UNIVERSITY OF MAIDUGURI
Maiduguri, Nigeria
CENTRE FOR DISTANCE
LEARNING
EDUCATION
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
EDU 120 MATHEMATICS FOR EDUCATION
STUDENTS
UNIT: 2
CDL, University of Maiduguri, Maiduguri
ii
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
Published
UNIT: 2
2010©
All rights reserved. No part of this work may be reproduced in
any form, by mimeograph or any other means without prior
permission in writing from the University of Maiduguri.
This text forms part of the learning package for the academic
programme of the Centre for Distance Learning, University of
Maiduguri.
Further enquiries should be directed to the:
Coordinator
Centre for Distance Learning
University of Maiduguri
P. M. B. 1069
Maiduguri, Nigeria.
This text is being published by the authority of the Senate,
University of Maiduguri, Maiduguri – Nigeria.
ISBN:
978-8133-
CDL, University of Maiduguri, Maiduguri
iii
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
PRE FA CE
This study unit has been prepared for learners so that they can
do most of the study on their own. The structure of the study unit
is different from that of conventional textbook. The course writers
have made efforts to make the study material rich enough but
learners need to do some extra reading for further enrichment of
the knowledge required.
The learners are expected to make best use of library facilities
and where feasible, use the Internet. References are provided to
guide the selection of reading materials required.
The University expresses its profound gratitude to our course
writers and editors for making this possible. Their efforts will no
doubt help in improving access to University education.
CDL, University of Maiduguri, Maiduguri
iv
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
Professor M. M. Daura
Vice-Chancellor
CDL, University of Maiduguri, Maiduguri
v
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
HOW TO STUDY THE UNIT
You are welcome to this study Unit. The unit is arranged to
simplify
your
study.
In
each
topic
of
the
unit,
we
have
introduction, objectives, in-text, summary and self-assessment
exercise.
The study unit should be 6-8 hours to complete. Tutors will
be available at designated contact centers for tutorial. The center
expects you to plan your work well. Should you wish to read
further you could supplement the study with more information
from the list of references and suggested readings available in the
study unit.
PRACTICE EXERCISES/TESTS
1. Self-Assessment Exercises (SAES)
This is provided at the end of each topic. The exercise can
help you to assess whether or not you have actually studied and
understood the topic. Solutions to the exercises are provided at the
end of the study unit for you to assess yourself.
2. Tutor-Marked Assignment (TMA)
CDL, University of Maiduguri, Maiduguri
vi
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
This is provided at the end of the study Unit. It is a form of
examination type questions for you to answer and send to the
center. You are expected to work on your own in responding to the
assignments. The TMA forms part of your continuous assessment
(C.A.) scores, which will be marked and returned to you. In
addition, you will also write an end of Semester Examination,
which will be added to your TMA scores.
Finally, the center wishes you success as you go through the
different units of your study.
CDL, University of Maiduguri, Maiduguri
vii
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
INTRODUCTION TO THE COURSE
CDL, University of Maiduguri, Maiduguri
1
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
EDU 120: MATHEMATICS FOR EDUCATION
STUDENT
UNITS: 2
TABLE OF CONTENTS
PAGES
PREFACE
-
-
-
-
HOW TO STUDY THE UNIT
-
-
-
-
-
iii
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
iv
INTRODUCTION TO THE COURSE
1
TOPIC:
1:
ELEMENTS OF SET THEORY
3
2:
INDICES AND LOGARITHMS
12
3:
NUMBER SYSTEMS -
-
17
CDL, University of Maiduguri, Maiduguri
2
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
4:
SURDS
5:
RATIONALIZATION OF SURD EXPRESSIONS
-
6:
-
-
-
-
-
-
UNIT: 2
-
23
-
27
EQUATIONS
-
-
-
-
-
-
-
-
-
29
7:
SIMPLE AND COMPOUND INTERESTS
-
44
SOLUTION TO EXERCISES
CDL, University of Maiduguri, Maiduguri
3
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
1.0
TOPIC;
1.1
DEFINITION
UNIT: 2
ELEMENTS OF SET THEORY
A set is a collection of well defined objects or a collection of distinct and
distinguishable objects or things e.g. {set of accountancy students}, {set of Library
science students}, {set of PHE students}, {letters of the English alphabets}. Sets are
normally denoted by capital letters while its members normally referred to as its elements
are denoted by small letters.
1.2
NOTATIONS
X  A means A belongs to or X is an element of set A.
X  B means X does not belong to B.
Sets are defined by listing the elements e.g.
A = {a,b,c,d,} or by partially listing the elements e.g.
B = {2,4,6,…} or by defining the properties common to all elements e.g.
C = {x:x is a prime number less than 10}.
1.3
CLASSIFICATION OF SETS
1.
Finite set contains a finite number of elements e.g.
A = {Vowels in the English alphabets}
2.
Infinite set contain infinite number of elements e.g.
N = {Natural numbers}.
3.
Singleton set contains only one element e.g.
B = {p}, C = {Abuja}
4.
Empty set contains no member or element and is sometimes called Null set e.g.
G = { }, 
1.4
SPECIAL SETS
1.
Equal sets:
If A and B are two sets such that every element of A is also
an element of B and every element of B is also an element of A, then both the
sets are equal i.e. A = B.
CDL, University of Maiduguri, Maiduguri
4
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
Example:
If A = {3,5,7},
UNIT: 2
B = {7,3,5}
The A = B or {3,5,7} = {7,3,5}, since each of 3, 5, 7, which are elements of
A, belong to B also and each of 7, 3, and 5, (elements of B) belong to A.
2.
Equivalent sets: Two sets A and B are equivalent if they have the same
number of elements. Obviously, equal sets are equivalent but equivalent sets
are not necessarily, equal.
Example:
If A = {1,2,3,4,5} and B = {1,2,3,7,8}, then A ≠ B, but
each has five (5) elements. We call such sets as equivalent sets.
Note: The number of elements in a set is called the CARDINAL
NUMBER OF THE SET. Cardinal number of equivalent sets is the same.
3.
Sub-sets: If A and B are two sets such that all the elements of B belong to A,
then B is a sub-se of A.
Example: If A = {1,2,3,4} and B = {2,3,4}, the B is a sub-set of A since each
of the elements 2, 3, 4, of B is a member of A.
B is a sub-set of A is written symbolically as:
B  A.
We may also read as ‘B is contained in A’
If ‘B is not a sub-sets of A’, it is written symbolically as:
B  A.
4.
Proper sub-set: If B  A and B ≠ A, we say that B is a proper sub-set of A
i.e. A contains B properly.
Definition:
The proper sub-set of a given set is such a sub-set which
does not contain all the elements of the given set.
Example:
Write all the sub-sets of A = {a, b, c,}.
Solution:
All the sub-sets of A are:  , {a}, {b}, {c}, {a, b}, {a, c},
{b. c}, {a, b, c}.
Note: All the sub-sets here except {a, b, c} are proper sub-sets.
5. Difference set: The set of those elements of A which do not belong to B is
known as difference set A – B.
A – B = {x: x  A and x  B}
Similarly,
B – A = {x: x  B and x  A}
CDL, University of Maiduguri, Maiduguri
5
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
Example:
If A = {1,2,3,4,5} and B = {3,5,7,9}, find A – B and B – A.
Solution:
B are
A – B = {1, 2, 4}, since the elements of A not belonging to
1, 2 and 4.
Similarly, B – A = {7, 9}, since the elements of B not belonging to A are 7 and 9.
6. Universal set: In set theory, we face situations when all the sets under
consideration are sub-sets of a set A, then A is known as the universal set.
Generally, universal set is denoted by U. In short, it is the largest set under
consideration.
Let U = {1,2,3,4,5, …}, and
A = {1,3,5, …}
B = {2,4,6, …}
C = {1,4,9,16, …}
Here each of A, B and C is a sub-set of U, hence U is the universal set for
A, B and C.
Note: The universal set is such that all the elements of the sets under
consideration belong to it.
Example:
7.
Complementary set:
If A is a set and U is its universal set, then
the set of those of U which do not belong to A is known as complementary
set of A. It is denoted as A’ or Ac .

A’ = {x: x  U, x  A}.
Example:
If U = {1, 2,3, …} and A = {2,4,6, …}, then find A’.
Solution:
A’ = {1,3,5, …}
A’
A
8.
Disjoint sets: If no element of the sets A and B is common, then A and B
are disjoint sets.
Example:
If A = {1,3,5,7, …} and B = {2,4,6, …}, then A and B are
disjoint as no element is common to both.
CDL, University of Maiduguri, Maiduguri
6
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
VENN DIAGRAM: Diagrams make the study of sets very easy and interesting. We can
easily understand the inter-relationship of sets, their important properties and many
theorems with the help of diagrams. Swiss Mathematician named Euler first
represented a set by a closed curve. Later on British Mathematician named Venn
developed this method further.
The diagrams which represent some properties of sets are known as ‘Venn Euler
Diagrams’ or simply ‘Venn Diagrams.
Generally, the sets are expressed by circles or closed curves. Sometimes the
elements of sets are also written inside the closed curves.
Example: If B  A and A ≠ B, this relation is represented by Venn diagrams as
follows:
A
B
This diagram represents two sets A and B such that A contains B.
Note:
A Venn diagram is the systematic representation of set. In this the
universal set is always represented by a rectangle. For instance, the Venn diagram
below represent the sets A, B and the universal set  .
CDL, University of Maiduguri, Maiduguri
7
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
U
A
1.5
1.
B
SET OPERATION
Union of sets (U): If A and B are any two sets; AUB is the set containing all
elements of A and all elements of B (Common elements written only once)
e.g. if A = {2,4,3} and B = {4,5,6} then AUB = {2,3,4,5,6}. The Venn
diagram can be used to illustrate the union between two sets. A and B as
shown by the shaded area below.
A
B
2.
Intersection of sets (∩): If A and B are two sets; A∩B is the set of elements
common to both A and B. Example. Given A = {2, 4,3} and B = {4,5,6}, then
A∩B = {4}. Using Venn diagram below, the shaded area shows the intersection.
A
B
Note the following properties of Union of sets and Intersection of sets
CDL, University of Maiduguri, Maiduguri
8
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
1.
2.
3.
4.
5.
Commutative law for Union of sets: If A and B are two sets, then A U B
= B U A.
Associative law for Union of sets: If A, B and C are three sets then
(AU B)U C = AU(BUC)
Commutative law for Intersection of sets: If A and B are two sets, then
A  B  B  A.
Associative law for Intersection of sets:
If A, B and C are three sets, then
( A  B)  C  A  ( B  C )
Distributive laws of Union over Intersection and Intersection over Union
a. AU(B∩C) = (AUB) ∩(AUC)
b. A∩(BUC) = (A∩B)U(A∩C)
6.
De-Morgan Laws
a.
b.
(AUB)’ = A’∩B’
(A∩B)’ = A’UB’
EXAMPLE 1: In a certain examination 72 candidates offered mathematics, 64
English language and 62 French language, 18 offered both mathematics and
English, 24 mathematics and French language, 20 English language and French
language, 8 candidates offered all the three subjects. How many candidates were
there for the examination?
SOLUTION:
Let  denotes all candidates for the exam.
M denotes candidates who offered mathematics.
E denotes candidates who offered English language
F denotes candidates who offered French language.
Since there are 8 candidates who offered all subjects
n{M∩E∩F} = 8
There are 18 who offered both maths and English so
n{M∩E∩F’} = 18-8 = 10
n{M’∩E∩F} = 20-8 = 12
n{M∩E’∩F} = 24-8 = 16
Since 72 candidates offered maths,
N{M∩{E’UM’}} = 72-10-16-8 = 38
CDL, University of Maiduguri, Maiduguri
9
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
i.e. 38 offered mathematics only.
Also, n{E∩{F’UM’}} = 64-10-12-8 = 34
 34 offered English language only.
The problem can be represented by a Venn diagram as follows:
M
38
10
8
12
34
26
F
E
n{F∩{M’UE’}} = 62-12-16-8 = 26
i.e. 26 offered French only.
Now, 38+34+26+8+10+12+ 16 = 144
CDL, University of Maiduguri, Maiduguri
10
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
Therefore, there were 144 candidates in the examination.
Example 2
1. A survey in a class shows that 15 of the pupils play cricket, 11 play tennis and
6 play both cricket and tennis. How many pupils are there in the class if every
one plays at least one of these games?
CDL, University of Maiduguri, Maiduguri
11
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
Solution:
From the Venn diagram below we note that:
C
T
9
6
5555
55
5
n(C) = 15
n(T) = 11
n(C∩T) = 6
n(CUT) = 9+6+5 = 20 pupils
Notes:
1. For any two sets A and B
n(AUB) = n(A) +n(B) – n(A∩B)
2. For any three sets A, B and C
n(AUBUC) = n(A) + n(B) + n(C) – n(A∩B)-n(A∩C)-n(B∩C)
2. In a class there are 50 students, 15 students study mathematics, 35 study
chemistry. How many of them study both subjects?
Solution:
Where M and C represent the sets of students studying mathematics and
chemistry respectively.
Then n(MUC) = 50, n(M) =30, n(C) = 35
But n(MUC) = n(M) + n(C) – n(M∩)
50 = 30+35 – n(M∩C)
 n(M∩C) = 30+35-50 = 65-50 = 15
Therefore, only 15 students study both mathematics and chemistry.
CDL, University of Maiduguri, Maiduguri
12
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
EXERCISES
1. Verify De-Morgan’s laws using the sets
U = {1,2,3,4,5,… 16}
A = {1,2,3,4,5,6,7,8,9}
B = {8,9,10,11,12}
2. Lt U = {1,2,3,4, … 16} be a universal set and
A = {x:x  U and x is even}
B = {x:x  U and x is less than 10}
C = {x:x  U and x is a multiple of 3}
D = {x::  U and x is a prime number}
Find (i) AUB (ii) A∩B (iii) CUD (iv) B∩(CUD)’ Note: The symbol n is called the
cardinal number of a set.
3. If A = {1,2,3,4,5,6}, B = {4,5,6,7,8}
C = {4,5,6,8,10,12},
Show that ( A  B)  C  A  ( B  C )
4. If A = {a, b, c, d, e,}, B = {c, e, f, g, a,}
Show that ( A  B)  C  A  ( B  C )
5. If A = {1, 2, 3, 4, 5,},
and
B = {1, 3, 5, 7, 9}
C = {b, c, e, h, k, f},
C = {2, 3,4, 7},
Show that A  ( B  C )  ( A  B)  ( A  C )
6. If A = {a, b, c, d, e, f,},
B = {a, b, e, f, g, h}
Prove that A  ( B  C )  ( A  B)  ( A  C )
CDL, University of Maiduguri, Maiduguri
C = {b, e, f, h, k, l},
13
EDU 120
2.0
MATHEMATICS FOR EDUCATION STUDENTS
TOPIC:
UNIT: 2
INDICES AND LOGARITHMS
In expression such as 34 , the base is 3 and the 4 is called the power or index or
exponent (the plural is indices). Working with indices involves using some properties
which apply to any base, these properties are known as rules of indices. Using ‘a’ as
a general base we have:
Rule 1:
ap +aq = ap+q e.g. 23 x 24 = 23+4 = 27
Because 23 means 2x2x2 and 24 means 2x2x2x2
 23 x 24 = 2x2x2x2x2x2x2 = 27
Rule 2:
ap /aq = ap-q e.g. 34 / 32 = 34-2 = 32
Because 34 means 3x3x3x3 and 32 means 3x3
3 x3 x3 x3
 34 / 32 =
= 3x3 = 32
3 x3
But by the same rule, a3 / a5 = a3-5 = a-2
i.e. a3 / a5 = axaxa/axaxaxaxa
 a-2 = 1/a
In general ap means 1/p, i.e. a-p means the reciprocal of ap . Also
axaxaxa
Consider a4 / a4 =
=1
axaxaxa
But by the same rule:
a4 / a4 =a4-4 = a0 . Therefore, a0 = 1, i.e. any base to power 0 equals 1.
Rule 3:
(ap ) q = apq e.g. (a3 )3 = a2x3 = a6
Rule 4:
This rule explains the meaning of a fractional index. From the first rule,
½
we have a x a½ = a½+½ = a i.e. a = a½ x a½, but a = a x a
 a½ means a i.e. the square root of a.
Similarly, a⅓x a⅓ x a⅓ = a⅓+⅓+⅓ =a
Thus a⅓ means
3
a i.e. cube root of a.
In general, the fractional index p/q, the third rule shows that ap/q =
For example,
a
3
4
= (a3 )¼ or (a¼ )3 =
4
a
3
i.e
a
3/4
q
a
p
represents either the fourth roots
of a3 or the cube of fourth root of a. The rules above are generally used to simplify a
wide range of expressions containing indices provided that the terms all have the
same base.
CDL, University of Maiduguri, Maiduguri
14
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
Examples: Simplify:
1.
25
((
)
9
3 / 2
=
1
(25 )
9
(p⅓ )2x (p2 )⅓ ÷
2.
2/3
p p
2/3

1/ 3
p

p
3
3/ 2
=
1
1
27
1
=
=
=
125
125
25
( 5 )3
3
(
)3
27
9
p=
2 / 3 2 / 3

1/ 3
p

p
4 / 3 1 / 3

p
3/3
 p
COMMON LOGARITHM
Consider the statement 102 = 100. If this is expressed in words, we have the base
10 raised to the power of 2 gives 100. Now this relationship can be rearranged to
give the same information but with a different emphasis i.e. the power to which the
base 10 must be raised to give 100 is 2. In form the power is called a logarithm (log).
The whole relationship can then → log3 81= 4 etc.
Although we have so far used only 10, 2 and 3, the base of a logarithm can be any
positive number or even any unspecified number represented by a letter. For example
b = ac ↔ log a b= c.
Note that the symbol ↔ means that each of facts implies the other.
EVALUATING LOGARITHMS
It is generally easier to solve a simple equation in index form than in log form so
we often use an index equation for instance.
If x = log 497 then 49x = 7 → x = ½.
In particular for any base b if x = logb 1→bx = 1 →x = 0, i.e. the logarithm to any
base of 1 is zero.
EXERCISE:
Evaluate the following:
1. Log3 81
2. Log 27 3
3. Log64 4
CDL, University of Maiduguri, Maiduguri
15
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
THE LAWS OF LOGARITHMS
When working with indices, earlier in this chapter, we found certain rules that
powers obey in multiplication and division of numbers. Because logarithm is just
other word for index or power or exponent, it is to be expected that logarithms too,
obey certain laws and these we are now going to consider.
Consider x = log a b and y = log a c →ax =b and ay= c
Now bc = (ax )(ay ) →bc = ax+y, therefore log a bc = x + y i.e.
Log a bc = log a b + log a c
This is the first law of common logarithms and as ‘a’ can represent any base this
law applies to the log of any product provided that the same base is use for all the
logarithms in the formula.
Using x and y again a law for the log of a fraction can be found.
b

c
a
a
x
y

b
x y
therefore log a b = x – y i.e. log a b = log a b – log a c
a
c
c
c
A third law allows us to deal with expressions of the type
log a bn . Using x = log a bn → x = n log b i.e. log a bn = n log a b
S o we now have the three (3) most important laws of common logarithms.
Because they are true for any base, it is unnecessary to include a base in the formula
but in each of these laws every logarithm must be to the same base. Thus:
a. log bc = log b + log c
b. log b/c = log b – log c
c. log bn = n log b
In addition to the three (3) most important laws of common logarithms, we have
the following:
1.
2.
log a a = 1 and
log a 1 = 0 i.e. any log to the same base is 1and on the other hand log 1 to
any base is 0.
EXAMPLES: Evaluate the following:
1.
2.
3.
4.
log 12 given that log 2 = 0.3010 and log 3 = 0.4771
log 2 1/8
log 1018 + log 10 5 – log10 9
Express log pq2 r in terms of log p, log q and log r.
CDL, University of Maiduguri, Maiduguri
16
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
Solution
1.
log 12 = log (4x3) = log 4 + log 3
= log 22 + log 3
= 2 log 2 + log 3
Substituting the value of log 2 and log 3 we have
= 2 (0.3010 )+0.4771
= 0.6020 + 0.4771
= 1.0791
2.
log 2 1/8= log2 1-log2 8 = 0 – log2 23 = 0 – 3 log2 2 = 0 – 3x1 = -3.
3.
log10 18 + log10 5 – log10 9 = log10 (18x 5) / log10 9 = log10 90/9 = log10 10 = 1
2
4.
log pq2 r = log p + log q2 + log r = log p + log q2 + log r = log p + 2 log
q + ½ log r.
Exercise
Given that log 2 = 0.3010, log 3 = 0.4771 and log 5 = 0.6990, evaluate the following:
1.
2.
3.
4.
5.
log 1.2
log 5
log 9/4
log 3 3
log 45
NOTE:
10.
If a logarithm is given without a be, it is generally assumed to be in based
However, another way of presenting the rules of common logarithms is as follows:
For any M and N ,
Let M = 10x and N = 10y then log10 M = x and log10 N = y,
we have the following rules of common logarithms.
M x N = 10x x 10y
M x N = 10x+y
Applying the definition of logarithm, we
Rule 1:
Log10 (M x N) = x + y
Substituting or x and y, we have
Log10 (M x N) = Log M + Log N
Rule 2:
Similarly,
CDL, University of Maiduguri, Maiduguri
17
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
x
M 10
x y

 10
y
N 10
By definition of logarithm, we have
M
)=x–y
N
Substituting for x and y, we have
Log10 (
Log10 (
M
) = log M – log N
N
Rule 3:
For any positive number n, we have
Log Mn = n Log M.
CDL, University of Maiduguri, Maiduguri
18
EDU 120
3.0
MATHEMATICS FOR EDUCATION STUDENTS
TOPIC:
UNIT: 2
NUMBER SYSTEMS
There are five number systems. There are as follows:
1.
NATURAL NUMBERS
Every day we use many different types of numbers to represent quantities. We
are so accustomed to using them that some people think of them as being part of
nature. Actually, different kinds of numbers were created by man at various stages of
his mathematical development to meet changing needs.
Numbers are classified into different kinds according to their properties.
When you are asked: “How many students are there in your class?”, you would
naturally start counting1, 2, 3, and so on until all the students are counted. These
numbers 1, 2, 3, etc which are obtained by counting are called COUNTING
NUMBERS OR NATURAL NUMBERS.
NOTE 1: All the natural numbers 1, 2, 3, … when put together form the natural
number system.
2.
INTEGERS
When you are asked: “How many students in your class are below the age of
15?”, you would probably find that there is no student below this age. In this case,
we introduce the number zero, i.e. 0.
In other cases, w may even need to count beyond zero. For example, the
temperature of the freezer is 8 degrees below zero on the Celsius scale. Then we use
negative counting numbers and say that the temperature is -80 C. Negative quantities
are regarded as less than zero.
Thus, the numbers … -2, -1, 0, 1, 2, … are called INTEGERS. The negative
counting numbers -1, -2, -3, … are called NEGATIVE INTEGERS, and the natural
numbers 1, 2, 3, … are called POSTIVE INTEGERS. Zerois neither positive nor
negative.
NOTE 2: The natural numbers, zero and negative counting numbers together
form a number system known as the integer system.
NOTE 3: An integer which is divisible by 2 is an even number. Otherwise, it is an
odd number. Any even integer can be written as 2n where n is an integer.
For example,
14 = 2 x 7
8=2x4
0=2x0
-6 = 2 x (-3)
CDL, University of Maiduguri, Maiduguri
19
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
Conversely, 2n where n is an integer must be even because it is certainly divisible
by 2.
As the number following an even integer is odd, every odd integer is the sum of
an even integer and 1.
For example:
7=6+1
-11 = (-12) + 1
43 = 42 + 1
It follows that any odd integer can be written as 2n + 1, where n is an integer.
Thus,
7 = 6 + 1 = 2x 3 +1
-11 = (-12) + 1 = 2 x (-6) + 1
43 = 42 + 1 = 2 x 21 + 1
RATIONAL AND IRRATIONAL NUMBERS
Sometimes, we meet quantities which cannot be represented by integers. For
example, when 1 litre of milk is shared equally among 4 children, each child gets ¼ litre
of milk.
Division is not always possible if we only the integers. Overcome this difficulty,
we introduce the rational numbers.
Definition:
A rational number is a number which can be expressed as
p
where p
q
and q are integers and q ≠ 0.
In other words, a rational number is a ratio-nal number, i.e. a number expressible
as a ratio.
The following are examples of rational numbers.
¼, ½, ¾, ⅓
Note that a rational number can be expressed in many different ways as ratios.
For example,
3 6 21  3  12
 


4 8 28  4  16
and
 2 6 8
4
10




5
15
20  10  25
CDL, University of Maiduguri, Maiduguri
20
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
In general, two rational numbers
UNIT: 2
a c
 are equal if ad =bc.
b d
As a matter of fact, all the integers are rational numbers. Example,
4=
4 8 12  4  8
 


 ...
1 2 3
1  2
-7 =
0=
 7  14
7
14



 ...
1
2
1  2
0
0
0

  ...
1 1 2
NOTE:
All the rational numbers put together form the rational number system.
Therefore, the rational number system includes the integer system.
EXAMPLE:
Express the following rational numbers as decimals:
i)
5
4
ii)
7
8
iii)
5
9
iv)
5
6
Solution:
i)
5
 5  4  1.25
4
ii)
7
 (7)  8  (7  8)  0.875
8
iii)
.
5
 0.555...  0.5
9
iv)
.
5
 0.8333...  0.83
6
.
In these examples, 1.25 and -0.875 are terminating decimals, whereas 0. 5 and .
.
0.8 3
are repeating or recurring decimals.
CDL, University of Maiduguri, Maiduguri
21
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
NOTE:
terminating
Any rational number when expressed in decimals form, is either a
decimal or a repeating decimal.
IRRATIONAL NUMBERS
Perhaps, you would think that decimals are either terminating or repeating
decimals. But observe the following numbers carefully:
0.101001000100001…
0.252255222555…
0.370337033370…
These are neither terminating nor repeating. You can invent more! They are
called NON-TERMINATING NON-REPEATING DECIMALS.
p
These numbers are not rational numbers. That is, they cannot be expressed as
q
where p and q are integers because otherwise, their decimal forms can either be
terminating or repeating. Thus they are called irrational numbers.
DEFINITION:
An irrational number is a number which cannot be expressed as
where p
p
q
and q are integers with q≠0.
Numbers like:
2  1.41421356...
∏= 3.14159265…
3
9  2.08008382...
sin 15  0.25881904...
0
are all non-terminating non-repeating decimals. They are irrational numbers.
Note: The square root of any number which is not a perfect square is an irrational
number. e.g.
2, 3, 5, etc.
In addition, there are also many irrational numbers which cannot be expressed as
roots such as ∏, log 10 2, etc.
REAL NUMBERS
CDL, University of Maiduguri, Maiduguri
22
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
Rational numbers and irrational numbers are used in elementary mathematics,
scientific calculations and in everyday situations to represent real quantities.
The figure below shows a circle with radius r= 2.4cm (a rational number). Its
circumference is 15.07964… cm (an irrational number).
2.4cm
DEFINITION:
A real number is any number that is either a rational number or an irrational
number. All the real numbers put together form the real number system.
NOTE:
A very important property of the real numbers is that every real is either positive,
negative or zero.
The real number Pizza below illustrates the relationship between the different kinds of
numbers we have learned so far.
Irrational numbers,
2,∏
Rational numbers,
1 5 1
,
,
2 4 3
Integers
Negative
Integers,
-1, -3, -2
Zero
o
Positive
integers
(natural Nos)
1, 19, 97
THE REAL NUMBER PIZZA
CDL, University of Maiduguri, Maiduguri
23
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
EXERCISES
1.
Solve the following equations. Put a tick against the kind of numbers to which
the solution(s) belongs.
Solution(s)
of
equations
Natural
numbers
Integer
system
Rational
number
system
Irrational
numbers
Real
number
system
i) 2x – 8 = 0
ii) 3x + 21 =
0
iii) x2 -2x +
1=0
iv) 6x2 + x –
2=0
2.
If a is rational and b is irrational , determine which of the following statements
is/are always true:
i)
3.
a + b is irrational
ii)
a – b is irrational
iii)
a x b is irrational
iv)
a / b is irrational
State whether each of the following real numbers is rational or irrational:
i)
v)
-0.76 ii)
16 vi)
5
4+
iii)
10.01001001 iv)
∏ = 3.14
2
CDL, University of Maiduguri, Maiduguri
24
EDU 120
4.0
MATHEMATICS FOR EDUCATION STUDENTS
TOPIC:
UNIT: 2
SURDS
We learned that
2and 5 are irrational numbers. It is not true that the square
root of any number is irrational. E.g.
16 =4 ,
0.25 = 0.5 and
9 3
 . These roots
4 2
are rational.
But, the square root of any rational which is not a perfect square is irrational. E.g.
5
are irrational numbers. These are called surds.
7,
11
DEFINITION:
A surd is an irrational number in the form of roots of rational numbers.
Thus,
7
2,
NOTE:
14 , ½ 3 are surds.
a denotes the positive square root of a.
16 = 4, but not -4.
ORDER OF SURDS
A surd is said to be order n if it is written as the nth root of a rational number.
e.g.
3 and
2,
3
4,
NOTE:
E.g.
3
5 are surds of order 2 (or called quadratic surds).
5 and 3 18 are surds of order 3
A surd of any order may be transformed into a surd of a different order.
3=
1
2
3
2
 34  4
2
3
=
4
9
1
i)
n
ii)
n
a  an
a
m
m
 an
Note: For any two positive real numbers a and b, and a positive integer n, if a < b, then
n
a  n b.
Exercises:
CDL, University of Maiduguri, Maiduguri
25
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
1.
Arrange the following surds in ascending order of magnitude without using
tables or calculators. 21, 13, 5, 15
2.
Express 2 , 3 4 , 4 5and 6 10 as surds of order 12 and arrange them in
descending order of magnitude.
SIMPLICATION OF SURDS
A surd is said to be in its simplest form if the number under the root sign is a
positive integer as small as possible. Usually, we give answers of surds in their
simplest forms.
NOTE:
i)
ab  a  b
ii)
a

b
a
b
Examples:
1
Express the following surds in their simplest forms:
i)
12
180
iii)
2 x3  2 x
32 3
ii)
5
9
iv)
4
7
Solution:
i)
12 
ii)
180 
iii)
5

9
5 1

5
9 3
iv)
4

7
4
2
2 7
2



7
7
7
7 7 7
3.
2
2
2 x 3 x5  2 x 3 x
2
2
2
2
5  2 x3 x 5  6 5
3  1.7321 , without using tables or
Given that
2  1.4142 and
calculators, evaluate
the values of the expressions correct to 3 significant
figures.
CDL, University of Maiduguri, Maiduguri
26
EDU 120
i)
27
MATHEMATICS FOR EDUCATION STUDENTS
ii/)
3
4
iii)
UNIT: 2
6
2
Solution
i)
fig.)
ii)
iii)
27  9 x3  9 x 3  3 3  3x1.7321  5.1963  5.20 (correct to 3 sig.
3

4
3 1
1

3  x1.7321  0.86605  0.866 (correct to 3 sig. fg.)
2
4 2
6
6
2 6

x

2  3 2  3x1.4142  4.2426  4.24 (correct to 3
2
2
2 2
sig. fig.)
Exercises:
1. Transform the following surds into surds of the 12th order and arrange in
ascending order of magnitude. 3 , 3 4 , 4 6 , 6 10
2. Express the following surds in their simplest forms:
144
243
i)
iv)
27 ii)
50 iii)
27
5
OPERATIONS OF SURDS
A SURD EXPRESSION is an expression which involves surds with or without
1
2  3 , 8 + 6 7 , 2 3  3 5, and
rational numbers. E.g.
are surd
2 3
expressions.
LIKE AND UNLIKE SURDS
Surds are called like surds if they have the same irrational factors when they are
1
2 are like surds.
expressed in their simplest forms. E.g.3 2 , 5 2 and
4
Two surds are called unlike surds if they are not like surds, i.e. having different
1
7 are unlike surds.
irrational factor. E.g. 3 5 and
4
CDL, University of Maiduguri, Maiduguri
27
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
Note:
Like surds can be combined (by adding or subtracting) into a single surd,
while unlike surds cannot. Unlike surds when added together give surd expressions.
Examples:
Simplify the following expressions:
7 2 7 5 7
i)
27  300 
ii)
49
3
Solution:
i)
7 2 7 5 7 =
ii)
27  300 
7 (1  2  5)  4 7
7
49
= 3 3  10 3 
3
3
7
7
32
3  (13  ) 3 
3
= 13 3 
3
3
3
Exercises:
Simplify the following expressions:
i)
3 34 32 3
ii)
4 36  216  2 81  5 24
iii)
2 ( 6  2 8)
CDL, University of Maiduguri, Maiduguri
28
EDU 120
5.0
MATHEMATICS FOR EDUCATION STUDENTS
TOPIC:
UNIT: 2
RATIONALIZATION OF SURD EXPRESSIONS
DEFINITION:
The process of making the denominator of a surd or surd expression rational is
2
2
7 1
called rationalization. E.g.



14 . The factor used in this process such
7
7
7 7
as
7 in this example, is called a rationalizing factor.
As observed, the product of two surd expressions may be rational. For example,
(3 2  2 3 )(3 2  2 3 )  18  12  6
Such expressions like (3 2  2 3 )and (3 2  2 3 ) which differ only in the sign
connecting the two terms are said to be CONJUGATE SURDS. The product of two
conjugate surds is always rational. Therefore, we have:
( a  b )( a  b )  (
Examples
i)
a)
2
(
b)
2
 a b
Rationalize the following expressions:
1
2 3 5
ii)
5 11
72 6
Solution:
1
=
2 3 5
2 3 5
2 3 5 2 3 5


2
2
12

5
7
(2 3)  ( 5
i)
ii)
1
2 3 5
x
2 3 5
=
2 3 5
7  2 6 5 11x(7  2 6 5 11x(7  2 6)
5 11
5 11


=
x
2
2
49  24
72 6 72 6 72 6
7  (2 6)
=
5 11x(7  2 6) 7 11  2 66

25
5
Exercises:
CDL, University of Maiduguri, Maiduguri
29
EDU 120
(1)
MATHEMATICS FOR EDUCATION STUDENTS
Given that
3  1.7321and 5  2.236 , simplify the expression
UNIT: 2
15
5 3 3 5
and evaluate the value of the expression correct to 3 significant figures.
2.
Rationalize the following expressions:
i)
2 2
3 5
ii)
1
2 3
iii)
1
1

7 2
7 2
iv)
6
6 3
CDL, University of Maiduguri, Maiduguri
30
EDU 120
6.0
MATHEMATICS FOR EDUCATION STUDENTS
TOPIC:
UNIT: 2
EQUATIONS
One of the most basic tools used in mathematics is an equation. An open
mathematical sentence in which two expressions are connected by an equal sign ‘=’ is an
equation. E.g 3x + 5 = -x is an equation.
In this equation, the letter x is called the unknown or variable. The 3x, 5 and –x
are the terms of the equation. Any term that does not contain a variable is called a
constant term, or simply a constant. Hence 5 is a constant. In the term 3x, the number 3
is called coefficient of x. In the term –x the coefficient of x is -1.
An equation which contains only term in x and constant is called LINEAR
EQUATION.
We have learnt that in solving simple equations, we make use of the following
four properties.
For any three numbers a, b and c, if a = b, then
i.
ii.
iii.
iv.
a+c=a+c
a–c=b–c
axc=bxc
a/c = b/c (c≠0)
EQUATIONS INVOLVING LIKE TERMS
When an equation involves two or more like terms on the same side, these like
terms should be combined before solving the equation.
Example 1:
Solve 2x + 6x + 25 = 1
Solution:
2x + 6x +25 = 1
8x + 25 = 1
8x + 25 – 25 = 1 – 25
8x = -24
8x/8 = -24/8
x = -3
If like terms appear on opposite sides of an equation, they should be brought
together on the same side of the equation so that they can be combined.
Example:
Solve 2x + 5 = 3x + 12
Solution:
2x + 5 = 3x + 12
Add -3x to both sides
2x + 5 – 3x = 3x + 12 -3
CDL, University of Maiduguri, Maiduguri
31
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
-x + 5 = 12
Add -5 to both sides
-x + 5 -5 = 12 – 5
-x = 7
Exercises:
3x = -24
→ x = -7
Solve the following equations 1)
2x + 5 = 9 + 4
2)
5x
+
EQUATIONS INVOLVING BRACKETS
When an equation involves brackets, these brackets should be removed first. Then if any
like terms are present, they should be combined.
Example :
Solve 3(x + 2) -4(x – 1) = 5
Solution:
3(x +2) -4(x – 1) = 5
Remove brackets
3x + 6 -4x + 4 = 5
3x – 4x + 6 + 4 = 5
-x + 10 = 5
Add -10 to both sides
-x + 10 – 10 = 5 – 10
-x = -5
multiply both sides by (-1)
x=5
Exercises:
1)
2)
Solve the following equations
3x – 2(x + 1) = 5(x + 2)
0.3t + 0.2(t + 10) = 5
CHANGE OF SUBJECT FORMULA
If we are given say for finding the volume of a solid right circular cylinder V=
∏r2h. The subject of the above formula is ‘V’ and we wish to change it so that ‘h’ is the
subject of the formula, we carry out the ordinary processes of algebra. Since V = ∏r 2h,
V
dividing both sides of the equation by ∏r2, h =
.
2
r
Examples:
1.
Make ‘W’ the subject of the formula T-W = WV2 /32x
CDL, University of Maiduguri, Maiduguri
32
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
Solution:
T-W = WV2 /32x
32xT-32xW = WV2 (cross multiplication)
32xT = WV2 + 32xW (open brackets)
32xT = W(V2 + 32x) (like terms)
dividing both sides by V2 + 32x
→
W=
32 xT
V
2
 32 x
1
1 1
Make x the subject of the formula and hence calculate x if y = 2½


R X Y
and R = 1½
2.
3.
A = P + PTR/100, make ‘P’ the subject of the formula
SOLUTION OF EQUATION INVOLVING FRACTIONS
When we solve an equation that contains fractions, we an clear off the fractions
by multiplying both sides of the equation by the L.C.M. of all the denominators of the
fractions. The following examples will illustrates this.
EXAMPLES:
1
Solve the equation (x-2)/3 – (3x-4)/4 = 1
Solution:
Multiply both sides by 12 which is the LCM of 3 and 4.
4(x – 2) – 3(3x – 4) = 12
Remove the brackets 4x -8 – 9x + 12 = 12
Collect like terms
4x – 9x – 8 + 12 = 12
→
→
→
→
2.
- 5x + 4 = 12
- 5x = 12 – 4
- 5x = 8
x = - 8/5
Solve x/2 + x/3 = 10
Solution:
x/2 + x/3 = 10
Multiply both sides by 6 which is the lcm of 2 and 3.
6 x (x/2 + x/3 ) = 10 x 6
6 x x/2 + 6 x x/3 = 60
CDL, University of Maiduguri, Maiduguri
33
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
3x + 2x = 60
5x = 60
divide both sides by 5
x = 12.
3.
Solve the following equations:
a)
x + x /3 = 4
b)
(p+1)/7 – 3(p -2)/14 = 1
Exercises:
Solve the following simple linear equations:
2y/5 – 3y/4 = 3/10
3(x -2)/2 – (x -3)/4 = 2
(t + 1)/(t- 1) = ¾
2(p -1)/5 – 3(p _ 1)/10 = p
1.
2.
3.
4.
PROBLEMS LEADING TO SIMPLE EQUATIONS
In solving some word problems, the following 4 steps should be followed.
1.
2.
3.
4.
Choose a letter for the unknown and say what it represents.
Set up an equation.
Solve the equation
Write the answer.
Note: If in an equation a letter has already be chosen for the unknown, we start directly
with step (2).
Example:
becomes
The original price of a carpet is N94x + 1). If it is reduced by 20%, it
N(3x + 3). Find the original price.
Solution:
Selling price = N(4x + 1) x (100% - 20%)
= N(4x + 1) x (1 – 0.2)
= N(4x + 1)(0.8)
Thus, (4x + 1)(0.8) = 3x + 3
Multiply both sides by 10
(4x + 1)(8) = (3x + 3) x 10
32x + 8 = 30x + 30
collect like terms
32x – 30x = 30 – 8
2x = 22
x = 11
CDL, University of Maiduguri, Maiduguri
34
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
Exercises:
1.
2.
3.
4.
A rectangular room, with length double its width, has a perimeter of 24cm.
Find its width and length.
A stick 24cm long is cut into pieces such that one is 6cm longer than the other.
Find the length of the shorter piece.
One number is 10 more than the other number, when the larger number is
subtracted from twelve times the smaller number, the difference is 45. Find
the smaller number.
In a bag, there are two kinds of coins: N1.00 coins and 50k coins. If there are
two more N1 coins than 50k coins in the bag and the total amount of money is
N115, find the number of N1 coins in the bag.
SIMULTANEOUS EQUATIONS
Equations such as x – y = 1 and 2x + y = 5 are linear equations in two unknowns
x and y.
For the equation x – y = 1, there are many ordered pairs in the form of (x, y) that
satisfy the equation. They are (3, 2), (2, 1), (1, 0), (0, -1), (-1, -2), etc.
Each ordered pair is a solution of the equation x – y = 1.
Another equation 2x + y = 5, there are also many ordered pairs that satisfy the
equation. They are: (3, -1), (2, 1), (1, 3), (0, 5), (-1, 7), etc.
Each ordered pair is a solution of the equation 2x + y = 5.
However, only (2, 1) is a common ordered pair that satisfies both equations at the
same time or simultaneously.
We call x – y = 1 and 2x +y = 5 a pair of simultaneous linear equations. The
ordered pair (2, 1) or x = 2, y = 1 is called the common solution or simply the solution of
the above simultaneous linear equations.
Note: To solve a system of simultaneous linear equations means to find their common
solution.
There are three methods that are commonly used to solve a system of simultaneous
equations.
CDL, University of Maiduguri, Maiduguri
35
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
They are:
1.
2.
3.
method of substitution,
method of elimination and
graphical method.
We will discuss the first 2 in the following sections.
METHOD OF SUBSTITUTION
The following examples illustrate how to solve a system of simultaneous equation
by the method of substitution.
Example:1
Solution:
then
Solve the simultaneous equations:
x + 2y = 6
3x – y = 11
(i)
(ii)
From (i), x = 6- 2y
(iii)
Express x in terms of y and
number the equation
Substitute (iii) in (ii), we have
3(6 -2y) – y = 11
18 – 6y –y = 11
18 – 7y = 11
18 – 11 = 7y
7 = 7y
y=1
Substitute y = 1 in (iii), we have
x = 6 – 2(1)
x=6–2
x=4
→
The solution is x = 4, y = 1
Example: 2
Solve the simultaneous equations:
4x – 3y = 20
(i)
6x + y = 8
(ii)
Solution:
From (ii), y = 8 – 6x
(iii)
Substitute (iii) in (1), we have
4x – 3(8-6x) = 20
4x-24 + 18x = 20
CDL, University of Maiduguri, Maiduguri
36
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
4x + 18x – 2 4 = 20
22x = 20 + 24
22x = 44
→
x=2
Substitute x = 2 in (iii), we have
y = 8 – 6(2)
y = 8 – 12
y = -4
The solution is x = 2, y = - 4
Exercises:
Solve the following simultaneous equations by substitution method.
1.
3x – 2y = 5
(i)
2x + 3y = 12
(ii)
2.
2x + y = 9
x – 3y = 8
(i)
(ii)
METHOD OF ELIMINATION
Consider the following equations:
2x = ½
3y = ¼
(i)
(ii)
What is 2x + 3y?
The result is:
2x + 3y = ½ +¼
Since 2x = ½ and 3y = ¼
From this, we can see that the sum of the left hand sides of equation (i) and
equation (ii) is the same as the sum of right hand sides of the two equations.
Similarly, 2x – 3y = ½ - ¼ .
Hence, the difference between the left hand sides of the two equations is the same as the
difference between the right hand sides of the two equations.
We now apply this idea to solve a system of simultaneous equations.
Example:
Solve the simultaneous equations:
3x + 2y = 19
3x – 2y = 11
(i)
(ii)
Adding the two equations:
CDL, University of Maiduguri, Maiduguri
37
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
(i)
(ii)
UNIT: 2
3x + 2y = 19
3x – 2y = 11
(i) + (ii)
3x + 2y + 3x – 2y = 19 + 11
6x = 30
x =5
(iii)
When we add the two equations, the
terms in
y are eliminated, leaving a simple
equation in x.
Substitute (iii) in (i), we have
3(5) + 2y = 19
15 + 2y = 19
2y = 4
y=2
The solution is x = 5, y = 2
The method used above in solving simultaneous equations is called the method of
elimination.
Example 2:
Solve the simultaneous equations:
2x + 3y = 3
x – 2y = 12
(i)
(ii)
Solution:
[If we multiply equation (ii) by 2, (i.e. each term on both sides of equation
(ii) is multiplied by 2) and subtract the result from equation (i), the term in x will
be eliminated].
(i)
(ii) x 2
(i) – (iii)
2x + 3y = 3
2x -4y = 24
7y = - 21
y = -3
(iii)
(iv)
Substitute (iv) in (i), we have
2x + 3(-3) = 3
2x – 9 = 3
2x = 3 + 9
2x = 12
x=6
The solution is x = 6, y = -3
CDL, University of Maiduguri, Maiduguri
38
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
Exercises:
Solve:
1.
2x – y = 4
2x + y = 8
(i)
(ii)
2.
7x – 3y = 18
3x - 7y = 2
(i)
(ii)
UNIT: 2
QUADRATIC EQUATIONS
SOLUTION BY FACTORIZATION
A quadratic equation is an equation of the form ax2 + bx + c = 0, in which a, b and
c are constants and a ≠ 0. A quadratic equation is readily solved if the algebraic
expression can be factorized. The method of factorization is based on the followed fact.
If a and b are real numbers such that a x b = 0 then a = 0 or b = 0.
For if both a and b are non-zero, the product a x b cannot be zero. Thus at least one of
them must be zero.
Example 1:
Solve x2 – 5x – 6 = 0
Solution:
x2 – 5x – 6 = 0

x
-6
x
-6x
+1
+x = -5x
x
-3
3x
-9x
+2
+2x = -7x
After factorization,
(x -6)(x + 1) = 0
Thus, (x – 6) = 0 or (x + 1) = 0
x = 6 or x = -1
Example 2:
Find the roots of (3x – 1)(x + 2) = 4)3x + 1).
Solution:
(3x – 1)(x + 2) = 4)3x + 1).
Expanding both sides:
3x2 + 5x – 2 = 12x + 4
3 x2 – 7x – 6 = 0
CDL, University of Maiduguri, Maiduguri
39
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
(x – 3)(3x + 2) = 0
Thus, x -3 = 0 or 3x + 2 = 0

Exercises:
1)
x = 3 or x = 2/3
Solve the following quadratic equations by factorization:
x2 + 7x + 12 = 0
2)
6x2 – 11x + 5 = 0
SOLUTION BY COMPLETING THE SQUARE
Some quadratic equations such as x2 – 4x + 2 = 0 cannot be solved by the method
of factorization. Apart from using the graphical method to get approximate solutions, we
can apply a new method which gives exact answers. This method is based on the fact
that any quadratic equation can be rearranged in the form
(x + m) 2 = k, in which m and k are
constants.
Before discussing the method of rearranging equations, le us try solving equations
of this form first.
Example:
1)
2)
Solution:
Solve the following quadratic equations:
(x + 1) 2 = 36
(x – 2) 2 = 7, leaving the answers in surds.
1)
For (x + 1) 2 = 36,
Taking the square root of each side, we have:
x + 1 = 6 or x + 1 = -6

x = 5 or x = -7
3)
For (x – 2) 2 = 7
Taking the square root of each side, we have:
x–2= 

x=2+
7
7 or 2 -
7
CDL, University of Maiduguri, Maiduguri
40
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
Exercises:
Solve the following quadratic equations:
i)
(x + 3)2 = 16
ii)
UNIT: 2
(x – 5)2 = 121.
Now, we can see that a quadratic equation of the form (x + m)2 = k can be readily solved
by taking square roots. Before we can rearrange equations into this form, note the
following two identities:
i)
ii)
(x + a ) 2 ≡ x2 + 2ax + a2
(x – a) 2 ≡ x2 -2ax + a2
Consider the following identity:
x2 + 10 + ?? ≡ (x + ??) 2
Comparing this with x2 + 2ax + a2 ≡ (x + a) 2
We have 2a = 10, so that a = 5.
Therefore, x2 + 10x + 52 ≡ (x + 5) 2
In general, the expression x2 + 2ax or x2 – 2ax can be made into perfect squares by the
addition of a2 , i.e. the square of half the coefficient of x.
Exercises:
values in
Complete the following expressions into perfect squares by filling suitable
the blanks:
x2 + 4x + ……….. = (x + ….)2
x2 + 6x + ……….. = (x + ….)2
i)
ii)
Now, we shall apply this method to solve quadratic equations as follows:
Example:
1.
Solution:
Solve the equation x2 – 4x + 2 = 0
Step 1:
Put the constant term on the RHS:
x2 – 4x = -2
Step 2: Complete the square on the LHS and add the same value on
the RHS
4
x2 – 4x + ( )
2
2
4
 2  ( )
2
CDL, University of Maiduguri, Maiduguri
2
41
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
(x -2) 2 = 2
Step 3:
Take the square root of each side:
x–2=  2

x=2+
2 or x = 2 -
2
The process of transforming a quadratic equation into the form (x + m) 2 = k as in
step 2 above is called the method of completing the square.
Solve the equation 3x2 = 8x + 1, giving the answers correct
decimal places.
Example:
to 2
2.
Solution:
3x2 = 8x + 1
3x2 - 8x = 1
Since the coefficient of x2 is not 1 in this case, we have to reduce it to
unity by dividing both sides by 3.
x2 -
8
1
x
3
3
Note:
2
4 1 4
8
x2 - x + ( )   ( )
3
3
3
1 8 4
x 
2 3 3
2
3
(x -
4
)
3
2

19
9
Take the square root of each side, we have:
(x -
Thus, x =
=
4
19
)
3
9
4
19

3
9
4
19

3
3
CDL, University of Maiduguri, Maiduguri
42
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
4  19 8.359

 2.786 or
3
3
4  19  0.359
x=

 0.1197
3
3
x=

Exercises:
i)
x = 2.79 or -0.12 (correct to 2 decimal places).
Solve the following equations by the method of completing the square:
x2 - 10x + 20 = 0
ii)
2 + 3x – 4x2 = 0
CDL, University of Maiduguri, Maiduguri
43
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
SOLUTION BY THE QUADRATIC FORMULA
To solve a quadratic equation every time by completing the square is quite
tedious. In this case, we shall derive a formula for the roots of a general quadratic
equation.
Given the quadratic equation
ax2 + bx + c = 0 where a ≠ 0.
To reduce the coefficient of x to unity, divide the equation throughout by a:
2
x2 +
b
c
x 0
a
a
x2 +
b
c
x
a
a
2
2
b
1 b
c 1 b
1 b
Add ( x ) to both sides: x2 + x  ( x )    ( x )
a
2
a 2
2
a
a
2
a
2
(x +
b
)
2a
(x +
b  b  4ac
)
4a
2a
2
 b
4a
2
2

c
a
2
2
Take the square root of each side:
b
x+

2a
b

2a
formula.
x =
Example:
b
2
 4ac
2a
=
b
b
2
b
 4ac
2a
2
 4ac
2a
This is known as the quadratic equation
Solve the quadratic equation
3 2 5
 x 1  0
2x 2
CDL, University of Maiduguri, Maiduguri
44
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
Solution:
As some coefficients are fractions, we convert them to integers for
convenience.
Multiply the equation throughout by 2
3x2 – 5x – 2 = 0. Thus, a = 3, b = -5, c = -2
By the quadratic equation formula
x=
b
b
2
 4ac
2a
 (5) 
=
(5)
2
 4 x3 x(2)
2 x3
=
5  49
6
57
,
= 6
x = 2 or -1/3
Exercises:
i)
ii)
iii)
iv)
Solve the following equations by formula:
2x2 – 3x – 2 = 0
3x2 – 8x – 3 = 0
3x2 + 8x + 1 = 0
¼x2 = -2x – 4
CDL, University of Maiduguri, Maiduguri
45
EDU 120
7.0
MATHEMATICS FOR EDUCATION STUDENTS
TOPIC:
UNIT: 2
SIMPLE AND COMPOUND INTERESTS
SIMPLE INTEREST
If you deposit money in a bank, the bank pays you interest for the use of your
money. Similarly, if you borrow money from a bank, the bank charges you interest. The
money deposited or borrowed is called the PRINCIPAL. The sum of the principal and
the interest is called AMOUNT.
i.e. Amount = Principal + Interest.
In symbols,
A = P + I.
Where A, P and I denote the amount, principal and interest respectively.
Interest is usually paid at regular intervals say, yearly, half-yearly or monthly and
it is calculated as a percentage of the principal. E.g.
If the principal is N1000 and the interest rate is 5% per annum, then
For 1 year, interest = N1000 x 5% = N50
For 2 years, interest = N1000 x 5% x 2 = N100
For 3 years, interest = N1000 x 5% x 3 = N150
The interest calculated as above is called the simple interest.
Thus,
Simple interest = principal x interest rate x time.
Suppose a principal of P is invested at an interest rate of R% p.a. for T years, then the
simple interest, I, is given by
PxTxR
I=
100
NOTE:
Since A = P + I
A=P+
Example:1
PxTxR
RT
 P( I 
)
100
100
Find the simple interest on N2000 invested for 9 months at4% per annum.
Solution:
We have P = N2000, R = 4, T = 9/12
Since I =
PxTxR
100
CDL, University of Maiduguri, Maiduguri
46
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
200 x 4 x
I= N
100
UNIT: 2
9
2  N 60
Example: 2 What sum of money will produce N15 interest in 73 days at 5% per
annum? (Take
1 year = 365 days).
Solution:
We have I = N15, R = 5, T =
Since I =
73 1

365 5
PxTxR
100
P=
100 xI
TxR
change the subject of the formula.
100 x15
 N1500
1
5x
5
How long will N5000 become N5500 at 4% per annum simple interest?
P= N
Example:3
Solution:
We have P = N5000, A = N5500, R = 4.
Interest, I = A – P
= N5500 – N5000
= N500
Since I = N
T=
Exercises:
for 5
simple
2.
PxTxR
100
100 xI 100 xN 500

 2.5 years
PxR
N 5000 x 4
1.
A man invests N800 at 10% per annum and N600 at 5% per annum
years. Find the total simple interest received by him.
2.
For how long would N2,500 have to be deposited at 5% per annum
interest to gain N1,250 simple interest?
Find the principal which amounts to N208 in 2 years at 2% per annum
simple interest.
CDL, University of Maiduguri, Maiduguri
47
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
COMPOUND INTEREST
If N10,000 is deposited in a bank which pays interest at a rate of 5% per annum,
then the interest after 1 year is N500. The amount in the bank at the end of the first year
will be N10,500.
If this amount is not taken out from the bank in the second year, then the principal
for the second year will be N10,500. At the end of the second year, the interest will be
5
N10,500 x
 N 525. The corresponding amount will be (N10,500 + 525) = N11,025.
100
The total interest received at the end of the second year is N11025 – N10,000 = N1025.
The total interest calculated in this way is called the COMPOUND INTEREST.
In short,
Compound interest = total amount – original principal.
Note: In general, the total amount for compound interest is different from the total
amount for
simple interest since the principal for compound interest increases yearly.
Example 1:
Find the compound interest on N1000 for 3 years at 10% per anum.
Solution:
For the first year, principal = N1000
Rate = 10
1000 x10 x1
 N100
100
…………………………………………………………………………………..
Interest =N
For the 2nd year, principal = N1000 + N100 = N1100
Rate = 10
Interest = N
1100 x10 x1
 N110
100
………………………………………………………………………………………………
.
For the 3rd year, principal = N1100 +N110 = N1210
Rate = 10
Interest =
1210 x10 x1
 N121
100
CDL, University of Maiduguri, Maiduguri
48
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
UNIT: 2
Total amount for 3 years = N1331

Compound interest = N1331 – N1000 = N331.
Note: Alternatively, the compound interest can also be calculated by adding together the
interests
obtained in the 1st, 2nd and 3rd years.
The compound interest = N100 + N110 + N121 = N331.
Example 2: Find the compound interest on N5000 for one year at 2% per annum,
interest being
calculated half-yearly.
Solution:
For the 1st half-year, principal = N5000
Rate = 2
5000 x 2 x
Interest =
1
2  N 50
100
……………………………………………………………………………………………..
For the 2nd half-year, principal = N5000 + N50 = N5050
Rate = 2
5050 x 2 x
Interest =

100
1
2  N 50.50
Amount for 1 year = N5100.50

Compound interest = N5100.50 – N5000
= N100.50
………………………………………………………………………………………………
.
Exercises:
1)
a)
10% per
b)
annum,
2)
for the
year is 5% p.a.
Find the simple interest on N1000 for 1 year 6 months at
annum.
Find the interest on N1000 for 1 year 6 months at 10% per
compounded half-yearly.
Find the compound interest on N3000 for 3 years if the interest rate
first year is 7% p.a., for the second year is 6% p.a. and for the third
CDL, University of Maiduguri, Maiduguri
49
EDU 120
MATHEMATICS FOR EDUCATION STUDENTS
CDL, University of Maiduguri, Maiduguri
UNIT: 2
50