 − 5x and g(x) = 3x − 7 .Find the value of x such that
f (x) = g(x)
9. Set A maps onto set B by the operation “divide by 2 add 3".If set A = {4, 8, 12, 16, 20, 24}
(a) list the elements of set B.
(b) Find the ordered pairs.
(c) Draw the graph of the relation.
5x
10. If f (x) = x2 −9
,find t:
(i) f (2)

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(ii) the values of x for which f (x) is meaningless

11. Find the unknown values in the arrow diagram for the mapping x −→ 2x + 3
x → 2x + 3

−1 p

a 11

1 c

q 13
D R

End

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 Chapter 2

VECTORS AND TRANSLATION

Learning objectives

By the end of this topic,the learners should be able to

• Define translation with a vector
• Identify scalars and vectors ♣
• Use vector notation
• Represent vectors both single and combined geometrically

Introduction
In chapter 10 of Book one ,we saw how objects can be reflected on different mirror lines.In this
chapter we will look at another type of transformation.

2.1 Translations

A translation is a transformation that moves every point in a figure the same distance in the same
direction.Translation deals with movement of an object to a new position.This can be in form of
moving a shape up, down or from side to side but it does not change its appearance in any other
way.
Activity
1. Trace the figures below on a tracing paper
B1

A1
C1

A C

2. Draw the line segment joining A to A1 as shown above.

3. Slide the tracing using line AA1 as a guide line, to ensure that A moves onto A1 in a
straight line.(You can as well make a cut out)

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2.1. TRANSLATIONS

4. When A coincides with A1 , stop the slide. What do you notice about the positions of
B and C?
5. What do you notice about the new position of the triangle ABC? What can you say
about the two triangles?

From the activity we notice that each point on triangle ABC has moved the same distance and
in the same direction. The process that moves triangle ABC onto triangle A1 B1 C1 is called
translation.

Properties of Translation

1. All the points on the object move the same distance.

2. All the points move in the same direction.
3. The object and the image are identical(same) and they face the same direction.Hence,
♣
they are directly congruent(same shape,size,and angles )
4. A translation is fully defined by stating the distance and direction that each point
moves.

2.1.1 Translation in the Cartesian plane

Activity
y-axis

4
B
3

1
A C
0
x-axis
-5 -4 -3 -2 -1 0 1 2 3 4 5 6
-1
P
-2

-3

-4
Q R
-5

1. Move the vertex A of triangle ABC ,4 steps to the left and 2 steps up.Name the new
position A1
2. Write the coordinates of the image A1

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3. Move the vertex B and C of triangle ABC ,4 steps to the left and 1 step up.Name the
new positions B1 and C1 respectively.
4. Join A1 ,B1 and C1 to form the image of triangle ABC.
5. Move each vertex of triangle PQR ,3 steps in the x -direction and −1 step in the y
direction.
6. Write the coordinates of the triangle P1 Q1 R1 formed

NOTE
• If each point moves distance a in the x-direction(horizontal distance)
  and distance b
in the y-direction(vertical distance),we use the ’vector’ notation ab to describe the
translation.
 
• A translation T= ab means that an object is moved a distance a in the x-direction and
a distance b in the y-direction
 
−4
• From the previous activity ,the Translation of triangle ABC is 2
 
• A translation T= ab moves point A(x, y) to a new position A1 (x + a, y + b) Thus
Translation + object = image .

Activity
y-axis

2
U
1

0
x-axis
-5 -4 -3 -2 -1 0 1 2 3 4 5 6
-1
T
-2

-3

-4

1. What is the vector that translates T to U

2. What is the vector that translates U to T

Example 2.1
(a) Draw the triangle ABC with corners at the points with coordinates (1,1),(2,3) and
(4,1) respectively.

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−4
(b) The triangle is translated along the vector 1
. Draw the new triangle obtained by
the translation.

SOLUTION
For this translation each point should be moved 4 steps to the left(-4 spaces in the x -
direction) and 1 step up(1 space in the y- direction) .
y-axis

5
1
B
4
B
3

2
A1 C1
1
A C
0
x-axis
-5 -4 -3 -2 -1 0 1 2 3 4 5 6
-1

-2

Example 2.2
(a) Draw the square ABCD with vertices A(1,1),B(4,1) , C(4,4) and D(1,4)respectively.
 
(b) The square is translated by the displacement vector 52 . Draw the new triangle
obtained by the translation and write down the coordinates for the images.

SOLUTION
For this translation each point should be moved 5 steps to the left(5 spaces in the x -direction)
and 2 step up(2 spaces in the y- direction) .
y-axis

7
D1 C1
6

5
D C
4

3
A1 B1
2

1
A B
0
x-axis
0 1 2 3 4 5 6 7 8 9 10

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The coordinates of the images are A1 (6,3),B1 (9,3),C1 (9,6) and D1 (6,6)

Example 2.3
 
5
A translation T= 2
maps the points P(3, 7) onto the point P1 . Find the coordinates of P1 .

SOLUTION
Image = Translation + Object
! !
51 3
P = +
2 7
!
5+3
P1 =
2+7
!
1 8
P =
9
1
The coordinate of the image is P (8,9)

Example 2.4
A triangle
  with vertices A(1, 1) B(2, 3) and C(4, 1) is mapped onto its image by a translation
T= −41
Find the coordinates of the image of the triangle ABC.

SOLUTION
Image = Translation + Object

A1 = T + A B1 = T + B C1 = T + A
! ! ! !
−4
!
1
!
1 −4 2 1 −4 4
A1 = + B = + C = +
1 1 1 3 1 1
! !
−4 + 1
! −4 + 2 −4 + 4
A1 = B1 = C1 =
1+1 1+3 1+1
! !
−3
!
1 −2 1 0
1
A = B = C =
2 4 2

The coordinates of the image of the triangle are A1 (−3, 2),B1 (−2, 4),C1 (0, 2)

Example 2.5
 
−4
A translation T= 1
maps point P onto P1 (0, 2) . Find the coordinates of point P

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SOLUTION
Translation + Object = Image
T + P = P1
! !
−4 0
+P=
1 2
! !
0 −4
P= −
2 1
!
0 − −4
P=
2−1
!
4
P=
1
The coordinate of the object is P(4,1)

Example 2.6

The diagram below shows the shapes A, B, C and D . Along what vector would you translate:

(a) A to B (c) D to B (e) D to C

(b) A to C (d) C to A (f) A to D
y-axis

6
A B
5

0
x-axis
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
-1
D
-2
C
-3

-4

-5

-6

-7

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SOLUTION
y-axis

 7
10
0
6
A B
5

0
x-axis
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
-1
D
-2
C
-3

-4

-5

-6

-7

(a) A to B. (d) C to A
10 steps to the right !and 0 step up. 9 steps to the left and! 8 steps up.
10 −9
0 8
(b) A to C (e) D to C
9 steps to the right and
! 8 steps down. 7 steps to the right and
! 1 step down.
9 7
−8 −1
(c) D to B (f) A to D
8 steps to the right and
! 7 steps up. 2 steps to the right and
! 7 steps down.
8 2
7 −7

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2.1 Exercise Set

1. (a) Draw the triangle which has corners at the points with coordinates (5, 3), (9, 3) and
(7, 6).

 
−2
(b) Translate the triangle along the vector −1

(c) Write down the coordinates of the corners of the translated triangle.

 
2. The square PQRS is translated by the vector −2
−3
.Copy the diagram and show on it the
1 1 1 1
square P Q R S under this translation hence write the coordinates for the images of the
square.

y-axis

7
S R
6

3
P Q
2

0
x-axis
0 1 2 3 4 5 6 7 8 9 10

3. The following diagram shows the shape A which is translated to give the shapes B, C, D
and E:Write down the vector that describes the translation from:

(a) A to B (d) A to E (d) B to A

(b) A to C (e) E to A (e) C to A
(c) A to D (f) E to D (f) E to B

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y-axis

5
B
4

2
C A
1

0
x-axis
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
-1

-2
D
-3

-4

-5

-6
E
-7

4. (a) Join the points with coordinates (1, 1), (2, 3) and (5, 4) to form a triangle. Label this
triangle A.

(b) Translate the triangle A along the vector:

   
2 −1
(i) 1
,to obtain triangle B (iv) −2
,to obtain triangle E
   
1 0
(ii) 3
,to obtain triangle C (v) −1
,to obtain triangle F
   
2 2
(iii) −5
,to obtain triangle D (vi) 0
,to obtain triangle G

5. (a) Draw the triangle, A, that has corners at the points with coordinates (−7, −2), (−5, −5)
and (−4, −2).
 
4
(b) Translate this shape along the vector 8
to obtain B.

(c) Describe the translation that would take B to A

6. Triangle ABC has vertices A (0, 0),B (5, 1) and C ( 1, 3). Find the coordinates of the points
A1 , B1 and 1
 C ,the images of A, B and C respectively, under a translation with displacement
vector 25

7. The image of a rectangle ABCD has vertices A1 , B1 , C1 and 1

  D at the points (5, 10), (9,
10) (9, 8), (5, 8) respectively. If the translation vector is 25 , find the coordinates of the
object rectangle.

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8. A parallelogram has corners at the points A, B, C and D. The points A, B and C have
coordinates (1, 2), (2, 5) and (5, 3) respectively.
(a) Draw the parallelogram.
(b) State the coordinates of the fourth corner, D.
(c) Describe the translation that moves A B onto D C.
(d) Describe the translation that moves A D onto B C.
9. The shape A has corners at the points with coordinates (4, 2)(4, −1)(6, −3) and (6, 0).
(a) What is this shape?
 
4
(b) The shape is translated along the vector −2
to give shape B and then shape B is
 
−5
translated along the vector 2
to give C.Draw A, B and C.
(c) What translation would take A straight to C ?
10. A translation T maps point P(2, 5) onto P1 (3,2) Find the image of Q(5, 7) under translation
T

2.2 Vectors and Scalars

Activity
1. A tourist arrived at Entebbe international airport and is to visit the source of the nile
in jinja town.
(a) What two aspects of the journey must he know?
(b) What is the name of the quantity that has these two aspects?
2. In pairs, roughly estimate the following :
(a) The distance between your home and the nearest shopping centre.
(b) The direction of your home from the nearest shopping centre.
(c) How did you estimate the direction in (b) above?
3. The distance from Kampala to Jinja is about 80km. Stephen drove from kampala to
Jinja and back.
(a) What is the total distance covered by Stephen?
(b) What is the total displacement?
(c) Are your answers for part(a) and (b) the same .If yes why and if they are different
explain the cause.

From the activity we have noticed that vectors are described using length and direction.

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Summary
1. A vector is a physical quantity that is described by magnitude(size or length) and
direction.
2. Examples of vectors include.displacement,velocity,force,acceleration etc
3. A scalar is a physical quantity that is described by only magnitude(size or length)
4. Examples of scalars include.distance,speed,time,etc

2.2.1 Notation and representation of vectors

• Geometrically, a vector is represented using a directed line segment, whose length is pro-
portional to the magnitude(size,length) of the vector and with an arrow indicating the
direction.The direction of the vector is from its tail to its head.

Tail Head

• In mathematics Vectors are denoted in any of the following ways.

−→ −−→
– Capital letters with over right arrows e.g AB, CD.
– Bold letters e.g AB, CD, a
– underaccent (Tilde)e.g a, b
˜ ˜
−→
• AB means a vector starting at A ending at B.So the arrow points towards B

A B

2.2.2 Position Vector

This is a vector that defines/describes the position of a point with reference to the origin.e.g
−→
OA = xy is the position vector of point A(x,y).

y-axis

4 A(5,4)

0 x-axis
1 2 3 4 5 6

From the origin, A is 5 units in the x- direction and 4units in the y -direction. Thus, A has
−→ −→ 5

coordinates (5 , 4) and OA has a position vector OA = 4
NOTE:All position vectors have O as their initial point.

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Activity
In pairs write the position vectors of point A,B,C and D
y-axis

B 5

3 A

0 x-axis
-5 -4 -3 -2 -1 1 3 5
-1

-2 D

-3

C -4

2.2.3 Column Vector

Column vectors  are vectors that

 awritten
 in a column form representing a change in x and y
x −4 5
directions.i.e y for example. 1 , 2

Activity
Describe each translation using a column vector
y-axis
C
7
B
6

3 D

2 A

1 E

0
x-axis
0 1 2 3 4 5 6 7

−→ −→ −→
1. AB 4. DE 7. DA
−→ −→ −→
2. BC 5. EA 8. CA
−→ −→ −→
3. CD 6. AE 9. DB

2.2.4 Displacement Vector

A displacement vector defines the distance that is moved in a specified direction.A displacement
vector is represented by a directed line segment and doesnot start from the origin.e.g AB,PQ,CD

Expressing displacement vectors in terms of position vectors

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y-axis

4 B

2 A

0
x-axis
0 1 2 3 4 5 6

AB is the displacement vector between A and B

y-axis

0 x-axis

−→ −→ −−→
OA + AB = OB
−→ −→ −→ −−→ −→
OA − OA + AB = OB − OA
−→ −−→ −→
AB = OB − OA
−→ −−→ −→
AB = OB − OA

2.2.5 Operations on vectors

Addition of vectors

• To add two vectors we add the corresponding numbers

   
x1 x2
• If a= y1
and b= y2
,then
! !
x1 x2
a+b= +
y1 y2
!
x1 + x2
a+b=
y1 + y2

Example 2.7
   
3 4
If a= 4
and b= 6
,find a+b

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2.2. VECTORS AND SCALARS

SOLUTION
! !
3 4
a+b= +
4 6
!
3+4
a+b=
4+6
!
7
a+b=
10

Subtraction of vectors

• To subtract two vectors we subtract the corresponding numbers

   
x1 x2
• If a= y1
and b= y2
,then
! !
x1 x2
a−b= −
y1 y2
!
x1 − x2
a−b=
y1 − y2
Example 2.8
   
3 4
If a= 4
and b= 6
,find a−b

SOLUTION
! !
3 4
a−b= −
4 6
!
3−4
a−b=
4−6
!
−1
a−b=
−2

Scalar Multiplication and Division of vectors

 
x
• A scalar k multiplied by vector a= y
is treated as follows:
!
x
ka = k
y
!
kx
ka =
ky
 
x
• A scalar k divided by vector a= y
is treated as follows:
!
1 1 x
a=
k k y
x
!
1 k
a= y
k k
 
1
• For a= 2
,we can represent 2a on a diagram as shown below.

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y-axis

3 2a

2 a

0
x-axis
0 1 2 3 4 5 6 7

Note: A scalar is just a number(constant value)

Example 2.9
   
4 2
If a= 6
and b= −3
,find

(a) 2a 1 (c) 2a + 3b
(b) 2
a

SOLUTION
(a) 2a !
4
2a = 2
6
!
2×4
2a =
2×6
!
8
2a =
12
1
(b) 2
a !
1 1 4
a=
2 2 6
4
!
1 2
a= 6
2 2
!
1 2
a=
2 3

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(c) 2a + 3b ! !
4 2
2a + 3b = 2 +3
6 −3
! !
2×4 3×2
= +
2×6 3 × −3
! !
8 6
= +
12 −9
!
8+6
=
12 + −9
!
14
2a + 3b =
3

Magnitude(modulus,length) of a vector

Activity
An insect moved from a point A, 10cm due East to point B. At point B,it turned north and
moved 8cm to point C . However, there is a direct route from A to C.
(a) Sketch the diagram showing the insect’s movement.
(b) Find the shortest distance from A to C.
(c) If the distance from A to B is x units and the distance from B to C is y units. Express
the distance A to C in terms of x and y.
 
x
• Consider the vector vector OB= y

y-axis
B
OB 2 = OA2 + AB 2 Pythagoras theorem
OB 2 = x2 + y 2
a y
q
OB = x2 + y 2
The magnitude orlength or modulus of
√
the vector OB= xy is |OB| = x2 + y 2
0 x A x-axis

Example 2.10
   
3 3
If a= 4
and b= 6
,find
(a) |a|
(b) |a − b|
(c) |4b|

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SOLUTION
(a) |a| q (c) |4b| !
|a| = x2 + y2 3
√ 4b = 4
6
= 32 + 42
√
!
4×3
= 9 + 16 =
√ 4×6
= 25 12
!

|a| = 5units 4b =
24
q
(b) |a − b| |4b| = x2 + y 2
3 3
! !
√
a−b= − = 122 + 242
4 6 √
! = 144 + 576
3−3 √
a−b= = 720
4−6
|4b| = 26.8328units
!
0
a−b=
−2
q
|a − b| = x2 + y 2
q
= 02 + (−2)2
√
= 0+4
√
= 4
|a − b| = 2units

Example 2.11
(a) Plot the points A (2, 3) and B(7,4) and show vector AB .
(b) Write down the column vector AB .

SOLUTION
y-axis

8
B
7
 
5
6 4

5 4

4
A
3
5
2

0 x-axis
1 2 3 4 5 6 7 8

 
5
There fore AB = 4

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Example 2.12
 
−3
Given r= 4
,find
(a) |r|
(b) |2r|
(c) 2|r|

SOLUTION
(a) |r| q (b) |2r| !
|r| = x2 + y2 −3
2r = 2
q 4
= (−3)2 + 42 !
√ 2 × −3
= 9 + 16 =
√ 2×4
= 25
!
−6
=
|r| = 5units 8
q
|2r| = x2 + y 2
q
= (−6)2 + 82
√
= 36 + 64
√
= 100
|2r| = 10units
(c) 2|r| q
2|r| = 2 (−3)2 + 42
√
= 2 9 + 16
√
= 2 25
2|r| = 2 × 5
2|r| = 10units

NOTE:|2r| = 2|r| ,therefore in general, |kr| = k|r|.

Example 2.13
Given the points A(2, 2) and B(6, 4), find the column vector AB

SOLUTION
Method 1

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y-axis

4   B
4
2
3 2

2 A
4
1

0 x-axis
1 2 3 4 5 6

 
4
There fore AB = 2

Method 2
y-axis

4 B

2 A  
6
 
2
2 4
1

0 x-axis
1 2 3 4 5 6

AB = OB − OA
! !
6 2
AB = −
4 2
!
6−2
AB =
4−2
!
4
AB =
2

Example 2.14
 
a
Write each of the following vectors in the form b
−→
(a) AB
−−→
(b) BC
−→
(c) AC
HINT :Draw a line to join the points

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y-axis
B
9

8
A
7

2
C
1

0 x-axis
1 2 3 4 5 6 7 8 9

SOLUTION
y-axis y-axis y-axis
B B B
9   9 9
7
2
8 2 8 8
A 7 A A
7 7 7

6 6 6
 
−4
5 5 5 −7 −7
 
3
4 4 −5 −5 4

3 3 3

2 2 2
C 3 C C −4
1 1 1

0 x-axis 0 x-axis 0 x-axis

1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9

−→  
(a) AB = 72
−−→  
(b) BC = −4
−7
−→  3 
(c) AC = −5

Example 2.15
   
3 4
If OA= 4
and OB= 6
,find the column vector AB

SOLUTION
AB = OB − OA
! !
4 3
AB = −
6 4
!
4−3
AB =
6−4
!
1
AB =
2

Example 2.16
   
2 −5
Given that vectorPQ= 6
and OP= 1
,find the column vector OQ

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SOLUTION
PQ = OQ − OP
OQ = PQ + OP
! !
2 −5
OQ = +
6 1
!
2 + −5
OQ =
6+1
!
2−5
OQ =
6+1
!
−3
OQ =
7

Example 2.17
   
3 −1
Given that vectorOF= 8
and OG= −4
,find the column vector

(a) FG (b) |FG| (c) GF (d) |GF|

SOLUTION
(a) FG (c) GF
FG = OG − OF GF = OF − OG
! ! ! !
−1 3 3 −1
FG = − GF = −
−4 8 8 −4
! !
−1 − 3 3 − −1
FG = GF =
−4 − 8 8 − −4
! !
−4 3+1
FG = GF =
−12 8+4
!
4
GF =
12
(b) |FG| q (d) |GF| q
|FG| = x2 + y2 |GF| = x2 + y 2
q √
= (−4)2 + (−12)2 = 42 + 122
√ √
= 16 + 144 = 16 + 144
√ √
= 160 = 160
|FG| = 12.649units |GF| = 12.649units

NOTE:In vectors FG6=GF

Example 2.18
   
2 3
If a= 1
and b= −4
,solve the equations below to find the column vector x

(a) a + x = b (b) 2x + a = b

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SOLUTION
(a) a + x = b
a+x=b
! !
2 3
+x=
1 −4
! ! ! !
2 2 3 2
− +x= −
1 1 −4 1
!
3−2
x=
−4 − 1
!
1
x=
−5
(b) 2x + a = b
2x + a = b
! !
2 3
2x + =
1 −4
! ! ! !
2 2 3 2
2x + − = −
1 1 −4 1
!
3−2
2x =
−4 − 1
!
1
2x =
−5
!
2x 1 1
=
2 2 −5
1
!
2
x 2
= −5
2
2
!
0.5
x=
−2.5

2.2 Exercise Set

1. State the position vectors of the following points.

(a) A(2, 3) (b) B(−6, 8) (c) P(−3, −4) (d) Q(2, 3)

2. Plot the positions of the points A, B, C, D, E and F relative to a point O if:

−→   −→   −−→  
(a) OA = 32 (c) AC = −3
2
(e) CE = −4
−1
−−→  5  −−→   −−→  
(b) OB = −2 (d) BD = −2
−2
(f) DF = 14

Write the vector EF as a column vector.

3. Using the diagram below describe the following translations using column vectors .

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y-axis
B
9

8
A C
7

4 F
D
3

2
E
1

0 x-axis
1 2 3 4 5 6 7 8 9

−→ −−→ −−→ −−→

(a) AB (d) BE (g) CD (j) F B
−→ −−→ −−→ −→
(b) AC (e) EB (h) DC (k) EF
−−→ −−→ −−→ −→
(c) DE (f) AD (i) BC (l) BA

4. A displacement starts from A(3, 3) and ends at B(-2,4).Find the column vector AB
5. Given the points A(2, 3) and B(4, −1).Find
(a) the position vectors for each of the points
(b) the column vector AB
(c) the column vector BA
     
1 4 −2
6. If a= 3
,b= −1
and c= −2
,solve the equations below to find the column vector d

(a) d−c=a (d) 3a + 2d = c

(b) d + b = c (e) 3a + 2d = 4b
(c) a − 2d = 4c (f) 3b + 2d = c
   
5 −3
7. Given that vectorOP= 3
and OQ= −2
,find the column vector PQ
     
1 4 −2
8. If OM= 3
,ON= −1
and OP= −2
,Find the column vector

(a) MN (b) MP (c) NM (d) MN

   
4 2
9. Given that vectorOA= −3
and AB= 5
,find the position vector OB
     
4 3 0
10. If a= 7
,b= −5
and c= 4
,find

(a) a + b (e) c − a (i) |2a + 3b|

(b) a + b + c (f) 3a
(j) |2b − a|
(c) a − b (g) −2b
(d) b − a (h) 4c (k) |5c − 3a|
 
4
11. Given that a = −5
and b = 3a,find |a + b|
˜ ˜ ˜ ˜ ˜
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12. Given the points A(4, 1) and B(12, 16), find the:
(a) column vector AB
(b) length of AB
−→   −→  5 
13. Given the vectors OP = 37 and P Q = −1 .Find the
−→
(a) column vector OQ
−→
(b) the length of |OQ|
−→ −→ −→
14. Given the points P(-2,3) and Q(3,6) ,find the coordintaes of R ,if OR = 3OP + 13 OQ
−→ −→ −→
15. Given the points A(3,4) and B(9,2) .Find the coordinates of T ,if OT = 3OA + 12 AB

Chapter Summary

1. Translation− This is a transformation that moves every point in a figure the same
distance in the same
  direction
2. A translation T= ab means that an object is moved a distance a in the x−direction
and a distance b in the y−direction
3. Translation + object = image
4. Vector−This is a physical quantity that has magnitude(size or length) and direction.e.g
displacement,velocity
5. A scalar −This is a physical quantity that has magnitude(size or length)only.e.g distance
,speed  
6. When a displacement vector is written as AB= xy is called a Column vector
7. Column Vector describes the movement of an object in both the x-direction and the
y-direction ♣
→
−
8. A Null vector has no magnitude and direction. It is denoted as O or O .
9. All position
   O as their initial
vector have  position.

x1 x2 x1 +x2
10. If a= y1 and b= y2 ,then a + b = y1 +y2
     
11. If a= x1
y1
x2
and b=,then a − b = xy11 −x
−y2
2
y2  
12. When a vector a= xy is multiplied by a scalar k, we obtain kx
ky
 
x √
13. The magnitude of the column vector a= y is given by |a| = x2 + y2
14. The vectors AB and BA are equal in length but opposite in direction i.e BA = −AB
15. AB =6 BA
16. When a vector is multiplied by a position scalar, its direction does not change. However,
when a vector is multiplied by a negative scalar, its direction is reversed.

 
ASSESSMENT
3
1. A translation T= 4
maps the points A(2, 5) onto the point P1 . Find the coordinates
of P1
     
1 3 2
2. If r= 3
,s= −4
and t= 4
,find

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(a) r + s (c) |2r|

(b) s − t (d) |3s + 2t|

3. Write down the translation vector that would take A to B.

y-axis

7
B
6

2
A
1

0
x-axis
0 1 2 3 4 5 6 7 8 9 10

−−→   −→  
4. Given that OB = 19 and AB = −2
5
.Find the:
(a) coordinates of A
(b) |OA|
5. Triangle ABC has vertices A (−3, 0),B (0, 3) and C (3, 0). Find the coordinates of the
points A1 , B1 and C1,the
 images of A, B and C respectively, under a translation with
2
displacement vector 3
6. Find the translation that maps point K(2, 6) onto K1 (3,8)
7. Given the points A(−2, 3) and B(3, 6), find the coordinates of C, if OC = 3OA + 31 OB
−→   −→  5 
8. Given the vectors OP = 37 and P Q = −1 .Find the
−→
(a) column vector OQ
−→
(b) the length of |OQ|
   
9. Given that a = −2
−9
,b = 72 ,and m=a + 2b
˜ ˜   
˜ ˜  
−2 8 4
10. Given the vectors a = 7 ,b = 11 and c = −5 .Find the length of a + 2b + c
˜ ˜ ˜ ˜ ˜ ˜
End

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 Chapter 3

NUMERICAL CONCEPTS (INDICES ,

LOGARITHMS AND SURDS)

Learning objectives

By the end of this topic,the learners should be able to

• Give approximate answers to calculations
• Write numbers to a given number of significant figures
• Differentiate between significant figures and decimal places
• Express numbers in standard form ♣
• Identify base number and index
• State and apply the laws of indices in calculations
• Use a calculator to find powers and roots
• State and apply the laws of logarithms in calculations

3.1 Approximation

3.1.1 Rounding off

• To round off whole numbers, find the place value of the rounding digit and look at the digit
just to the right of it. if the digit is less than 5(e.g 4,3,2,1,0),do not change the rounding
digit but change all the digits to the right of the rounding digit to zero. If the digit is greater
than or equal to 5(e.g 5,6,7,8,9), add one to the rounding digit and change all the digits to
the right of the rounding digit to zero.
• To round off decimals, find the place value of the rounding digit and look the digit right
of the digit, if it is less than 5(e.g 4,3,2,1,0), do not change the rounding digit but drop all
digits to the right of it. If that digit is more or equal to 5(e.g 5,6,7,8,9), add one to the
rounding digit and drop all digits to the right of it

46
 3.1. APPROXIMATION

Example 4.1
Round to the nearest ten

(a) 52 (b) 56 (c) 524 (d) 528

SOLUTION
(a) (c)
5 2 5 2 4
+ 0 + 0
5 0 5 2 0

(b) (d)
5 6 5 2 8
+ 1 + 1
6 0 5 3 0

Example 4.2
Round to the nearest hundred

(a) 528 (b) 563 (c) 5328 (d) 7356

SOLUTION
(a) (c)
5 2 8 5 3 2 8
+ 0 + 0
5 0 0 5 3 0 0

(b) (d)
5 6 3 7 3 5 6
+ 1 + 1
6 0 0 7 4 0 0

3.1 Exercise Set

1. Round off the numbers to the nearest number indicated in the bracket

(a) 489(10) (c) 6489(1000) (e) 48999(100) (g) 45234489(10000)

(b) 489(100) (d) 949(10) (f) 125789(10000) (h) 9989967(1000)

2. A company was reported to have made a profit of sh 93 678 563 .New vision gave the figure
to the nearest 1000,000 and Daily monitor to the nearest 10000.What was the difference
between the rounded off figures

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3. Taibah international school paid sh 1155800 for the online lessons.Round off to the nearest
10000
4. The ministry of Health has recorded the following covid 19 cases in some districts.

District Buikwe Kampala Wakiso Elegu

Covid 19 cases 49 395 256 299

(a) Write the cases for each district to the nearest 10

(b) Obtain the total number of Covid 19 cases and write them to the nearest 100

3.1.2 Decimal Places

• Look at the first digit after the decimal point if rounding to one decimal place or the second
digit for two decimal places
• Draw a vertical line to the right of the place value digit that is required
• Look at the next digit
• If it’s 5 or more, increase the previous digit by one
• If it’s 4 or less, keep the previous digit the same
• Remove any numbers to the right of the line
Example 4.3
Write the following numbers to the decimal places(dp) indicated in the bracket

(a) 12.451(2dp) (b) 556.4567(3dp) (c) 0.45623(2dp) (d) 0.2432(1dp)

SOLUTION
(a) 12.45(2dp) (b) 556.457(3dp) (c) 0.456(3dp) (d) 0.2(1dp)

3.2 Exercise Set

1. Write the numbers correct to one decimal place(1d.p)

(a) 0.99 (c) 0.556 (e) 19.53 (g) 0.095

(b) 556.899 (d) 0.212 (f) 9.111 (h) 19.53

2. Write the numbers correct to two decimal places (2d.p)

(a) 5.391 (c) 0.0724 (e) 9.015 (g) 0.46666

(b) 0.414 (d) 8.0062 (f) 88.044 (h) 11.0482

3. Complete the following table by collecting to a given number of decimal places

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Digit 0 dp 1dp 2 dp 3dp 4 dp

0.3567856
8.3257
0.687432
7.2301
0.562789
9.2346
7.4466
0.002489
56.987

4. Use a ruler to measure the dimensions of the rectangle below

(a) Write down the length and width in cm correct to 1d.p

(b) Work out the area of the rectangle and give your answer to 1d.p

3.1.3 Significant figures

Are digits in a number that are known with certainty plus the first digit that is uncertain.The
Significant Figures of a number have meaning in reference to a measured or specified value. Cor-
rectly accounting for Significant Figures is paramount while performing arithmetic so that the
resulting answers accurately represent numbers that have computational significance or value.

Rules governng significant figures

1. All non zero digits in a number are significant e.g 4362(4 S.F),
1241(4 S.F),1.26(3 S.F),1.2(2 S.F)
2. All zeroes occuring between non zero digits are significant.They are also known as trapped
zeroes.e.g 1.004(4 S.F),1002(4S.F)
3. Law of trailing zeroes:All zeroes to the right of the last non zero digit are
(a) Significant if they are not as a result of rounding off.e.g 720(3 S.F)
(b) Not significant if they are obtained as aresult of rounding off e.g 6259 to 3S.F is
6260(3S.F).so this zero is not significant
4. Zeroes before a non zero digit are not significant.They are known as leading zeroes.e.g
0.06(1S.F),0.000000054(2 S.F).

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Example 4.4
Write the following numbers to the significant figures(sf) indicated in the bracket

(a) 12.471(3sf) (b) 12600(2sf) (c) 0.45623(1sf) (d) 12.0058(4sf)

SOLUTION
(a) 12.5(3sf) (b) 13000(2sf) (c) 0.5(1sf) (d) 12.01(4sf)

3.3 Exercise Set

1. Write 5894 to 3 significant figure.

2. Write the numbers correct to four significant figures

(a) 486.82 (c) 3.88888 (e) 587.55 (g) 588852

(b) 600.36 (d) 4.04053 (f) 0.071542 (h) 2345.44

3. Complete the following table by writing to the required number of significant figures

Digit 1s.f 2s.f 3s.f 4s.f 5s.f

0.78667
8.3257
0.687432
7.04423
1.999978
9.2346
26.5673
0.00248992
56.987

3.2 Indices

3.2.1 Index Notation

Activity
1. Write the following numbers as products of their prime numbers.

(a) 4 (c) 9 (e) 25 (g) 125

(b) 8 (d) 16 (f) 27 (h) 2401

2. Express the factors of the numbers in simple form.

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From the activity considering the number 8. Writing it as a product of its prime numbers we get
8 = 2 × 2 × 2,We notice that in this format, the factor 2 is repeated 3 times.i.e 23 .The raised
numeral is called an index (plural indices), power or exponent. Representing a number in this
short form is known as index notation.
• Index refers to the power to which a number is raised
• 34 is read as 3 raised to power 4
• 3 is known as the base and 4 is known as the index(power)
Example 4.5
Write the following expanded forms in index form:

(a) 3 × 3 × 3 (c) 5 × 5 × 5 × 5
(b) 2 × 2 × 2 × 2 × 2 × 2 × 2 (d) 7 × 7 × 3 × 3 × 3

SOLUTION
(a) 33 (b) 27 (c) 54 (d) 72 × 33

Example 4.6
Calculate the value of

(a) 32 (b) 73 (c) 103 (d) 72 × 33

SOLUTION
(a) 32 (c) 103
32 = 3 × 3 103 = 10 × 10 × 10
=9 = 1000
(b) 73 (d) 72 × 33
73 = 7 × 7 × 7 72 × 33 = 7 × 7 × 3 × 3 × 3
= 343 = 1323

Example 4.7
Write the following numbers in index form

(a) 125 (b) 32

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3.2. INDICES

SOLUTION
(a) 125 32

125
2 16

5 25
2 8

5 5
2 4

125 = 5 × 5 × 5 2 2
3
=5
32 = 2 × 2 × 2 × 2 × 2
(b) 32 = 25

3.4 Exercise Set

1. Calculate:

(a) 102 (b) 23 (c) 53 (d) 32

2. Write the following expanded forms in index form:

(a) 19 × 19 × 19 (c) 10 × 10 × 10 × 10 × 10 × 10 × 10
(b) 11 × 11 × 11 × 11 × 11 × 11 × 11 (d) 7 × 7 × 7 × 7 × 7

3. Write each of the following index forms in expanded form:

(a) 105 (b) 93 (c) 55 (d) 22

4. Write the following numbers in index form

(a) 81 (c) 10000 (e) 49 (g) 169

(b) 625 (d) 216 (f) 256 (h) 243

5. Calculate:

(a) 32 + 33 (c) 103 + 25 (e) 23 × 34 (g) 43 × 63

(b) 32 × 33 (d) 52 − 22 (f) 32 × 43 × 24 (h) 22 × 20

6. Writing your answers in index form, calculate:

(a) 103 × 102 (b) 32 × 33 (c) 105 ÷ 103 (d) 45 ÷ 42

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3.2.2 Laws of Indices

When working with numbers involving indices there are three fundamental laws which can be
applied .These are
. Multiplication Law
1. am × an = am+n
. Quotient (or Division) Law
2. am ÷ an = am−n
. Power Law
3. (am )n = amn
Example 4.8
Simplify

(a) 32 × 34 (b) 75 ÷ 72 (c) 103 × 102 × 104 (d) (112 )3

SOLUTION
(a) 32 × 34 (c) 103 × 102 × 104
32 × 34 = 32+4 103 × 102 × 104 = 103+2+4
= 36 = 109
(b) 75 ÷ 72 (d) (112 )3
75 ÷ 72 = 75−2 (112 )3 = 112×3
= 73 = 116

Example 4.9
Simplify

(a) a2 × a5 × a3 (b) y 6 ÷ y 3 (c) b3 × b2 × b4 × b2 (d) (y 4 ) 2

1

SOLUTION
(a) a2 × a5 × a3 (c) b3 × b2 × b4 × b2
a2 × a5 × a3 = a2+5+3 b3 × b2 × b4 × b2 = b3+2+4+2
= a10 = b11
1
(d) (y 4 ) 2
1 1
(y 4 ) 2 = y 4× 2
2
(b) y 6 ÷ y 3
4 ×
1
1
6 3 6−3
y ÷y =y =y 2


= y3 = y2

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3.2. INDICES

Example 4.10
Simplify

22 ×24 63 ×62 ×64

(a) 23
(b) 62 ×63

SOLUTION
22 ×24 65 ×66 ×64
(a) 23
(b) 62 ×63
22 × 24 22+4 65 × 66 × 64 65+6+4
= =
23 23 62 × 63 62+3
26 615
= 3 = 5
2 6
= 26−3 = 615−5
= 23 = 610

3.5 Exercise Set

1. Simplify the following using indices

(a) 32 × 35 (d) 45 × 52 × 46 × 53 (g) 72 × 73 × 71 × 75 × 74

(b) 95 × 94 (e) 24 × 57 × 53 × 62 × 66 (h) a2 × a3 × a8
(c) 52 × 53 × 55 (f) 192 × 194 × 192 10 × 10 (i) y 2 × y 7

2. Simplify the following using indices

65 197
(a) 35 ÷ 33 (c) 57 ÷ 53 (e) 64
(g) 193
312
(b) 95 ÷ 94 (d) 45 ÷ 42 (f) 37
(h) y 9 ÷ y 7

3. Simplify the following using indices

1
(a) (33 )3 (b) (95 )4 (c) (57 )3 (d) (45 ) 5

4. Simplify the following

y 2 ×y 5 ×y 3
 2 4y ×8y−2
(a) 32 ×35 (l)
y4 (f) 33 32y−1

3 5y ×125y+1
(b) 56 ×57 (g) (x4 y 3 ) (m) 625y+1
53 ×59

44 ×28 ×43
(h) 8a4 ÷ 4a2 (n)

62 ×68
4
(c) 27 ×46
63
(i) (x3 )3 ÷ x2  5
(62 )3 ×62 78
(d) a7 ×(ab4 )3 (o) 72 ×73
67 (j) b3
 4
2 2y ×8y−1 32 ×9
(e) (26 × 28 ) (k) 16y−1
(p) 33

5. Simplify the following using indices

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(a) 3y 2 × 4y 4 (b) 25x4 ÷ 5x2 (c) x5 × 2x4 (d) y 5 ÷ y − 2

3.2.3 The Zero index

Activity
1. Using the law of indices, find the solution of the following:

(a) 54 ÷ 54 (b) a3 ÷ a3 (c) 32

32

2. What is your observations

• Any non zero number raised to power 0 is equal to 1

• a0 =1

• For example 100 = 1,10000000 = 1,(ab)0 = 1,c0 = 1,40 = 1

3.6 Exercise Set

1. Calculate

(a) 30 + 40 (c) 20 − 40 (e) b0 + z 0

(b) 60 × 80 (d) 50 ÷ 150 (f) 10000000 ÷ 2340

2. Simplify

24 x2 y 3
(a) 53 × 50 (c) (100 )2 (e) 2×2×2×2×y 2 ×y×x×x

z 2 ×x4 ×b2
(b) 73 × 7−3 (d) 62 × 6−2 (f) x2 z×b2 zx2

3.2.4 Negative Indices

Activity
1. Using the quotient law of indices, solve the following:

(a) 23 ÷ 22 (b) 22 ÷ 23 (c) a3 ÷ a4

2
2. Express 223 in a standard form,what relationship exists with the answer obtained in
number 2.

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3.2. INDICES

From the activity we can observe that

a2 a×a
=
a 3 a×a×a
1
=
a
Also on applying the law of indices
a2
= a2−3
a3
= a−1
Therefore
1
a−1 =
a
In general
1
a−n =
an
For positive integers of n

Example 4.11
Calculate, leaving your answers as fractions:

(a) 7−2 (b) 3−1 + 4−1

SOLUTION
(a) 7−2 (b) 3−1 + 4−1
1 1 1
7−2 = 3−1 + 4−1 = + 1
72 31 4
1 1 1
= = +
7×7 3 4
1 7
= =
49 12

Example 4.12
Simplify:(5−2 )3

SOLUTION
(5−2 )3 = 5(−2×3)
= 5−6
1
= 6
5

Example 4.13
Simplify:

y3 22 25
 −2
(a) y4
(b) 25
(c) 2−5 (d) 3
2

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SOLUTION
y3 25
(a) y4
(c) 2−5
y3 25
= y 3−4 = 25−−5
y4 2−5
= y −1 = 210
1 = 1024
= 1
y
1  −2
= (d) 3
y 2
 −2
22
3 1
(b) =  2
25 2 3
22 2
5
= 22−5 1
2 = 9
= 2−3 4
1 9
= 3 =1÷
2 4
1 4
= =1×
2×2×2 9
1 4
= =
8 9

Example 4.14
−2
Simplify: 225×5
64× 1
42

SOLUTION
125 × 5−2 53 × 5−2
=
64 × 412 43 × 4−2
53+−2
= 3+−2
4
51
= 1
4
5
=
4

3.7 Exercise Set

1. With out using a calculator ,evaluate the following

(a) 3−2 (d) 5 × 10−3 (g) 3

2−2
(j) (a−2 )3

(b) 64 × 2−4 (e) 100 × 10−1 (h) 5

4−2
(k) (b−4 )−2
 −2
8−6
(c) 100 × 10−2 (f) 36 × 6−3 (i) 8−8
(l) 1
3

2. Simplify the following

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8−4 ×4−4
(a) 2x5 y 3 ÷ 4x−1 y (d) (a5 )3 × a−5 (g) 4−2 ×8−4

(b) 81 × 106 ÷ (9 × 10−4 ) (e) b12 × b−2 ÷ b−5 (h) 2−3 × 45 + 32 × 3−4
(c) (y 3 )2 ÷ y 1 3 (f) 2a−5 × a (i) 6−1 ÷ 6−4

3.2.5 Same power Law

When the index of the terms is the same then the following rules should be applied:
•
an × bn = (a × b)n
= (ab)n

•
an ÷ bn = (a ÷ b)n
OR
an
 n
a
=
bn b

•
(am bn )p = amp bnp

• Any number raised with a negative power, is the reciprocal of the number with a positive
power. !n
 −n
a b
=
b a

•
axm × bxn = (a × b)xm+n

•
axm ÷ bxn = (a ÷ b)xm−n

Example 4.15
Evaluate

 −2
(a) 42 × 32 (b) 153 ÷ 53 (c) (2xy)5 (d) 2
3

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SOLUTION
 −2
(a) 42 × 32 (b) 153 ÷ 53 (c) (2xy)5 (d) 2
3
42 × 32 = (4 × 3)2 153 ÷ 53 = (15 ÷ 5)3 (2xy)5 = 25 x5 y 5  −2
2
 2
3
= 122 = 33 = 32x5 y 5 =
3 2
= 144 = 27 32
=
22
9
=
4

3.8 Exercise Set

1. Simplify each of the following

 3
(a) b2 (c) (5x4 )2 (f) 7x4 × 2x (i) (54 )3
c4

(d) (4y 3 )2 (g) 5n3 × n4 (j) 2a2 × 3a3

2y 4
 
2
(b) (5a3 bc4 ) (e) (3a2 )3 (h) 6y 2 × 2 (k) y 3

2. Simplify each of the following

(a) 3y 2 × 4y 3 (c) 12b3 ÷ 3b5 (e) 2a × 5a (g) 6y 4 ÷ y −2

16x3
(b) 4a3 × 5a3 (d) 7y 6 × 8y −2 (f) 49b5 ÷ 7b3 (h) 4x−2

3.2.6 Fractional Power (or Root) Law

• The square root of any number is that number raised to power a half.
√ 1
a = a2

• A number raised to the power of one -nth is equal to the nth root of the same number
1 √
an = n a

• In general
m √
a n = ( n a)m

Example 4.16
Find the value of the following index forms:

1 2 1  −3
(a) 49 2 (b) 8 3 (c) 125 3

16 4
(d) 81

Use a calculator to find the roots of the numbers

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SOLUTION
1  −3
(a) 49 2

16 4
1 √ (d) 81
49 =
2 49  −3 3
16 81 4
 
4

= ±7 =
81 16
 3
2 81 4
(b) 8 3 √ = 3
2
8 3 = ( 8)2
3 16 4
 √ 3
= (2)2 4
81
=  √ 3
=4 4
16

(c) 125 3
1 33
√ =
1
125 3 =
3
125 23
27
=5 =
8

Example 4.17
Simplify the following:
1 2
(a) 3a 2 × 4a 3
  −1
1 25 2
(b) 2
− 4

SOLUTION
1 2
(a) 3a 2 × 4a 3
1 2 1 2
3a 2 × 4a 3 = (3 × 4)a 2 + 3
7
= 12a 6
  −1
1 25 2
(b) 2
− 4
 −1 1
1 25 1 4 2
 
2
− = −
2 4 2 25 
1
1  42 
= − 1
2 25 2
√ !
1 4
= − √
2 25
1 2
= −
2 5
1
=
10

Example 4.18
−1 −1
 2
8 3
Simplify the following:16 2 + 32 5 − 125

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SOLUTION
 2

2 1 1
8 1 1 83 
  
−1 −1 3 2 5
16 2 + 32 5 − = +− 2
125 16 32 125 3
 √ 2
3
1 1 8
= 1 + 1 −  √ 2
16 2 32 5 3
125
1 1 22
=√ +√ −
16 5
32 52
1 1 4
= + −
4 2 25
3 4
= −
4 25
59
=
100

3.9 Exercise Set

1. Evaluate each of the following:
1 1 2
(a) 27 3 (c) 16 4 (e) 729 3
1 3 2
(b) 81 4 (d) 32 5 (f) 216 3

2. Given that a = 36 and b = 64 ,evaluate

1 1 1 −2 −3
(a) 3a 2 (b) 4a 2 − 3b 3 (c) 2b 3 ÷ 6a 2

3. Calculate:
 1  1  1   −1
4×8 2 125×5 2 27×9 4 625 3
(a) 2
(b) 25
(c) 3 (d) 5

4. Simplify
1 −1 1 1
(a) (x9 ) 3 (b) (a10 ) 2
(c) a3
a
(d) a3
−1
a 2
q
125
5. Without using a calculator, evaluate the expression 3
1000

6. Calculate the value of

−1 q 2
(a) y×y
1
2
when y = 64 (c) x
y
when x = 64 3 and y = 3−2
y3
−1
p÷p 2 5 1
(b) 1 when p = 64 (d) (y − 1) 2 + (y + 6) 2 + 5y 0 .When y = 10
p3

7. Simplify
−1 1 1  1  2
(a) (27) 3 (25) 2 − 5 (0.008) 3 (c) 16 4
− 8 3
81 27
1 1 1
(b) 32 5 × 25 2 × 27 3

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3.2.7 Solving equations involving indices

• If am = an then m = n then Similarly, if am = bm , then a = b, provided both the bases are

positive numbers.
• To solve equations involving indices,express both sides of the equation with a common base
and simplify
Example 4.19
Solve for x in the following equations:

(a) 4x = 16 (b) 2x = 1
8
(c) 81x = 3 (d) 22x+1 = 32

SOLUTION
(a) 4x = 16 (c) 81x = 3
x
4 = 16 81x = 3
(22 )x = 24 (34 )x = 3
22x = 24 34x = 31
2x = 4 4x = 1
2x 4 4x 1
= =
2 2 4 4
x=2 1
x=
4
(d) 22x+1 = 32
22x+1 = 32
22x+1 = 25
x 1
(b) 2 = 8 2x + 1 = 5
x 1
2 = 2x + 1 − 1 = 5 − 1
8
x 1 2x = 4
2 = 3
2 2x 4
2x = 2−3 =
2 2
x = −3 x=2

3.10 Exercise Set

1. Solve for the value of x

(a) 2x = 32 (f) a4 ÷ a1 2 = ax (k) 2x+1 = 4x (p) 10x = 1000

(b) 4x+1 = 8x (g) 3x+1 = 9x (l) x2 = 9
(q) 7x = 1
x 3x+2 x+5 3
(c) 81 = 1 (h) 2 =4 (m) 10x = 640
1 (r) 5x = 25
(d) 5x = 25
(i) 32x−1 = 243 (n) 3x = 3
1
(e) 128x = 2 (j) 5x+3 = 25x (o) 10x = 0.1 (s) 4x = 4

2. Solve for the value of y :4 × 23y−3 = 2−y

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3. Solve for the value of n

(a) 52n+3 = 125n+5
(b) 0.5n = 8

3.3 Standard Form

• Standard form is a convenient way of writing very large or very small numbers in terms of
powerss of 10
• Standard form is also known as Scientific notation or Standard Index Form
• In standard form, numbers are written as a × 10 n where 1 ≤ a < 10 and n is an integer.
• To express a number in standard form, we shift the decimal point until the digit part a is
between 1 and 10. This digit a has a decimal point placed after the first digit.
• The power part (10n ) shows how many places to move the decimal point
• The rules of indices apply to calculations in standard form
Example 4.20
Write the following numbers in standard form

(a) 82000 (b) 40000 (c) 3000000 (d) 1230

SOLUTION
(a) 8.2 × 104 (b) 4 × 104 (c) 3 × 106 (d) 1.23 × 103

Example 4.21
Write the following numbers in standard form

(a) 0.063 (b) 0.0000567 (c) 0.0004 (d) 0.000621

SOLUTION
(a) 6.3 × 10−2 (b) 5.67 × 10−5 (c) 4 × 10−4 (d) 6.21 × 10−4

Example 4.22
Express the following numbers in standard form:

(a) 234 × 103 (b) 7680 × 10−7 (c) 34.5 × 10−4 (d) 0.0876 × 105

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SOLUTION
(a) 234 × 103 (c) 34.5 × 10−4
234 × 103 = 2.34 × 102 × 103 34.5 × 10−4 = 3.45 × 101 × 10−4
= 2.34 × 102+3 = 3.45 × 101−4
= 2.34 × 105 = 3.45 × 10−3
(b) 7680 × 10−7 (d) 0.0876 × 105
7680 × 10−7 = 7.68 × 103 × 10−7 0.0876 × 105 = 8.76 × 10−2 × 105
= 7.68 × 103−7 = 8.76 × 10−2+5
= 7.68 × 10−4 = 8.76 × 103

Example 4.23
By expressing each of the numbers in standard form, evaluate the following:

(a) 0.0003 × 0.002 800 0.009×8000

(c) 0.004
(e) 0.002×0.3
0.81
(b) 80000 × 0.0005 (d) 0.02 × 0.0015 (f) 0.0027

SOLUTION
(a) (d)
−4 −3
0.0003 × 0.002 = 3 × 10 × 2 × 10 0.02 × 0.0015 = 2 × 10−2 × 1.5 × 10−3
= (3 × 2) × 10−4+−3 = (2 × 1.5) × 10−2−3
= 6 × 10−7 = 3 × 10−5
(e)
0.009 × 8000 9 × 10−3 × 8 × 103
=
0.002 × 0.3 2 × 10−3 × 3 × 10−1
(b)
(9 × 8) × 10−3+3
80000 × 0.0005 = 8 × 104 × 5 × 10−4 =
(2 × 3) × 10−3+−1
= (8 × 5) × 104−4 72 × 100
= 40 × 100 =
6 × 10−4
= 4 × 101 × 100 = (72 ÷ 6) × 100−−4
= 4 × 100+1 = 12 × 104
= 4 × 101 = 12 × 104
= 1.2 × 101 × 104
= 1.2 × 105
(c) (f)
800 8 × 10 2
0.81 8.1 × 10−1
= =
0.004 4 × 10−3 0.0027 2.7 × 10−3
= 2 × 102−−3 = (8.1 ÷ 2.7) × 10−1−−3
= 2 × 105 = 3 × 102

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Example 4.24
Without using a calculator, determine:

(a) (4 × 103 ) × (2 × 104 ) (b) (9 × 105 ) ÷ (3 × 10−3 )

SOLUTION
(a)
(4 × 103 ) × (2 × 104 ) = (4 × 2) × 103+4
= 8 × 107
(b)
(9 × 105 ) ÷ (3 × 10−3 ) = (9 ÷ 3) × 10(5−−3)
= 3 × 108

Example 4.25
Without using a calculator, determine:

(a) (4 × 103 ) + (2 × 104 ) (b) (9 × 105 ) − (3 × 103 )

SOLUTION
(a)
(4 × 103 ) + (2 × 104 ) = 4000 + 20000
= 24000
= 2.4 × 104
Method 2.Change the indices such that they have the same index
(4 × 103 ) + (2 × 104 ) = (4 × 103 ) + (20 × 103 )
= (4 + 20) × 103
= 24 × 103
= 2.4 × 104
(b)
(9 × 105 ) − (3 × 103 ) = 900000 − 3000
= 897000
= 8.97 × 105
Method 2.Change the indices such that they have the same index
(9 × 105 ) − (3 × 103 ) = (900 × 103 ) − (3 × 103 )
= (900 − 3) × 103
= 897 × 103
= 8.97 × 105

3.11 Exercise Set

1. Write each of the following numbers in standard form:

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(a) 4000 (c) 6 000 000 (e) 870 000 (g) 81 900 000 000
(b) 9000 (d) 87 000 (f) 24 000 000 000 (h) 7 35.234

2. Write each of the following numbers in standard form:

(a) 0.05 (c) 0.5682 (e) 124.00688 (g) 245.12

(b) 210.00856 (d) 0.00004356 (f) 0.00000098 (h) 0.00000000000023

3. Without using a calculator, evaluate the following give your answer in standard form:

(a) (5 × 105 ) × (2 × 10−5 ) (f) (2 × 103 ) + (3 × 103 )

(b) (9 × 105 ) × (4 × 103 ) (g) (5 × 105 ) − (2 × 102 )
(c) (8 × 106 ) ÷ (4 × 10−3 ) (h) (7 × 102 ) − (5 × 101 )
(d) (6 × 102 ) ÷ (3 × 103 ) (i) 4000 × 8000 ÷ 640
(e) (9.8 × 103 ) + (2.5 × 103 ) (j) (6.5 × 108 ) ÷ (1.3 × 104 ) × (5 × 103 )

4. If x = 3 × 103 and y = 2 × 102 ,work out the value of

(a) xy (c) x − y
(b) x + y (d) x ÷ y

0.0035
5. Without using a calculator or tables, evaluate: 0.07×0.2

6. Without using a calculator or tables, evaluate: 0.42×0.35×0.0015

0.049×0.003

7. The radius of the earth is 6.4 × 106 m. Giving your answers in standard form, correct to 3
significant figures, calculate the circumference of the earth .Take π = 314 × 10−2

3.4 Logarithms

• Logarithm is a derived term from two Greek words, namely: logos (expression) and
arithmos (number) . Thus, logarithm is a technique of expressing numbers.

• Logarithm is a system of evaluating multiplication, division, powers and roots by appropri-

ately converting them to addition and subtraction.

• The logarithm of b to the base a is written as loga b

• A logarithm to the base of 10 is called a common logarithm .It is written as log10 b or

log b

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3.4.1 Rules of Logarithm

1. Addition-Product Law
logc a + logc b = logc ab

2. Subtraction-Quotient Law
a
 
logc a − logc b = logc
b

3. Power Law
logc am = m logc a

4. (Logarithm to the) Same Base Law

logc c = 1

5. Unity Law (or Log of Unity Law)

logc 1 = 0

6. logc a = logc b implies a=b

7. logc a = b implies a = cb
Example 4.26
Without using a calculator simplify:

(a) log10 10 (c) log3 3 (e) log10 1000

(b) log10 100 (d) log3 81 (f) log2 64

SOLUTION
(a) (c) (e)
log10 10 = 1 log3 3 = 1 log10 1000 = log10 103
= 3 log10 10
=3×1
=3
(b) (d) (f)
2 4
log10 100 = log10 10 log3 81 = log3 3 log2 64 = log2 26
= 2 log10 10 = 4 log3 3 = 6 log2 2
=2×1 =4×1 =6×1
=2 =4 =6

Example 4.27
Without using a calculator simplify:

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1 1
(a) log4 4
(b) 3
log3 729

SOLUTION
(a) (b)
1 1 1
log4 = log4 4−1 log3 729 = log63
4 3 3
1
 
= −1 log4 4 = × 6 log3 3
3
= −1 × 1
=2×1
= −1
=2

Example 4.28
Without using a calculator simplify:

(a) log10 5 + log10 2 (b) log2 4 + log2 8 (c) log3 9 + log3 3 (d) log5 5 + 3 log5 5

SOLUTION
(a) (c)
log10 5 + log10 2 = log10 (5 × 2) log3 9 + log3 3 = log3 (9 × 3)
= log10 (10) = log3 27
=1 = log3 33
= 3 log3 3
=3
(d)
(b) log5 5 + 3 log5 5 = log5 5 + log5 53
log2 4 + log2 8 = log2 (4 × 8) log5 5 + 3 log5 5 = log5 (5 × 125)
= log2 32 = log5 625
= log2 25 = log5 54
= 5 log2 2 = 4 log5 5
=5 =4

Example 4.29
Without using a calculator simplify:

(a) log10 10000 − log10 100 (b) log4 64 − log4 4

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SOLUTION
(a) (b)
10000 64
   
log10 10000 − log10 100 = log10 log4 64 − log4 4 = log4
100 4
= log10 100 = log4 16
= log10 102 = log4 42
= 2 log10 10 = 2 log4 4
=2 =2

Example 4.30
Without using a calculator simplify:

(a) log10 40 + log10 50 − log10 20 (b) log2 18 + log2 32 − 2 log2 6

SOLUTION
(a)
log10 40 + log10 50 − log10 20 = log10 (40 × 50) − log10 20 Apply BODMAS
= log10 (2000) − log10 20
2000
 
= log10
20
= log10 100
= log10 102
= 2 log10 10
=2
(b)
log2 18 + log2 32 − 2 log2 6 = log2 (18 × 32) − log2 62
= log2 576 − log2 36
576
 
= log2
36
= log2 16
= log2 24
= 4 log2 2
=4

3.12 Exercise Set

1. Without using a calculator simplify:

1
(a) log10 10000 (d) log3 27 (g) 5
log5 625 (j) log 1000000
1 1
(b) log8 64 (e) log5 125
(h) log4 256
1
(c) log2 128 (f) 4 log9 9
(i) log 100000

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2. Without using a calculator simplify:

(a) log2 32 + 2 log2 8 (e) log4 256 − 2 log4 8

(b) log10 25 + log10 4 (f) log4 80 − log4 5
(c) log10 20 + 13 log10 125 (g) log6 12 + log6 18
(d) log5 625 − log2 25 (h) log2 56 − 12 log2 49

3. Without using a calculator simplify:

(a) log10 40 + log10 50 − log10 20 (c) log2 18 + log2 32 − 2 log6

(b) log 75 + 2 log 2 − log 3 (d) 2 log5 5 + 21 log5 81 − log5 45

4. Express the following as a single logarithm:

(a) log10 25 + log10 5 (c) log3 56 − log3 27

(b) log10 84 − log10 12 (d) 2 log10 8 + 13 log10 125 − 1

3.4.2 Using Logarithms for calculation

Rules of logarithms apply when we are multiplying or dividing numbers.

Example 4.31
Use logarithms to calculate 38 × 145
SOLUTION
x Standard form log10 x
38 3.8 × 101 1.5798
145 1.45 × 102 +2.1614

5511 5.511 × 103 3.7412

Therefore 38 × 145 ≈ 5511

Example 4.32
Use logarithms to evaluate 356 × 43.6
SOLUTION
x Standard form log10 x
356 3.56 × 102 2.5514
43.6 4.36 × 101 + 1.6395

15520 1.552 × 104 4.1909

Therefore 356 × 43.6 ≈ 15520

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Example 4.33
Use logarithms to evaluate 0.0417 × 0.00928
SOLUTION
x Standard form log10 x
0.0417 4.17 × 10−2 2̄.6201
0.00928 9.28 × 10−3 +3̄.9675

0.0003869 3.869 × 10−4 4̄.5876

Therefore 0.0417 × 0.00928 ≈ 0.0003869

Example 4.34
Use logarithms to evaluate 8.62 ÷ 3.457
SOLUTION
x Standard form log10 x
8.62 8.62 × 100 0.9355
3.457 3.457 × 100 −0.5387
2.49 2.49 × 100 0.3968
Therefore 8.62 ÷ 3.457 ≈ 2.49

Example 4.35
Use logarithms to evaluate 4565 ÷ 98.88
SOLUTION
x Standard form log10 x
4565 4.565×103 3.6594
98.88 9.888 × 101 −1.9951
46.16 4.616 × 101 1.6643
Therefore 4565 ÷ 98.88 ≈ 46.16

3.13 Exercise Set

1. Evaluate using logarithms

(a) 36.5 × 480.2 (c) 245 × 22.34 (e) 56.34 ÷ 5.86

(b) 8.21 × 516.4 (d) 12.9 × 3.135 (f) 24.46 ÷ 13.34

2. Evaluate using logarithms

(a) 75.6 × 0.8563 (c) 234.2 ÷ 12.34 (e) 1200 ÷ 12

(b) 0.0075 × 98 (d) 6.26 × 45.678 (f) 63.74 ÷ 8.46

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3.5 Surds
√ √ √ √
• 9 = 3, 16 = 4, 225 = 15.These expressions can be evaluated exactly.Thus 9 is not a
surd as it simplifies to 3
√ √
• 3 ≈ 1.732050808, 5 ≈ 2.236067978 etc.These expressions can not be evaluated ex-
actly.They are Irrational real numbers
√ √ √ √
• A surd is just a root that is Irrational .e.g 3, 7, 10, 11 etc
• Surds can also be called RADICALS
The Arithmetic of surds(radicals) or Surdic Operations
The following rules apply to surds:Assume that a,b and c are non-negative numbers i.e
a ≥ 0, b ≥ 0, c ≥ 0
1. The Product Rule
(a) √ √
√
a× b= ab
Example √ √ √
3× 2= 3×2
√
= 6
(b) √ √
√
c a × d b = cd ab
Example √ √ √
5 3 × 3 2 = (5 × 3) 3 × 2
√
= 15 6
2. The Divison Rule
(a)
√ √ r
a
a÷ b=
b
OR √
a a
r
= √
b b
Example s
√ √ 6
6÷ 2=
2
√
= 3

√ √ r
a
c a ÷ d b = (c ÷ d)
b
Example s
√ √ 6
15 6 ÷ 3 2 = (15 ÷ 3)
2
√
=5 3

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3. √ 2 √ √
b = b× b
√
= b×b
√
= b2
1
= (b2 ) 2
=b
Example √ 2 √ √
3 = 3× 3
√
= 3×3
√
= 9
=3
4. √ √ √
a2 b = a2 × b
√
=a b
5. Distributive rule
(a)
√ √ √
a c + b c = (a + b) c
Example √ √ √
3 5 + 4 5 = (3 + 4) 5
√
=7 5
(b)
√ √ √
a c − b c = (a − b) c
Example √ √ √
6 5 − 3 5 = (6 − 3) 5
√
=3 5
(c) √ √
√ √ 
a b+a c=a b+ c

Examples
Write as a single surd:

√ √ √
1. 5× 4 2. √30
6

SOLUTION
√ √ √
1. 5× 4√ √ 2. √30
6
√
q
5 × 4 = (5 × 4)
s
√ 30 30
√ =
= 20 6 6
q
= (30 ÷ 6)
√
= 5

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Examples
Write as a single surd or rational number:

√ √ √ 2
1. 3 5 × 3 4 2. 5

SOLUTION
√ √ √ 2
1. 3 5 ×√3 4 √ q 2. 5
3 5 × 3 4 = (3 × 3) (5 × 4) √ 2 √ √
√ 5 = 5× 5
= 9 20 q
= (5 × 5)
√
= 25
=5

Examples
Without using a calculator simplify:

√ √
1. 8 3. 50
√ √
2. 48 4. 72

SOLUTION
√
1. 8 √ √
8= 4×2
4 is the largest perfect square factor of 8 √ √
= 4× 2
√
=2 2
√
2. 48 √ √
48 = 16 × 3
16 is the largest perfect square factor of 48
√ √
= 16 × 3
√
=4 3
√
3. 50 √ √
50 = 25 × 2
25 is the largest perfect square factor of 50
√ √
= 25 × 2
√
=5 2
√
4. 72 √ √
72 = 9 × 8
9 is the largest perfect square factor of 72 √ √
= 9× 8
√
=3 8

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Examples
Without using a calculator simplify:

√ √ √ √
1. 3+ 3 4. 8+3 2
√ √ √ √
2. 2 7 + 3 7 5. 4 3 − 12
√ √ √ √ √
3. 3 5 − 8 5 6. 45 + 20 − 80

SOLUTION
√ √
1. 3+ 3 √ √ √
3+ 3 = (1 + 1) 3
√
=2 3
√ √
2. 2 7 + 3 7 √ √ √
2 7 + 3 7 = (2 + 3) 7
√
=5 7
√ √
3. 3 5 − 8 5 √ √ √
3 5 − 8 5 = (3 − 8) 5
√
= −5 5
√ √
4. 8+3 2 √ √ √
√
8+3 2= 4×2+3 2
√ √  √
= 4× 2 +3 2
√ √
=2 2+3 2
√
= (2 + 3) 2
√
=5 2
√ √
5. 4 3 − 12 √ √ √ √
4 3 − 12 = 4 3 − 4 × 3
√ √ √ 
=4 3− 4× 3
√ √
=4 3−2 3
√
= (4 − 2) 3
√
=2 3
√ √ √
6. 45 + −
20 √ 80 √ √ √ √ √
45 + 20 − 80 = 9 × 5 + 4 × 5 − 16 × 5
√ √  √ √  √ √ 
= 9× 5 + 4× 5 − 16 × 5
√ √ √
=3 5+2 5−4 5
√
= (3 + 2 − 4) 5
√
= (5 − 4) 5
√
= 5

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Exercise
1. Write the following as the simplest possible surds.

√ √ √
(a) 200 (e) 50 (i) 75
√ √
(b) 20 (f) 12 √
√ √ (j) 32
(c) 18 (g) 108
√ √ √
(d) 20 (h) 27 (k) 24

2. Write as a single surd or rational number:

√ √ √ √ √ 2
(a) 2× 5 (c) 4 3 × 2 3 (e) 8
√ √ √ √
(b) 3 2 × 2 (d) √12 (f) √18
2 2

3. Without using a calculator simplify:

√ √ √ √
(a) 3 2 + 2 (f) 2 27 + 2 12
√ √ √
18
(b) 10 5 − 5 5 (g) 3
√ √ √ √
(c) 27 − 5 3 (h) 3 2 − 8
√ √ √ √
(d) 3 7 + 7 − 2 7 (i) √6
3
√ √ √ √ √ √
(e) 2 + 8 − 32 (j) 6 3 − 12 + 48

Mixed Numbers√ √ √
• We cannot simplify expressions such as this a + b c or a + b c because there are no
common terms .
• When we multiply numbers like this together, we multiply just as we do in ordinary
algebra.

Examples
Expand the following and simplify:

√ √   √  √ 
1. 3 3 + 2 4. 2 + 5 3+ 5
√ √ √   √  √ 
2. 3 7 + 5 5. 2 + 5 2 − 5
√ √ √   √ 2
3. 2 6 + 3 2 6. 5 + 2

Solution
√ √ 
1. 3 3+ 2
√ √  √ √
3 3+ 2 =3 3+3 2

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√ √ √ 
2. 3 7+ 5
√ √ √  √ √ √ √
3 7+ 5 = 3 7+ 3 5
q q
= (3 × 7) + (3 × 5)
√ √
= 21 + 15
√ √ √ 
3. 2 6+3 2
√ √ √  √ √ √ √
2 6 + 3 2 = ( 2 × 6) + ( 2 × 3 2)
q q
= (2 × 6) + 3 (2 × 2)
√ √
= 12 + 3 4
√
= 12 + (3 × 2)
√
= 12 + 6
q
= (4 × 3) + 6
√
=2 3+6
 √  √ 
4. 2 + 5 3+ 5
 √  √  √ √ √
2+ 5 3 + 5 = 2(3 + 5) + 5(3 + 5)
√ √ √
= 6 + 2 5 + 3 5 + ( 5)2
√
= 6 + (2 + 3) 5 + 5
√
= 11 + 5 5
 √  √ 
5. 2 + 5 2− 5
 √  √  √ √ √
2+ 5 2 − 5 = 2(2 − 5) + 5(2 − 5)
√ √ √
= 4 − 2 5 + 2 5 − ( 5)2
=4−5
= −1
 √ 2
6. 5 + 2
 √ 2  √  √ 
5+ 2 = 5+ 2 5+ 2
√ √ √
= 5(5 + 2) + 2(5 + 2)
√ √ √
= (5 × 5) + 5 2 + 5 2 + 2 × 2
√
= 25 + 10 2 + 2
√
= 27 + 10 2

Rationalising the denominator

• The process of eliminating surds from the denominator is called rationalizing the de-
nominator.(Clearing it of irrational numbers)

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√
• To rationalize √b ,
multiply top and bottom by a i.e
a √
b b× a
√ =√ √
a a× a
√
b a
=
a
√ √
• The conjugate surd of a + b is a − b
c√
• To rationalize multiply top and bottom by the conjugate of the denominator.
a+ b √
c c × (a − b)
√ = √ √
a+ b (a + b) × (a + b)
√
ca − c b
=
a2 − b
• √ √ √ √ √
(a + b)(a − b) = a(a − b) + b(a − b)
√ √ √
= a2 − a b + a b − b)2
= a2 − b
Example √ √
(5 + 2)(5 − 2) = 52 − 2
= 25 − 2
= 23

Examples
Rationalise the denominator of the following fractions.
1. √1
3

2. √10
6−1
1√
3. 3− 2
√
2+√7
4. 3− 7

Solution
1. √1
3 √
1 1× 3
√ =√ √
3 3× 3
√
3
=q
(3 × 3)
√
3
=√
9
√
3
=
3

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2. √10
6−1 √
10 10 × 6 + 1
√ = √ √
6−1 ( 6 − 1) × ( 6 + 1)
√
10 6 + 10
= √ 2
( 6) − (1)2
√
10 6 + 10
=
6−1
√
10 6 + 10
=
√5
10 6 10
= +
√5 5
=2 6+2
1√
3. 3− 2 √
1 1 × (3 + 2)
√ = √ √
3− 2 (3 − 2)(3 + 2)
√
3+ 2
= q
(3)2 − ( 2)2
√
3+ 2
=
9 −√2
3+ 2
=
7
√
2+√7
4. 3− 7 √ √ √
2+ 7 (2 + 7)(3 + 7)
√ = √ √
3− 7 (3 − 7)(3 + 7)
√ √ q
2(3 + 7) + 7(3 + 7)
= √
32 − ( 7)2
√ √
6+2 7+3 7+7
=
√9 − 7
13 + 5 7
=
2

Exercise
1. Expand the following and simplify:

√ 
 √  √√ 2
(a) 2 + 3 3+ 3 (e) 5+ 3
 √  √   √  √ 
(b) 5 − 7 3 + 7 (f) 5 + 2 5 − 2
 √ 2  √ √  √ √ 
(c) 2 − 3 (g) 5 3 + 2 5 3 − 2
 √ 2 √ √  √ √ 
(d) 5 + 2 (h) 2+ 3 2− 3

√ √ √ √ √
2. Express (2 5 + 2)( 5 + 3 2) in the form a + b c,hence state the values of a, b and
c.

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3. Rationalise the denominator of the following fractions.

(a) √1
5

(b) √5
5−2
1√
(c) 7− 5
√
2+√5
(d) 3− 5

4. Rationalise the denominator of the following fractions.

√ √
(a) √5+√3
5− 3
√
(b) √3+2
3−2
1√
(c) 3− 2
5√
(d) 3− 5
8√
(e) 3− 5

(f) √ 1√
5− 2

Chapter Summary

1. During rounding off, if the digit to the right is 0, 1, 2, 3, 4 the digit being rounded to
remains the same i.e Rounded down
2. During rounding off,if the digit to the right is 5, 6, 7, 8, 9 the digit being rounded to
increases by one i.e Rounded up
3. In standard form, numbers are written as a × 10 n where 1 ≤ a < 10 and n is an integer.
4. The following rules apply to indices:
LAW Example
m n m+n
a ×a =a 2 × 24 = 27
3

am ÷ an = am−n 23 ÷ 24 = 2−1
(am )n = amn (23 )4 = 21 2
a0 =1 20210 = 1
a−n = a1n 2−3 = 18 ♣
n n n 3 3 3
a × b = (ab) 2 ×3 =6
5. The following rules apply to logarithm:
LAW Example
logc a + logc b = logc ab log3 4 + log3 5 = log3 20
a
logc a − logc b = logc ( b ) log3 10 − log3 2 = log3 5
m
logc a = m logc a log3 65 = 5 log3 6
logc c = 1 log4 4 = 1
logc 1 = 0 log3 1 = 0
logc a = logc b implies a=b log2 x = log2 3 implies x=3
b
logc a = b implies a = c log2 x = 3 implies x = 23
log10 a is sometimes written as log a log10 6 is the same as log 6

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ASSESSMENT
1. During the 2021 national elections ,Museveni obtained 5,300,831 votes.Round off to
the nearest 1000
2. Round the following to the specified degree of accuracy.

(a) 0.00621968 to 4 d.p (d) 460.762 to 2 d.p

(b) 3.95451 to 3 d.p (e) 0.571859 to 3 d.p
(c) 2174.12 to 1 d.p (f) 1.066564 to 2 d.p

3. Round the following to the specified degree of accuracy.

(a) 0.7862 to 1 significant figure. (d) 401 to 1 significant figure.

(b) 9371 to 2 significant figures. (e) 0.00967947 to 3 significant figures.
(c) 2.3608 to 3 significant figures. (f) 376.92 to 2 significant figures.

4. Write the following expanded forms in index form:

(a) 5 × 5 × 5 × 5 × 5 (b) 3 × 3 × 3 × 3 × 3 × 3

2.4×102
5. Simplify 6.0×10−3
.Give your answer in standard form
6. With out using mathematical tables or calculator,find the value of

(a) 2 log10 5 + log10 4 − log10 0.1 (c) log 120 − 2 log 6 + 13 log 27
(b) log3 9 + log3 21 − log3 7 (d) log 400 + log 500 − log 200 + 1

7. Simplify

5x+1 ×5−2x 3n ×9n−2

(a) 5−x
(c) 27n−1
1 3  −2
27 3 ×16 4

(b) 8 3
1 (d) 27
92

8. Express √ 1√ with a rational denominator.

7− 3
1
(y2 ) 6
9. Evaluate 1 when x=16 and y=8
(9x) 2
3 1
10. Solve for x in 32 5 ÷ x 2 = 2
11. Use logarithms to work out the following

(a) 67.44 × 34.5 (b) 57.4 ÷ 24.5

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√ √ √ √
12. Simplify 125 + 20 − 45 .Give your answer in the form a b where a and b are
constants.
End

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