BASIC MATHEMATICS

David S I Iiyambo

University of Namibia

November, 2008

0.1 Acknowledgements

Usage of questions from Mathematics 1A and Foundation Mathematics past question papersfor exercises is acknowledged. Some questions were also taken from a variety of basic algebratextbooks. The proofreader is thanked for the time he/she has put in.

i

Contents

0.1 Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . i

1 Sets 11.1 Definition of a Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Cardinality of Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Subset and Proper subset . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4 Operations on Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.5 Venn Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.6 Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.6.1 Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.6.2 Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.6.3 Operations on Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.8 Answers to Exercises in Unit 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2 Algebraic Expressions 212.1 Simplification and Expansion of Algebraic Expressions . . . . . . . . . . . . . . . 24

2.1.1 Addition and Subtraction . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.1.2 Multiplication and Division . . . . . . . . . . . . . . . . . . . . . . . . . . 272.1.3 Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342.1.4 Factoring Quadratic expressions . . . . . . . . . . . . . . . . . . . . . . . 35

2.2 Rational Exponents and Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . 382.2.1 Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402.2.2 Rationalization of Denominator and Numerator . . . . . . . . . . . . . . . 42

2.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452.4 Answers to Exercises in Unit 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

3 Equations 483.1 Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513.2 Simultaneous Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

ii

3.3 Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 583.3.1 Solving By Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603.3.2 Solving by Completing the Squares . . . . . . . . . . . . . . . . . . . . . . 613.3.3 Solving by using The Quadratic Formula . . . . . . . . . . . . . . . . . . 63

3.4 Simultaneous Nonlinear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 653.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 683.6 Answers to Exercises in Unit 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

4 Inequalities 704.1 Linear Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 734.2 Nonlinear Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 744.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 774.4 Answers to Exercises in Unit 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

5 Absolute Value 795.1 Properties of Absolute Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 815.2 Equations Involving Absolute Values . . . . . . . . . . . . . . . . . . . . . . . . . 835.3 Inequalities Involving Absolute Values . . . . . . . . . . . . . . . . . . . . . . . . 855.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 875.5 Answers to Exercises in Unit 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

6 Polynomials 896.1 Addition and Subtraction of Polynomials . . . . . . . . . . . . . . . . . . . . . . 926.2 Multiplication of polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 926.3 Polynomial Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 936.4 Real Zeroes/Roots of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . 98

6.4.1 Factorizing Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 996.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1026.6 Answers to Exercises in Unit 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

7 Partial Fractions 1047.1 Case 1: Denominator with Distinct Linear Factors. . . . . . . . . . . . . . . . . . 1077.2 Case 2: Denominator with Repeated Linear Factors. . . . . . . . . . . . . . . . . 1097.3 Case 3: Denominator with Irreducible Quadratic Factors. . . . . . . . . . . . . . 1107.4 Case 4: Improper Rational Expressions. . . . . . . . . . . . . . . . . . . . . . . . 1127.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1147.6 Answers to Exercises in Unit 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

iii

8 Functions 1168.1 Definition of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1188.2 Arithmetic with functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1278.3 Even and odd functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1288.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1318.5 Answers to Exercises in Unit 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

9 Trigonometry 1339.1 Angle Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1359.2 Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1389.3 Evaluating Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . 1419.4 Graphs of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 1469.5 Basic Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1489.6 Trigonometric equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1519.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1549.8 Answers to Exercises in Unit 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

10 Sequences 15710.1 The General Term . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16010.2 Recursive Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16010.3 Arithmetic Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16110.4 Geometric Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

10.4.1 Infinite Geometric Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . 16710.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16810.6 Answers to Exercises in Unit 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

11 The Binomial Formula 17011.1 Factorial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17111.2 The Binomial Coefficient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17111.3 The Binomial Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17311.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17411.5 Answers to Exercises in Unit 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

iv

Unit 1

Sets

Content

1.1 Definition of a Set

1.2 Cardinality of Sets

1.3 Subset and Proper subset

1.4 Operations on Sets

1.5 Venn Diagrams

1.6 Intervals

1.6.1 Real Numbers

1.6.2 Intervals

1.6.3 Operations on Intervals

Introduction

Collections of objects of all types are found practically everywhere in life. The moment you haveobjects that have common trait (characteristic), you can form what is known in mathematics asa set of these objects. In this Unit, we will look at the formal definition of a set and how setsare presented, and also at the terminology used in relation to sets. Since the number of objectsbelonging to a set is in most cases informative, we will also look at this concept briefly. Thereare some sections in this Unit that deal with operations on sets and how we deal with subsetsof a given set. Other sections put a scope on the Real number system and on the subsets of theset of Real numbers, referred to as intervals.

1

Objectives

By the end of this Unit, you should be able to:

define a set, and determine whether two given sets are equal or not.

represent a set properly, concisely and understandably.

distinguish finite sets from infinite sets.

find the cardinality of a given set (if it is finite).

identify subsets of a given set.

perform rigorous operations on sets.

represent sets with, and interpret Venn diagrams.

prove identities involving sets.

show an understanding of the basic structure of the real number system.

represent intervals as sets, and on the number line.

Activities and Study Times

All the activities and studying this Unit should not take you more than six (6) hours.

2

1.1 Definition of a Set

Definition 1.1.1 In mathematics, a set is a collection of well-defined objects. This means thatone must be able to tell whether a given object belongs to that set or not. The objects that belongto a set are called the elements or members of the set, and the notation x X denotes the factthat x is an element of set X, while the notation x / X denotes the fact that x is not an elementof set X.

The elements of a set can be anything, as long as they have a common characteristic. Forexample, it can be a set of numbers, a set of books, a set of cows, a set of people, etc. Tothis end, sets are then given names: in most cases sets are given capital letters as their names,A,B,C,X, Y , etc. Intuitively1, lets consider equal sets as those sets that have the same elements.i.e. X = Y if set X has the same elements as set Y .

There are different ways of representing/describing a set and they are as follow;

A set can be represented by simply listing all its elements, separated by a comma (,).e.g. A = {1, 2, 3, 4}. The arrangement/order of elements within the set does not matter,so set A above is the same set as B = {3, 1, 2, 4}. i.e A = B. Elements of a given setare unique within that set, so a repeated element in a set will just be counted once. e.g.C = {1, 2, 3, 1, 4} is the same set as set A above.

A set can also be represented by giving a description of the (characteristics of the) objectsbelonging to that particular set. e.g. D = all the positive integers less than 5. Note thatD contain exactly the same elements as A above.

A set can also be represented by using a mathematical notation called the set buildernotation. e.g. E = {x|x Z and 1 x 4}, which again has exactly the same elementsas A above.

Example 1.1.1 Determine whether the following statements are true or false. Give reasons.

a) 3 {1, 2, 3, 4}

b) 9 {x|x N and x 10}

c) {2, 4, 6, 8} 6= {x|x is even and 0 < x < 10}

1This is not the formal definition of the equality of sets. The formal definition will come later in the text.

3

Solutions

a) True, because 3 is really an element of {1, 2, 3, 4}.

b) False. Although 9 is a natural number, 9 < 10.

c) False. Because these two sets contain exactly the same elements.

1.2 Cardinality of Sets

When you are dealing with a collection of objects, its only natural that there is an interestin the size of that particular collection, for several reasons. One is that smaller collections areeasier to deal with than larger collections. For instance, a Grade 12 class with 20 students iseasier to teach than that of say, 35 students. Therefore in mathematics we have a concept ofthe cardinality of a set (which gives the size of that particular set).

Definition 1.2.1 (Set cardinality) Let X be a set. The cardinality of X, denoted by |X|or sometimes n(X), is defined as the number of the elements of set X. i.e. |X| = n means thatset X contains n elements.

Definition 1.2.2 A set X is said to be finite (countable) if its elements can be counted (i.e.the cardinality is a finite natural number). Otherwise the set is said to be infinite.

Remark 1.2.1 There exists a set with no elements called the empty set or the null set andit is denoted by {} or .

Example 1.2.1 Find the cardinality of the following sets.

a) X = {4, 6, 8, 10, 12}

b) Y = {x|x N}

c) C =

d) A = {}

4

Solutions

a) |X| = 5.

b) Sice we can not count the elements of the set of all the natural numbers, we say that Y isinfinite or |Y | =.

c) Since C is an empty set and the empty set does not have elements, then |C| = 0.

d) |A| = 1 because A has only one element, namely .

Definition 1.2.3 If two sets have the same cardinality, then they are said to be equipotent(equivalent). If sets A and B are equipotent we denote it by A B, otherwise A B.

Example 1.2.2 If A = {10, 12, 14}, B = {a, b, c} and C = {}, then A B, but A C because|A| = 3, |B| = 3 and |C| = 1.

This is where we will leave the concept of equipotent sets (at least for now). If you continuewith mathematics after first year, then you will get to explore this concept more in advancedcourses on set theory.

1.3 Subset and Proper subset

Consider the collection of all the science students at the University of Namibia (UNAM). Nowthis collection/group includes all the UNAM students majoring in mathematics among others.i.e. The collection of all the mathematics majors at UNAM is a subcollection of the collection ofall the UNAM science students because every mathematics major student at UNAM is a UNAMscience student. To this end, we define a subset of a given set as follows;

Definition 1.3.1 Let X and Y be two sets. Set X is said to be a subset of set Y , denoted byX Y , if and only if every element of set X is also an element of set Y . i.e. if x X thenx Y . One needs to note here that X and Y could be the same set. If X Y and X 6= Y thenwe say X is a proper subset of Y and we denote this by X Y or X $ Y . One can then deducethat, set X is said to be equal to set Y , denoted by X = Y , if and only if X Y and Y X.

Note that for two sets X and Y , X $ Y (X a proper subset of Y ) implies that, for all x X,x Y , and there exist some y Y such that y / X. To show that set X is not a subset of setY , we use the notation X * Y .

5

Remark 1.3.1 Let X be any set. Then,

a) X X.

b) X.

Consider the following example.

Example 1.3.1 Let A = {5, 3, 7} and B = {1, 3, 5, 7, 9}.

a) Is A a subset of B?

b) Is B a subset of A?

Solution

a) Yes, A B because all the elements of A are also elements of B.

b) No, B * A because not all the elements of B are also elements of A, e.g. 1, 9 B but1, 9 / A.

Lets look at another example.

Example 1.3.2 Consider the following sets; A = {4, 6, 8, 10, 12, 14}, B = {12, 9, 10, 8, 14, 6, 16, 4, 18},C = {4, 6, 8} and D = {12}. Determine whether the following statements are true of false.

a) A B

b) A $ B

c) D C

d) C B

e) A * C

Solutions

a) True. All the elements of A are contained in B (although not in the same order).

6

b) True. A B for the same reason given in a), and there are some elements of B that are notelements of A. e.g. 16.

c) False. The only element of D (12) is not an element of C.

d) True. All the elements of C are contained in B.

e) True. There are some elements of A that are not elements of C. Come up with an example.

Remark 1.3.2 Let A, B, and C be sets.

1. If A B, then |A| |B|.

2. If A $ B, then |A| < |B|.

3. If A = B, then |A| = |B|.

4. If A B and B C, then A C and hence |A| |C|.

From Remark 1.3.1 it follows that for any given set X, we can find at least one subset of X(more than one subset if X is nonempty, and exactly one if X = ). Therefore, for any givenset X, we are able to write down a set, whose elements are all the possible subsets of X, and wedefine such a set as follows;

Definition 1.3.2 Let X be a set. The powerset of X, denoted by P(X) is the set whose elementsare all the possible subsets of X. The cardinality of the powerset of X is given by |P(X)| = 2|X|.

Example 1.3.3 Let X = {1, 2}. Then |X| = 2 and |P(X)| = 2|X| = 22 = 4.Hence, P(X) = {, {1, 2}, {1}, {2}}.

In Example 1.3.3, |P(X)| = 4 means that X = {1, 2} has four subsets.

In most cases when you are considering a problem, the set of all the elements under considerationsis called the universal set and it is denoted by U . Therefore, all the sets you are considering areactually subsets of some much larger set, U .

7

Exercises 1.1

1. Describe (i) in words, and (ii) using the set boulder notation, the sets whose elements arelisted below.

a) X = {3, 9, 27, 81, . . .}b) Y = {5, 10, 15, 20, 25, . . . , 100}

2. List the elements, and give the cardinalities of the following sets.

a) A = The set of the powers of 6 with not more than 5 digits.

b) B = The set of three digit positive numbers whose digits are all the same.

c) C = {x N|x > 12} .d) D = {x|x = n2, n N}.

3. Given that X = {0, 1, 3, 7}, Y = {0} and Z = {0, 1, 2, 3, 4, 5, 6}. Determine whether eachof the following is true or false.

a) Y Xb) X Yc) X Zd) Y Z

4. Let X be a set defined by X = {x N|x3 < 50, x 6= 0}.

a) List the elements of X.

b) Find |P(X)|.c) Find P(X).

8

1.4 Operations on Sets

One can perform operations on sets (within a specified U) in a similar manner as one canperform operations, such as addition and multiplication, on numbers. We now define some basicoperations on sets.

Definition 1.4.1 Let X and Y be two sets.

a) The union of set X and set Y , denoted by X Y , is a set of all the elements that are eitherin set X or in set Y or in both.

b) The intersection of set X and set Y , denoted by X Y , is a set of all the elements thatare in set X and in set Y . If the intersection of set X and set Y is an empty set, then wesay that sets X and Y are disjoint.

c) The complement of set X relative to set Y , denoted by Y \X or Y X, is the set of all theelements of set Y that are not in set X.

d) The complement of set X, denoted by X or Xc, is the set of all the elements of the universalset U that are not elements of set X.

Remark 1.4.1 Also note the following:

(i) If X Y then X Y = X and X Y = Y .

(ii) X (Y Z) = (X Y ) Z and X (Y Z) = (X Y ) Z.

(iii) (X ) = X.

Study the following examples.

Example 1.4.1 Let A = {6, 8, 10, 12} and B = {3, 6, 9, 12}

a) Find A B.

b) Find A B.

c) Find A \B and B \A.

9

Solutions

a) A B = {6, 12}.

b) A B = {3, 6, 8, 9, 10, 12}.

c) A \B = {8, 10} and B \A = {3, 9}.

Theorem 1.4.2 Let X and Y be subsets of the universal set U . Then de Morgans laws for setsstate that

(i) (X Y ) = X Y ,

(ii) (X Y ) = X Y .

Example 1.4.2 Let U = {x Z|2 x 12}, X = {x Z|x is a prime number} and Y ={x Z|x is a factor of 12}.a) Find the following sets.

(i) X

(ii) Y

(iii) X Y (iv) (X Y )

b) Find the following values.

(i) |X|(ii) |X Y |

Solutions

a) (i) X = {4, 6, 8, 9, 10, 12}(ii) Y = {5, 7, 8, 9, 10, 11}(iii) X Y = {2, 3, 5, 7, 11} {5, 7, 8, 9, 10, 11} = {5, 7, 11}(iv) (X Y ) = X Y = {4, 6, 8, 9, 10, 12}{5, 7, 8, 9, 10, 11} = {4, 5, 6, 7, 8, 9, 10, 11, 12}

b) (i) |X| = 5(ii) |X Y | = 9

10

1.5 Venn Diagrams

We use Venn diagrams2 to illustrate Sets and set operations diagrammatically. Venn diagramsare made up of a square/rectangle, which represents the universal set, and circles inside thesquare/rectange, which represent the sets under considerations. All the sets inside a universalset are subsets of it. For instance, consider the following scenario:

Let U = {1, 2, 3, 4, 5, 6}, X be the set of all factors of 6: X = {1, 2, 3, 6}, and Y be the set of allfactors of 4: Y = {1, 2, 4}. Then we can represent this using a Venn diagram as follows:

'

&

$

%

U X Y

&%'$&%'$

3

6

1

245

Now, in the above Venn diagram, X Y (which is the set of the natural numbers that arefactors of both 4 and 6) is represented by the overlapping region of the two sets, and X Y isrepresented by the region within the collective boundaries of X and Y .

We can also use Venn diagrams to indicate the number of elements in each set within theuniversal set. In terms of the number of elements in the indicated set, the above scenario isrepresented by a venn diagram as follows;

'

&

$

%

U X Y

&%'$&%'$

2 2 1

1

In the above diagram, the numbers appearing in each set are not elements of those particularsets, but rather they indicate the number of elements belonging to those sets. For instance,|X Y | = 2, |Y \X| = 1 and so on.

Example 1.5.1 In a group of 7 science students, 3 major in mathematics, 4 major in physics,and 2 do not major in mathematics and they do not major in physics.

a) Represent this scenario using a Venn diagram.2This method was invented by John Venn in 1880.

11

b) How many students in this group major in both mathematics and physics?

c) How many students major in maths but not in physics?

Solutions

a) Denote the set of the students majoring in mathematics with M and the set of the studentsmajoring in physics with P . Now, let x be the number of students that are majoring inboth mathematics and physics, i.e. |M P | = x. Then we have the following diagram;'

&

$

%

U M P

&%'$&%'$

3 x x 4 x2

b) Since the total number of students in the group is 7, from the venn diagram in a) we havethat

3 x+ x+ 4 x+ 2 = 79 x = 7

x = 2

c) The number of students that major in mathematics but not in physics is the cardinality ofthe compliment of set P relative to set M , i.e. |M \ P |. So,

|M \ P | = 3 x= 3 2= 1

1.6 Intervals

Before we give a definition of, and discuss intervals, it would be most helpful if we give a briefreview of the number system.

1.6.1 Real Numbers

We all first encountered numbers when we learned counting. So, the numbers we learned firstare the whole (counting) numbers 0, 1, 2, 3, 4, . . . i.e. the set {0, 1, 2, 3, 4, . . .}. This set is called

12

the set of natural numbers, denoted by N. If one is only interested in the positive naturalnumbers (without the 0) then we denote the set of all the positive natural numbers by N+ =N \ {0} = {1, 2, 3, 4, . . .}. So, we have N+ N.From there we moved onto fractions. For instance, one needed to represent dividing a wholebread among 3 or 4 people equally. After that, we were introduced to assigning a negative signto positive natural numbers, such as 2,7, and together with the natural numbers we obtainthe set of integers, denoted by Z = {. . . ,3,2,1, 0, 1, 2, 3, . . .}. Note that N Z.Now, if we do the same to the positive fractions to obtain negative fractions, the set of integerstogether with the set of positive and negative fractions give us the set of what is called therational numbers, denoted by Q. So, rational numbers are those numbers that can be repre-sented as fractions of two integers. The reason integers are also rational numbers is that theyare fractions of themselves and 1. e.g. 5 = 51 =

51 , 3 = 31 = 31 . Again, note that Z Q.

Furthermore, there are numbers that cannot be represented as fractions of two integers. Forinstance,

5, pi, etc. These are called the irrational numbers. The difference between rational

and irrational numbers is that rational numbers have terminating or repeating decimals, whileon the other hand, irrational numbers have nonterminating and nonrepeating decimals. Now, ifwe put the rational and irrational numbers together, we get a set of what is called as the realnumbers, denoted by R. Again, we have Q R. The above story can be represented using aVenn diagram as follows;

'

&

$

%

R

&%'$Q

ZNj

The real numbers can be represented on a number line, by choosing a point, marking it 0. Thenthe positive real numbers are marked uniformly on the right side of 0, while the negative realnumbers are marked uniformly on the left side of 0. For example;

-3 2 21 0 12 1 2 73 3

13

Theorem 1.6.1 (The Density of Real Numbers Property) Between any two rational num-bers you can always find another rational number, and between any two irrational numbers youcan find another irrational number.

Theorem 1.6.1 implies that between any two real numbers there are infinitely many real numbers.

1.6.2 Intervals

For a given subset of the set of real numbers to be called an interval, it need to satisfy theconditions given in following definition.

Definition 1.6.1 Let I R, and a, b I such that a < b. Then I is called an interval if givenany real number c such that a < c < b, we have c I.

You can think of intervals as corresponding to line segments;

-I

I is an interval

-I

I is not an intervalf

Example 1.6.1

a) A = {a R| 1 a 1} is an interval because for any x, y A and c R such thatx < c < y, c A.

-Av v

1 1012 12 32

b) B = {1, 0, 1} is not an interval because for example 1, 1 B, and 12 R where 1 < 12 < 1but 12 / B.

c) R is an interval, but N is not an interval.

14

We shall use the following notations for different types of intervals. For a, b R such that a b,we have the following;

(a, b) = {x R|a < x < b}

-f fa b

[a, b] = {x R|a x b}

-v va b

[a, b) = {x R|a x < b}

-v fa b

(a, b] = {x R|a < x b}

-f va b

[a,+) = {x R|a x}

--va

(a,+) = {x R|a < x}

-fa

15

(, b] = {x R|x b}

-vb

(, b) = {x R|x < b}

-fb

(,+) = R

-

1.6.3 Operations on Intervals

Intervals are sets, and therefore we can perform operations on them the same way as we did inSection 1.4 on other sets.For example, consider the two intervals A = (, 1) and B = (1,). The intersectionof A and B is the set of all the real numbers that are less than 1 and grater than 1. i.e.A B = (, 1) (1,) = (1, 1). This can also be illustrated using the number line asfollows;

-f

ff f1 10

For the intervals A = (,1) and B = (1,), their union is the set of all the real numbersthat are either less than 1 or greater than 1. i.e. AB = (,1)(1,). This is illustratedusing the number line as follows;

16

-f ff f1 10

17

Exercises 1.2

1. Let A = {2, 4, 8, 16, 32}, B = {4, 8, 12, 16, 20, 24, 24, 32}, and C = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}.List the elements of the following sets.

a) A B b) A Cc) A B d) C (A B)e) B \ (A B)

2. Two sets X and Y are such that |X| = 30, |Y | = 24, |X Y | = 47, and |U| = 50.

a) Represent this information using a Venn diagram.

b) Find the following:

(i) |X Y |(ii) |X \ Y |(iii) |Y \X|

3. Determine whether each of the following sets is an interval and give sufficient reasoning.

a) R b) Zc) [3, 8] (5, 6) d) (9, 9)

4. Determine whether each of the following statements is true or false.

a) 0 (7, 0) b) 8.7 [8, 9]c) 13 (0, 13 ] d) 4 (4, 6)e) 5 / (3, 6) \ {4, 5}

1.7 Summary

1. {} is not the same as , and is not the same as {0}.

2. The order of the elements in a set does not matter and theres no need to repeat elementsin a set. i.e X = {1, 3, 5, 7} = {3, 5, 7, 1} = {1, 7, 5, 3} and Y = {1, 3, 1} = {1, 3}.

3. All the elements of the powerset are sets.

18

1.8 Answers to Exercises in Unit 1

Exercise 1.1

1. (i) a) X = The set of all the powers of 3.

b) Y = The set of all the positive multiples of 5 not greater than 100.

(ii) a) X = {x|x = 3n, n N+}b) Y = {y|y = 5k, k N+, y 100}

2. a) A = {6, 36, 216, 1296, 7776, 46656} and |A| = 6b) B = {111, 222, 333, 444, 555, 666, 777, 888, 999} and |B| = 9c) C = {13, 14, 15, 16, 17, . . .} and |C| =d) D = {0, 1, 4, 9, 16, 25, . . .} and |D| =

3.a) True b) Falsec) False d) True

4. a) X = {1, 2, 3}b) |P(X)| = 8c) {, X, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}}

Exercise 1.2

1.

a) A B = {2, 4, 8, 12, 16, 20, 24, 28, 32} b) A C = {2, 3, 4, 6, 8, 9, 12, 15, 16, 18, 21, 24, 27, 30, 32}c) A B = {4, 8, 16, 32} d) C (A B) = {12, 24}e) B \ (A B) = {12, 20, 24, 28}

2. a)

'

&

$

%

U X Y

&%'$&%'$

30 y y 24 y3

19

b) (i) |X Y | = 7(ii) |X \ Y | = 23(iii) |Y \X| = 17

3.

a) not an interval b) not an intervalc) interval d) not an interval

4.

a) False b) Truec) True d) Falsee) True

20

Unit 2

Algebraic Expressions

Content

2.1 Simplification and Expansion of Algebraic Expressions

2.1.1 Addition and Subtraction

2.1.2 Multiplication and Division

2.1.3 Factoring

2.1.4 Factoring Quadratic Expressions

2.2 Rational Exponents and Radicals

2.2.1 Radicals

2.2.2 Rationalization of Denominator and Numerator

Introduction

Most of algebra and advance mathematics are made up mainly of algebraic expressions. Knowinghow to deal with these expressions is thus very important. In this Unit, we are going to learnabout what constitute algebraic expressions, what like terms in an algebraic expression are andhow they can be dealt with. There is also a section on how to simplify and add algebraicexpressions.

Objectives

By the end of this Unit, you should be able to:

identify algebraic terms.

21

collect, add or subtract like terms.

multiply and divide algebraic expressions.

use the laws of exponents appropriately and well.

simplify rational expressions.

factorize all algebraic expressions.

rationalize numerators or denominators of fractional expressions involving indices.

Activities and Study Times

All the activities and studying this Unit should not take you more than eight (8) hours.

22

The ability to manipulate algebraic expressions correctly is an essential skill for advanced math-ematics. But before we look at some of the basic algebraic manipulation, we need to know justwhat algebraic expressions are.

Definition 2.0.1 Algebraic expressions are made up of variables (normally represented by let-ters), numerical constants and signs of operations.

For instance, 2x + y is an algebraic expression, where x and y are the variables (or variablefactors), 2 and 1 are the numerical constants, and addition and multiplication are the signsof operations. Such an expression represents one quantity. Just as the sum of 1 and 2 is onequantity, namely, 3, the sum of x and y is one quantity, namely, x+ y.

Other examples of algebraic expressions are as follows:

a) 3x2 2x+ y

b) ax2 + bx+ c

c)ax

b+ y

d) 4x+ 3 3

xx

When there is nothing between two variables (i.e. when there is no sign between two variables),it is taken to mean that those two variables are being multiplied by one another. e.g. ax = a x,and xyz = x y z. The numerical constants put in front of variables are called the coefficientsof those respective variables. For instance, for the algebraic expression in a) above, 3 is thecoefficient for x2, 2 is the coefficient for x and 1 is the coefficient for y. Note therefore that,when theres no numerical constant in front of a variable, the coefficient of that particularvariable is taken to be 1.

Theres another concept about algebraic expression that we need to define, and thats the conceptof terms of algebraic expressions. The terms of an algebraic expression are the parts of theexpression that are connected by plus or minus signs. For example, in the expression 3abx+cyk,the terms are 3abx, cy, and k. An expression containing only one term such as 2x4, is called amonomial, while the ones that contain two terms are called binomials and the one that containthree terms are called trinomials.

23

2.1 Simplification and Expansion of Algebraic Expressions

In order to be able to operate on algebraic expressions, there are fundamental properties thatyou should be able to follow. The operations include addition/subtraction, multiplication, fac-torizing, using laws of exponents and to cancel like terms in the numerator and denominatorswhere necessary.

Definition 2.1.1 (Like Terms) Terms containing variables can be combined only if their vari-ables parts are the same. Terms containing variable factors in which the same letters are raisedto the same power are called like terms.

For example, 5x and 9x are like terms because they have the same variable factor, namely x.Also, 3a2b and a2b are like terms and the variable factor is a2b.

2.1.1 Addition and Subtraction

Algebraic expressions are added or subtracted by adding or subtracting like terms. Like terms areadded or subtracted by adding or subtracting the constant (numerical) coefficients and placingthe result in front of the variable factor. For example, 5x+ 9x = (5 + 9)x = 14x.

You are expected to know the basic sign manipulations. i.e. Things like xy = x+(y) = y+x,or (x + y) = x y, and so on. You are also expected to know that multiplying a positivewith a negative number gives a negative answer, multiplying a positive number with a positivenumber gives a positive number, and so does multiplying a negative number with a negativenumber.

Example 2.1.1 Simplify the following expressions.

a) 3x+ 7x

b) 2ab+ 5ab

c) (5x y) (2x+ 3y)

d) 4z (2 + 3z)

e) x+ 4xy 2x+ y y2

Solutions

a) Since 3x and 7x are like terms, we can add them to have

3x+ 7x = (3 + 7)x = 10x.

24

b) The two terms 2ab and 5ab are like terms. So,

2ab+ 5ab = (2 + 5)ab = 7ab.

c) In this expression we first need to open the brackets, and then collect the like terms. Theminus sign in front of the second set of brackets means that you should multiply everythinginside the brackets with 1. If we do this we get(5x y) (2x+ 3y) = 5x y 2x 3y.Now that we have opened the brackets, we need to collect the like terms, i.e. write thelike terms together and add/subtract them. Therefore,

5x y 2x 3y = 5x 2x y 3y = (5 2)x+ (1 3)y = 3x 4y.This implies that (5x y) (2x+ 3y) = 3x 4y.Now, since 3x and 4y are not like terms, we cant add them and this last expression isas simple as it can get and therefore it is our answer.

d) We open the brackets and add/subtract the like terms.

4z (2 + 3z) = 4z 2 3z = (4 3)z 2 = z 2Since z and 2 are not like terms, we cant add/subtract them and therefore this is ouranswer.

e) We collect and add/subtract the like terms. Note that the only like terms in this expressionare x and 2x. So,x+ 4xy 2x+ y y2 = x 2x+ 4xy + y y2 = x+ 4xy + y y2.Now, since x, 4xy, y and y2 are not like terms, we cant simplify this last expressionany further, and therefore it is our answer.

Addition/Subtraction of Fractional Algebraic Expressions

We do addition and subtraction of fractional algebraic expressions in a similar way as additionand subtraction of fractional numbers. Remember that to add 38 to

512 , you need to first look for

the least common denominator (lcd) (which is just the least common multiple of the denomina-tors), make it the denominator for your answer. You then divide this lcd by the denominatorsand multiply the respective numerators with the quotients and then add/subtract the products

25

(numerators). i.e.

38+

512

=3 3 + 2 5

24(24 is the lcd)

=9 + 1024

=1924

In a similar way, we add/subtract fractional algebraic expressions as follows.

Example 2.1.2 Simplify the following algebraic expressions.

a)x

y+s

t

b) x 22x+ 1

c)1

x+ 1 3x 1

Solutions

a) We note that for these two algebraic fractions, the lcd is yt. Hence we have

x

y+s

t=

xt

yt+sy

yt

=xt+ syyt

b) For this example, first we need to realize that x is also a fractional expression because it canbe written as x1 . The lcd therefore is just 2x+ 1. Hence we have the following,

x 23x+ 1

=x

1 2

2x+ 1

=x(2x+ 1)2x+ 1

22x+ 1

=2x2 + x 2

2x+ 1

c) We note that the lcd here is (x+ 1)(x 1) = x2 1. So,1

x+ 1 3x 1 =

x 1x2 1

3(x+ 1)x2 1

=x 1 (3x+ 3)

x2 1=

2x 4x2 1

26

Its very important to know how to add/subtract these types of expressions

because youll need this knowledge later in the course when we get to discuss

partial fractions.

2.1.2 Multiplication and Division

Unlike in the case of addition and subtraction, we can multiply/divide two algebraic terms evenwhen they are not like terms. e.g. 2x y = 2xy, and 2xy x = 2xyx = 2y. Before we take a lookat the laws and properties that will help us multiply/divide algebraic expressions with ease, webriefly need to understand a few concepts.

Consider the monomial x2. In this monomial, x is called the base, and 2 is called the exponent(the power) which indicates how many times one needs to multiply x with itself. e.g. x2 = x x.Now, if one is multiplying monomials like the one above, say x2 and x3, we have

x2 x3 = (x x) (x x x) = x x x x x = x5 = x2+3

This means that if you are multiplying monomials which have the same base, then you just needto keep the base and add the exponents. Similarly, for other operations we have the followinglaws of exponents.

Theorem 2.1.1 (Laws of Exponents) Let x, y R and m,n Z. Then;a) xmxn = xm+n

b) xm

xn = xmxn = xmn

c) (xm)n = xmn

d) (xy)m = xmym; (xmy)n = xmnyn

e) (xy )m = x

m

ym ; (y 6= 0)

f) (xy )m = ( yx)

m; Also ( 1x)n = xn; xm = 1xm ; (x, y 6= 0)

g) xmyn =

yn

xm ; (x, y 6= 0)

h) If x 6= 0, then x0 = 1.

It is very important that you verify and convince yourself that the laws of exponents listed inTheorem 2.1.1 above are indeed true. It does not help memorizing them if you dont understandthem.

27

Example 2.1.3 Simplify the following expressions.

a) (2x)(3x4)

b) (2x23)(5x79)

c) (3s)(st)

d) 2x2(x+ 3) 5x(x2 + 3x)

e) x(x 1) x(2 x) 3(x+ 5) 2x2

Solutions

a) For this expression, we multiply the coefficients and then we use the laws of exponents forthe variable factor.

(2x)(3x4) = (2 3)(x x4)= 6x1+4

= 6x5

b) We do the same as in a) to get,

(2x23)(5x79) = (2 5)(x23 x79)= 10x23+79

= 10x102

c) For this expression, note that the two terms have factors that have the same base s. So, weapply the laws of exponents for this one and then just multiply with the other factors. i.e.

(3s)(st) = 3 s1+1 t = 3s2t.

d) For expressions like this one, you need to open the brackets first, and then collect andadd/subtract the like terms.

2x2(x+ 3) 5x(x2 + 3x) = 2x3 + 6x2 5x3 15x2

= (2 5)x3 + (6 15)x2

= 3x3 9x2

28

e) In this case, we do the same as in d).

x(x 1) x(2 x) 3(x+ 5) 2x2 = x2 x 2x+ x2 3x 15 2x2

= (1 2 3)x+ (1 + 1 2)x2 15= 6x+ 0x2 15= 6x 15

We divide algebraic expression in a similar way as multiplication, except that we should notethat 1x = x

1. Therefore, when we divide x with y, we are really just multiplying x with y1.i.e.

x

y= x y1

Example 2.1.4 Simplify the following expressions.

a)2x5

3x3

b)xy2z

xz

c)a2bc3

ab1

Solutions

a) You just need to apply the laws of exponents. You are dividing, and therefore since the baseis the same, you keep the base, divide the coefficients, and and subtract the exponents.

2x5

3x3= (2 3)x53

=23x2

b) xy2z

xz= x11 y2 z11= y2

c) a2bc3

ab1= a21 b1(1) c3= ab2c3

29

Often we encounter cases where algebraic expressions of more than one term are being multiplied.To multiply two algebraic expressions that have more than one term, multiply all the terms inone expression with all the terms in the other expression, then simplify by adding/subtractingthe like terms if and where necessary. For example, (a+ b)(c+ d) = ac+ ad+ bc+ bd

The following types of products occur frequently in mathematics, and hence we lay them downexplicitly. Let x and y represent any real numbers or algebraic expressions, then

a) (x+ y)(x y) = x2 y2 (Difference of squares)

b) (x+ y)2 = x2 + 2xy + y2 (Perfect square)

c) (x y)2 = x2 2xy + y2 (Perfect square)

d) (x+ y)3 = x3 + 3x2y + 3xy3 + y3 (Perfect cube)

e) (x y)3 = x3 3x2y + 3xy2 y3 (Perfect cube)

Lets study the following example.

Example 2.1.5 Multiply and simplify the following expressions.

a) (x+ 1)(x+ 2)

b) (3x+ 7)(4x2 x+ 3)

c) (2xy z)(z + 2)

d) (x2 x 3)(x2 + 3x 1)

Solutions

a) Remember, we multiply each term in the fist expression with each term in the second ex-pression. i.e.

(x+ 1)(x+ 2) = x(x+ 2) + 1(x+ 2)

= x2 + 2x+ x+ 2

= x2 + 3x+ 2

30

b) (3x+ 7)(4x2 x+ 3) = 3x(4x2 x+ 3) + 7(4x2 x+ 3)= 12x3 3x2 + 9x+ 28x2 7x+ 21= 12x3 + 25x2 + 2x+ 21

c) (2xy z)(z + 2) = 2xy(z + 2) z(z + 2)= 2xyz + 4xy z2 2z

d) (x2 x 3)(x2 + 3x 1) = x2(x2 + 3x 1) x(x2 + 3x 1) 3(x2 + 3x 1)= x4 + 3x3 x2 x3 3x2 + x 3x2 9x+ 3= x4 + 2x3 7x2 8x+ 3

Multiplication/Division of Fractional Algebraic Expressions

To multiply two (or more) fractional algebraic expressions, you multiply the numerators withone another, and then the denominators with one another. The (unsimplified) answer wouldbe a fraction with the numerator a product of the numerators of the two (or more) fractionalexpressions, and the denominator a product of the denominators of the two (or more) fractionalexpressions. Again, this method is similar to the method of multiplying fractional numbers.

Example 2.1.6 Multiply the following algebraic fractions and simplify.

a)x

y s

t

b)y

21x4 14x

y2

c)z + 14

3z + 12

Solutions

a)x

y s

t=

xs

ty

31

b)y

21x4 14x

y2=

14xy21x4y2

=2

3x3y

c)z + 14

3z + 12

=3(z + 1)4(z + 12)

=3z + 34z + 48

Division of fractional algebraic expressions is performed in an almost similar manner as multi-plication, but you need to know the following definition.

Definition 2.1.2 Let x be an algebraic expression. The reciprocal of x is defined as1xor x1.

Now, dividing an algebraic expression with x is the same as multiplying that particular expressionwith the reciprocal of x. If x is a fractional algebraic expression (i.e. x =

y

z), then the reciprocal

of x is1x

=1(yz )

=z

y. Therefore, dividing an algebraic expression with

y

zis the same as

multiplying it withz

y. Let us thus consider the following examples:

Example 2.1.7 Simplify the following expressions.

a)x

y s

t

b)y

21x4 14x

y2

c)z + 14

3z + 12

Solutions

a) Note that the reciprocal ofs

tis

t

s. Therefore we have

x

y s

t=

x

y ts=

xt

ys

b) The reciprocal of14xy2

isy2

14x. Hence

y

21x4 14x

y2=

y

21x4 y

2

14x

=y3

294x5

32

c) The reciprocal of3

z + 12isz + 12

3. Hence

z + 14

3z + 12

=z + 14

z + 123

=(z + 1)(z + 12)

12

=z2 + 13z + 12

12

Exercises 2.1

1. Simplify the following.

a) 3x+ 9x 2xb) 4xy + y + 7xy

c) 8x (4x+ 2)d) 7

x 3x+ 2x

e)3y+

2y

f)1y 3

4y + 1

g)8

x 1 +2

3x 1h)

1x 2 +

2(x 2)2

2. Simplify the following.

a) (4xy)(xy)(7xy)

b) (3xy)(2x2)(

13xy

)c)

254 421253 161

d)(3x2y3

y2x1

)2e) (2x 5)(x+ 7)

33

2.1.3 Factoring

Before we look at how an algebraic expression can be factored, we will briefly give a statementon what factoring actually is.

Remember from your elementary number theory that if a, b, and c are integers different fromzero such that c = ab, then a and b are called factors of c. We will give an informal definition,in a similar way, of what a factor is with reference to algebraic expressions.

Definition 2.1.3 Let x, y, and z be algebraic expressions. If z = xy then x and y are said tobe factors of z. The process of writing z as a product of its factors is called factoring z (orsometimes factorizing z).

For example, the algebraic expression 3x2+9x has factors 3, x, and x+3 since it can be writtenas 3x2 + 9x = 3x(x+ 3).

Common Factors

The first thing you need to know about factoring is the concept of the greatest common factor(gcf ). The greatest common factor (gcf ) is a greatest number or letter which is a factor of allthe terms in the algebraic expression. For example, for the expression 3x2 + 9x, the gcf forthe terms 3x2 and 9x is 3x. Now, factorizing an algebraic expression involves dividing everyterm in the expression by the gcf and writing the expression as a product of the gcf and thesum/difference of the resultant quotients. For instance, for the expression 3x2+9x, we said thatthe gcf is 3x, and therefore after we divide every term in the expression with 3x and writingthe expression as a product of 3x with the sum of the quotients we get 3x2 + 9x = 3x(x+ 3).

Note that sometimes not all the terms in an expression have a common factor, but one may stillbe able to do some factoring. Here are some examples for you to follow:

Example 2.1.8 Factorize completely the following algebraic expressions.

a) 6x3y 9x2y3 + 12xy

b) 10a2 + 5a+ 2ab+ b

c) x2 + 2xy + 5x3 + 10x2y

34

Solutions

a) Note that the gcf in the expression 6x3y 9x2y3 + 12xy is 3xy. Therefore we have

6x3y 9x2y3 + 12xy = 3xy(2x2) + 3xy(3xy2) + 3xy(4) dividing every term by the gcf= 3xy(2x2 + 3xy2 + 4) multiplying the gcf by the sum of the quotients

b) For the expression 10a2 + 5a+ 2ab+ b not all the terms have a common factor, but we cansee that the first two terms have a common factor which is 5a and the last two terms havea common factor which is b. Therefore we get the following

10a2 + 5a+ 2ab+ b = 5a(2a+ 1) + b(2a+ 1) now we have a common factor x = 2a+ 1, i.e.

= 5ax+ bx

= x(5a+ b) factoring x out

= (2a+ 1)(5a+ b)

c) We have a gcf x. So,

x2 + 2xy + 5x3 + 10x2y = x(x+ 2y + 5x2 + 10xy)

= x(1(x+ 2y) + 5x(x+ 2y)) factoring inside the brackets

= x(1 + 5x)(x+ 2y) factoring x+ 2y out of the brackets

2.1.4 Factoring Quadratic expressions

Quadratic expressions are algebraic expressions with the general form

ax2 + bx+ c,

where a, b, c R. Now, not all quadratic expressions can be factorized, but those that can befactorized must have two linear factors, i.e factors of the form dx+ e for d, e R.

Factoring x2 + bx+ c by Trial and Error

The process of finding the gcf sometimes is really a matter of trial and error. We are startingwith a quadrating expressions in which a = 1.

Consider the expression x2 + bx+ c. To factorize this expression, we need to find two factors ofc whose sum is b. i.e. We needs to find two real numbers d and e (by trial and error) such that

35

c = de and b = d + e. This implies also that bx = dx + ex. We then rewrite the expression inthe following form;

x2 + bx+ c = x2 + dx+ ex+ de since bx = dx+ ex and c = de

= x(x+ d) + e(x+ d) factoring as in the previous subsection

= (x+ e)(x+ d) factoring x+ d out

Hence, we have managed to write the quadratic expression x2 + bx + c as a product of its twolinear factors. One should see therefore that the main task really is finding the two factors of c,whose sum is b, and the rest is employing things we covered in the previous subsection.

Example 2.1.9 Factorize the following quadratic expressions.

a) x2 + 5x+ 6

b) x2 + x 2

c) x2 3x 4

Solutions

a) The factors of 6 are 1, 2, 3, and 6 itself. Now, the two factors of 6 whose sum is 5 are 2 and3. Therefore we have,

x2 + 5x+ 6 = x2 + 2x+ 3x+ 2 3= x(x+ 2) + 3(x+ 2)

= (x+ 2)(x+ 3)

b) The factors of 2 are 1, 1, 2, and 2 itself. The two factors of 2 whose sum is 1 are 1and 2. So,

x2 + x 2 = x2 x+ 2x 2= x(x 1) + 2(x 1)= (x 1)(x+ 2)

c) The factors of 4 are 1,2, 1, 2, 4, and 4 itself. The two factors of 4 whose sums is 3are 1 and 4. So,

x2 3x 4 = x2 + x 4x 4= x(x+ 1) 4(x+ 1)= (x+ 1)(x 4)

36

Factoring ax2 + bx+ c by Trial and Error

We are looking at quadratic expressions where a 6= 1. For such expressions, we should expectfactors of the form ix+ j and kx+ l, where i, j, k, l R. i.e

ax2 + bx+ c = (ix+ j)(kx+ l) (2.1.4.1)

= ikx2 + (il + jk)x+ lj (2.1.4.2)

Now, from equation 2.1.4.2 we have that ac = iklj and the numbers il and jk (whose sum is b)are factors of ac. Therefore to factorize the quadratic expression ax2 + bx + c, we need to findtwo real numbers il and jk such that they are factors of ac and their sum is b.

Example 2.1.10 Factorize the following quadratic expressions.

a) 3x2 + 5x+ 2

b) 2x2 + 11x+ 12

c) 8x2 + 10x 3

Solutions

a) The factors of 3 2 = 6 are 1, 2, 3 and 6. The two factors whose sum is 5 and whose productis 6 are 2 and 3. So,

3x2 + 5x+ 2 = 3x2 + 3x+ 2x+ 2

= 3x(x+ 1) + 2(x+ 1)

= (x+ 1)(3x+ 2)

b) The factors of 2 12 = 24 are 1, 2, 3, 4, 6, 8, 12 and 24. The two factors whose sum is 11 andwhose product is 24 are 3 and 8. So,

2x2 + 11x+ 12 = 2x2 + 8x+ 3x+ 12

= 2x(x+ 4) + 3(x+ 4)

= (x+ 4)(2x+ 3)

37

c) The factors of 8 (3) = 24 are 12,8,6,4,3,2,1, 1, 2, 3, 4, 6, 8, 12, 24 and 24.The two factors of 24 whose sum is 10 and product is 24 are 2 and 12. So,

8x2 + 10x 3 = 8x2 2x+ 12x 3= 2x(4x 1) + 3(4x 1)= (4x 1)(2x+ 3)

The following are some special factoring formulae. Let x and y be numbers or algebraic terms,then

a) x2 y2 = (x y)(x+ y) (Difference of squares)

b) x2 + 2xy + y2 = (x+ y)2 (Perfect square)

c) x2 2xy + y2 = (x y)2 (Perfect square)

d) x3 y3 = (x y)(x2 + xy + y2) (Perfect cube)

e) x3 + y3 = (x+ y)(x2 xy + y2) (Perfect cube)

Remark 2.1.2 In a) above, expressions x y and x + y are said to be conjugates of eachother.

2.2 Rational Exponents and Radicals

When we looked at the exponents in Theorem 2.1.1, the expression xn was defined for x a realnumber and n an integer. In this section we want to modify the assumptions of Theorem 2.1.1to take care of cases where n is any rational number. e.g. we want to look at the meaning ofexpressions like 3

12 , and write down laws of dealing with such expressions. Therefore we have

the following definition.

Definition 2.2.1

If n is an even positive integer and x 0, then x 1n is the nonnegative real number suchthat (x

1n )n = x.

If n is an odd positive integer, then x 1n is the real number such that (x 1n )n = x.

38

For all positive integers m and n such that mn is in its simplest form, and for all realnumbers x for which x

1n is a real number,

xmn = (x

1n )m = (xm)

1n .

For instance, 912 = 3 because 32 = 9; (64) 13 = 4 because (4)3 = 64; and 9 32 = (9 12 )3 =

33 = 27, or 932 = (93)

12 = 729

12 = 27.

Remark 2.2.1 Note that if n is an even positive integer and x < 0, the x1n is not a real number.

In the same way that we have laws of exponents when the exponent is an integer (Theorem2.1.1), we have laws of exponents when the exponent is a rational number. The following lawsshould be followed when one is working with rational exponents.

Theorem 2.2.2 (Laws of Rational Exponents) Let n and m be rational numbers, and xand y be positive real numbers. Then

a) xn xm = xn+m

b)xn

xm= xnm

c) (xn)m = xnm

d) (xy)n = xnyn

e)(x

y

)n=

xn

yn

f) xn =1xn

Example 2.2.1 Simplify the following expressions.

a) (81x12y16) 14

b)x

12 y

56

x32 y

16

c)(x2y4

x2y

) 12

39

Solutions

Note that an exponential expression is in its simplest form when no powers of powers or negativeexponents appear.

a) We use the laws of rational exponents in Theorem 2.2.2 to simplify the expression as follows.

(81x12y16) 14 = 81 14 (x12) 14 (y16) 14= 3x 124 y 164= 3x3y4

b) In this expression we use the fact that 1b = b1. So,

x12 y

56

x32 y

16

= x12 y

56 (x

32 y

16

)1= (x

12x

32 )(y

56 y

16 )

= x12 3

2 y56 1

6

= x1y46

=y23

x

c) Simplifying this we get,(x2y4

x2y

) 12

=(x2y4)

12

(x2y)12

=(x2)

12 (y4)

12

(x2)12 y

12

=xy2

x1y12

= x1(1)y212

= x2y32

2.2.1 Radicals

When you are working with rational exponents, you are bound to encounter expressions of theform n

x. We will briefly look at how such expressions are defined and how they are handled.

Definition 2.2.2 In the expression nx, the square root symbol is called the radical sign, x

is called the radicand, and the positive integer n is called the index of the radical.

40

Now, the relationship between rational exponents and radicals is that; if n is a positive integerand x is any real number such that x

1n is defined as a real number then

nx = x

1n .

For example, 364 = 64

13 = 4, and 4

81 = 81

14 = 3, but 4

16 = (16) 14 is not defined as a realnumber (according to Remark 2.2.1).

If the index of the radical is 2 then the radical is written without specifying the index, i.e.2x =

x, which is referred to as the square root of x.

x is the positive square root x while

x is the negative square root of x. e.g. 4 = 2 and 4 = 2.

From above, it then follows that if n is a positive integer, m any integer, and x any real numbersuch that n

x is defined as a real number, then

xmn = (x

1n )m = ( n

x)m = n

xm.

Example 2.2.2 Evaluate the following expressions.

a) ( 38)2

b) (3)2

Solutions

a) ( 38)2 = 8

23 = (8

13 )2 = 22 = 4

b) (3)2 = 3

22 = 3

Combining the fact that nx = x

1n and the laws in Theorem 2.2.2 we have the following laws,

which should be followed when one is dealing with radicals.

Theorem 2.2.3 (Laws of Radicals) Let n and m be integers greater than or equal to 2, andlet x and y be nonnegative real numbers. Then

a) nx ny = nxy

b)nx

ny= nx

y(y 6= 0)

41

c) m

nx = mn

x

d) ( nx)n = x

e) nxn = x

Simplification of algebraic expressions involving radicals is in line with these rules.

Example 2.2.3 Simplify the following expressions.

a)81x2

b)50x9y11

c) 316x4y5

Solutions

Note that a radical expression is said to be simplified (at least so far) if the following are met;

The powers of the radicand are less than the index.

The index of the radical is as small as possible.

The radicand is not a fraction.

a)81x2 =

(9)2x2 =

92

x2 = 9x

b)50x9y11 =

(25x8y10)(2xy) =

(2x4y5)2 2xy = 5x4y52xy

c) 316x4y5 = 3

(8x3y3)(2xy2) = 3

(2xy)3 3

2xy2 = 2xy 3

2xy2

2.2.2 Rationalization of Denominator and Numerator

Firstly, we need to remember that if you have a quantity and you multiply it with 1 you getthe same quantity. i.e. the value of ones quantity does not change when multiplied with 1.For example 1

2 1 = 1

2. Rationalizing denominator and numerator of an algebraic fraction

(involving radicals) is based on this fact.

Definition 2.2.3 Rationalization is a process of changing a radical denominator or numeratorto a perfect nth power without changing the value of the algebraic expression.

42

To rationalize, we multiply the algebraic fraction with an (radical) expression equivalent to 1,that will make the appropriate radicand a perfect nth power.

Example 2.2.4 Rationalize the denominator in each of the following expressions.

a)12

b)73a

Solutions

a) We need to find an expression equivalent to 1 that will turn the radicand in the denominatorinto a perfect square. So, we find

22= 1 and therefore,

12=

1222=222

=22.

b) Going the similar route as in a) we get3a2

3a2

= 1, So,

73a=

73a

3a2

3a2

=7 3a2

3a3

=7 3a2

a.

The following example is on rationalizing the numerators of fractional expressions.

Example 2.2.5 Rationalize the numerator in the following expressions.

a)

23

b)53

6

Solutions

a) We proceed in a similar manner as in Example 2.2.4, only we do it to the numerator. So,23

=2322=

22

32=

232

b)53

6=

53

633=

532

63=

5 363=

523.

43

We look at even more complicated examples on rationalization.

Example 2.2.6 Rationalize the denominator in the following expressions.

a)4

x+y

b)3

xx+ h

Solutions

a) We multiply the top and the bottom with the conjugate tox +

y, namely

x y to

get,

4x+

y

=4

x+yxyxy

=4(xy)

(x+

y)(xy)

=4(xy)x y difference of squares

b) The conjugate of our denominator isx+

x+ h. So,

3xx+ h =

3xx+ h

x+

x+ h

x+x+ h

=3(x+

x+ h)

(xx+ h)(x+x+ h)

=3(x+

x+ h)

x (x+ h) difference of squares

=3(x+x+ h)

h

44

Exercises 2.2

1. Factorize the following expressions completely.

a) a3b3 b3.b) 18x4y2 + 12x3y3 24x2yc) 4x2 2xy 6x+ 3yd) 3x2 2ax 3bx+ 2ab

2. Factorize the following expressions

a) x2 + 6x+ 5

b) 9x2 + 10x+ 1

c) 22x2 77x+ 33d) 3x2 5xy + 2y2

3. Find all the values of k such that the trinomial x2 + kx+ 16 is a perfect square trinomial.

4. Simplify the following expressions completely.

a) (81x12y20)34

b) 827x3 8

29x13

c)

12x3y520x4y

5. Rationalize the numerator/denominator of the following expressions.

a)3 6x2y3

7x 6b)

35 +

x

c)x+ hx

h

2.3 Summary

1. Always factor out the greatest common factor of all terms.

2. After each factorization, examine the new factors to see whether they can be factored.

45

2.4 Answers to Exercises in Unit 2

Exercise 2.1

1. a) 10x

b) 11xy + y

c) 4x 2d) 6

x

e)5y

f)y + 1

y(4y + 1)

g)26x 10

(x 1)(3x 1)h)

x

(x 2)2

2. a) 28x3y3

b) 2x2

c) 5

d) 9x6y2

e) 2x2 + 9x 35

Exercise 2.2

1. a) b3(a3 1)b) 6x2y(3x2y 2xy2 4)c) (2x 3)(2x y)d) (x b)(3x 2a)

2. a) (x+ 1)(x+ 5)

b) (9x+ 1)(x+ 1)

c) 11(2x 1)(x 3)d) (3x 2y)(x y)

3. k = 8 or k = 8.

46

4. a)1

27x9y15

b) 4x2

c)y2

10x

70x

5. a)3x2y3

(7x 6) 6(x2y3)5b)

3(5x)

5 xc)

1x+ h+

x

47

Unit 3

Equations

Content

3.1 Linear Equations

3.2 Simultaneous Linear Equations

3.3 Quadratic Equations

3.3.1 Solving by Factoring

3.3.2 Solving by Completing the Squares

3.3.3 Solving by Using the Quadratic Formula

3.4 Simultaneous Nonlinear Equations

Introduction

Most real life situations can be modeled using equations (linear or nonlinear). So, knowing howto solve equation is an essential tool, not only in mathematics, but also in life in general. Inthis Unit, we will look at methods of solving linear and nonlinear equations in one variable, andmethods of solving linear and nonlinear simultaneous equations in two variables.

Objectives

By the end of this Unit, you should be able to:

solve single linear equations.

solve linear simultaneous equations.

48

determine whether a quadratic equation has real solutions, and the type of solutions.

solve quadratic equations.

solve nonlinear simultaneous equations.

Activities and Study Times

All the activities and studying this Unit should not take you more than seven (7) hours.

49

An equation is a statement that states that two mathematical expressions are equal. For exam-ple,

(i) 4 + 5 = 9

(ii) 7x 1 = 12 + x

The above statements are equations. Example (ii) contains an unknown variable, and what onewants is to find the value of the unknown variable that would make the equation true. Thisvalue is called the solution of the equation, and the process of finding a solution to an equationis called solving the equation.

Equations with the same solutions are called equivalent equations and we use the symbol toindicate that two equations are equivalent. Now, if you are given an equation, you can obtainan equivalent equation by performing the same mathematical operation (addition, subtraction,multiplication, division, etc.) on both the left and the right hand side of the equation. Here aresome properties of equity: Let the letters A,B,C stand for any algebraic expressions,

a) A = B A C = B C

b) A = B CA = CB (C 6= 0)

Remark 3.0.1 Note that, whatever operation you perform on one side of the equal sign, youshould perform the same operation on the other side as well to obtain an equivalent equation.

Before we look at ways of solving equations, we say some brief statements about a very importantconcept in solving equations, Inverse operations.

Definition 3.0.1 (Inverse Operations) Two operations are called inverse operations if onereverses the effect of the other.

For example, addition and subtraction are inverse operations. Multiplication and addition arealso inverse operations. This is to say that, if you start with an algebraic expression x, and youadd 2 to x, you obtain x+ 2. Now, if you then subtract 2 from this new expression x+ 2, youobtain (x+2)2 = x+(22) = x+0 = x. So, you obtain an expression x which you started with.On the other hand, if you start with the same expression x and you multiply with 2, you obtain2x. Now if you divide this new expression with 2, you obtain 2x 2 = 2x2 = 22 x = 1 x = x,which is again what you started with.

50

3.1 Linear Equations

Definition 3.1.1 A linear equation in one variable is an equation of the form ax+b = 0, wherea and b are real numbers and x is the variable.

We solve linear equations by using inverse operations to rewrite the given equation into anequivalent equation of the form

variable = number.

Consider the following example.

Example 3.1.1 Solve the following equations

a) 2x+ 7 = 31

b) 4x+ 7 = 9x 3

Solutions

a) Remember, we want to obtain an equivalent equation with the variable x on one side, andeverything else on the other side, by performing appropriate inverse operations on bothsides of the equal sign, as follows.

2x+ 7 = 31

(2x+ 7) 7 = 31 7 subtract 7 from both sides to undo the effect of +7 on the left2x = 24

12 2x = 1

2 24 divide both sides by 2 to have x on its own on the left .

x = 12

If you want to confirm that the value that you obtained is indeed the solution to yourequation, just substitute that value for your variable in the equation, and you should obtainthe same values on both sides of the equal sign. i.e. LHS1= 212+7 = 24+7 = 31 =RHS2,which is what you have on the right hand.

1LHS stands for Left Hand Side2RHS stands for Right Hand Side

51

b) Note that this is also a linear equation because, by using the inverse operations, we canrewrite this equation into an equivalent equation of the form ax + b = 0. i.e. we collectthe like terms.

4x+ 7 = 9x 3(4x+ 7) 7 = (9x 3) 7 subtract 7 from both sides

4x = 9x 104x 9x = (9x 10) 9x subtract 9x from both sides

5x = 10(1

5) (5x) = (1

5) (10) divide both sides by 5 to have x on its own

x = 2

From Example 3.1.1 b) we saw that there are some linear equation which are not of the formax+ b = 0, but which can be rewritten into an equivalent equation of the form ax+ b = 0. Thefollowing are some of such examples.

Example 3.1.2 Solve the following equations.

a)x 2x+ 1

=23

b)5

2x+ 3=

3x 1

Solutions

a) We use inverse operations to write this equation into a linear form and solve.

x 2x+ 1

=23

x 2 = 23(x+ 1) multiply both sides by x+ 1

3(x 2) = 2(x+ 1) multiply both sides by 33x 6 = 2x+ 2 expand both sides

3x 2x = 2 + 6 collect like termsx = 8

52

b) We follow the same procedure as in a) above to obtain the following.

52x+ 3

=3

x 15(x 1) = 3(2x+ 3) multiply both sides by (x 1)(2x+ 3)5x 5 = 6x+ 9 expand both sides5 9 = 6x 5x collect like terms14 = x

x = 14

Equations in Example 3.1.2 can also be referred to as the rational equations. In general, onlysome of the rational equations have equivalent equations in linear form. We will look at thosethat do not have equivalent equations in linear forms later in the course.

3.2 Simultaneous Linear Equations

All the equations that we have considered so far only contain one variable. For instance 2x+7 = 3is an equation in one variable only, namely x. There are however cases/systems were the numberof variables involved is more than one.

Now, for the system to have a unique solution (i.e. exactly one value for each variable) thenumber of equations representing the system must be the same as the number of variables. Forexample, the equation x + y = 2 is an equation in two variables namely, x and y, and thevalues x = 1 and y = 1 form a solution to this equation because 1 + 1 = 2. But you shouldalso note that the values x = 4 and y = 2 also form a solution to this same equation since4 + (2) = 4 2 = 2.

This means that this equation does not have a unique solution. To have a unique solution, weneed to introduce a second equation in two variables, and then we solve this pair of equationtogether. This will give the values of x and y that satisfy both the two equations. Solvingmore than one equation together is referred to as solving simultaneous equations, or some times,solving the equations simultaneously.

If these equations are all of the form ax + by + c = 0 (for equations in two variables) then wecall them simultaneous linear equations. In this course, we will only restrict ourselves to solvingtwo equations in two variables simultaneously.

53

There are several methods of solving equations simultaneously. We will discuss two of themhere.

Substitution Method

To solve simultaneous equations by the substitution method, you make one of the variables thesubject of the formula3 in one equation, and then substitute it in the other equation. This givesyou an equation in one variable, which we now know how to solve.

Example 3.2.1 Solve the equations simultaneously in each case.

a) 2x y = 2x+ y = 7

b) 4x+ y = 62x+ 3y = 24

Solutions

a) So, we consider this pair of equation and try to solve it using the substitution method.

2x y = 2x+ y = 7

If we make y the subject of the formula in the second equation we get y = 7 x. Now, inplace of y in the first equation, we substitute with 7 x and solve for x to get;

2x (7 x) = 22x 7 + x = 2

3x = 2 + 7

x =93

x = 3

Now, since y = 7 x, we have y = 7 3 = 4.

Note that you can verify whether the solution youve obtained is correct by substitutingthe values of x and y into the equations and see if the equations are satisfied.

3Making a variable the subject of the formula means that you use the inverse operations to have only that

variable on one side of the equal sign and everything else on the other side.

54

b) We follow the same procedure here as we did in a).

4x+ y = 62x+ 3y = 24

This time we are making y the subject of the formula in the first equation to get y =6 4x. We substitute this into the second equation and solve to get;

2x+ 3(6 4x) = 242x 18 12x = 24

14x = 24 + 18x =

4214

x = 3

y = 6 4x = 6 4(3) = 6 + 12 = 6.

The Addition/Subtraction Method

This method works best when one of the variables has either the same coefficient in both equa-tions or its coefficient in one equation is minus its coefficient in the other equation. That way,when you add the two equations, that variable gets eliminated and you obtain one equation inone variable, which we know how to solve.

If you have a pair of equations where no variable has the same coefficient in both equation andno variable has a coefficient in one equation which is minus its coefficient in the other equation,then you will need to multiply one of the equations with an appropriate real number to obtainone of the two cases. We will illustrate this in the following examples.

Example 3.2.2 Solve the following pairs of equations simultaneously.

a) x+ 3y = 11x+ y = 3

b) 4x+ y = 53x 2y = 1

55

Solutions

a) So we have the pair of equations

x+ 3y = 11

x+ y = 3

And we note that the variable x has the same coefficient, 1, in both equations. We thenproceed by subtracting the second equation from the first one as follows:

x+ 3y = 11

x+ y = 3

2y = 8

y = 4

Now, having obtained the value of y, we can substitute it in any of the two equation tofind the value of x. If we, for instance, decide to substitute it into the first equation, weget the following;

x+ 3 4 = 11x+ 12 = 11

x = 1

b) In this case we have the following pair of equations.

4x+ y = 5

3x 2y = 1

Now, since none of the variables have the same coefficient is both the equations, and notone has a coefficient in one equation that is minus its coefficient in the other equation,we try multiplying one of the equation with an appropriate real number. Note that if wemultiply the first equation with 2, the coefficient of y is the first equation would be 2 andin the second equation is already 2, which is what we want. So, we get,

2(4x+ y = 5)

3x 2y = 1

Which works out to be

8x+ 2y = 10

3x 2y = 1

56

Now we add the second equation to the first one to get the following;

8x+ 2y = 10

3x 2y = 111x = 11

x = 1

Substituting this value of x in the first equation we get,

4 1 + y = 54 + y = 5

y = 1

You should get the same solution for a given pair of simultaneous equation regardless of whichmethod you use. The choice as to which method to use depends on your personal preference(and maybe on the one you think saves you some time). Unless the method is specified, you arewelcome to use any method you are comfortable with, as long as you apply it the right way andobtain the correct answer.

We now turn our attention to nonlinear equations. Although there are other nonlinear equationsthat are well worth our time, in this course we will restrict ourselves to the quadratic equations.

Exercises 3.1

1. Solve the following equations.

a) 2x+ 3 = x 4b)

2x+ 13x 5 = 0

c)2y+

5y=

149

d)2

x+ 1=

64x 3

2. Use the substitution method to solve the following equations simultaneously.

2x y = 13x y = 2

57

3. Use the addition/subtraction method to solve the following equations simultaneously.

3x+ 2y = 22x 3y = 16

3.3 Quadratic Equations

Definition 3.3.1 A quadratic equation is an equation of the form ax2+ bx+ c = 0, where a, b, care real numbers with a 6= 0.

While all the quadratic equations have solutions, only some of the quadratic equations have realsolutions. In this course, if a quadratic equation does not have real solutions we would say thatit does not have a solution4. So, before we start solving quadratic equations, it makes senseto first know which quadratic equations have solutions (and what type of solution) and whichquadratic equations do not have solutions.

The Discriminant

Definition 3.3.2 Let a, b and c be real numbers such that a 6= 0. Then the discriminant of thequadratic equation ax2 + bx+ c = 0, denoted by 4, is given by

4 = b2 4ac.

Now, for a given quadratic equation, the discriminant determines if it has real solutions and thetype of solutions it would have, as follows:

(i) If 4 > 0 then the quadratic equation ax2 + bx+ c = 0 has two distinct real solutions.

(ii) If 4 = 0 then the quadratic equation ax2 + bx+ c = 0 has one (repeated) real solution.

(iii) If 4 < 0 then the quadratic equation ax2 + bx+ c = 0 has no real solution.

The reasons for the above three statements will become clear a little later, but for know, letsjust take them as given.

4As long as you keep in mind that it does have non-real solutions.

58

Example 3.3.1 Determine the number of real solutions the following quadratic equations have.

a) 2x2 3x+ 1 = 0

b) x2 14x+ 49 = 0

c) 2x2 + 5x+ 7 = 0

Solutions

We use the discriminant to determine the number of solutions for each equation.

a) Note that for this equation a = 2, b = 3 and c = 1. So,

4 = b2 4ac= (3)2 4 2 1= 9 8= 1

Since 4 > 0, the equation has two distinct real solutions.

b) In this case a = 1, b = 14 and c = 49. Therefore,

4 = b2 4ac= (14)2 4 1 49= 196 196= 0

That is, 4 = 0 and hence the equation only has one (repeated) real solution.

c) For this equation a = 2, b = 5 and c = 7. So,

4 = b2 4ac= (5)2 4 2 7= 25 56= 31

Now, since 4 < 0, the equation does not have real solutions.

In this course, we will look at three ways of solving quadratic equations, for those that have realsolutions. The first one is, solving by factoring.

59

3.3.1 Solving By Factoring

Before we go any further, we need to state a very important tool in solving equations by factoring,namely The Zero-Product Property.

Theorem 3.3.1 (The Zero-Product Property) Let and be algebraic expressions. Then

= 0 if and only if = 0 or = 0.

Now, lets consider the quadratic equation ax2+ bx+ c = 0. The left hand side of this equationis just a quadratic expression. The methods and procedures of factoring all kinds of quadraticexpressions that can be factored are outlined in Section 2.1.4.

We factor the left side of our quadratic equation and get an equation of the following form;

(ix+ j)(kx+ l) = 0.

Applying Theorem 3.3.1 to our resultant equation we get that

ix+ j = 0 or kx+ l = 0.

Solving these two linear equations we get the two solutions x = ji or x = lk . So, lets considerthe following example.

Example 3.3.2 Solve the following quadratic equations using the factoring method.

a) x2 + 3x 10 = 0

b) 6x2 + 21x 12 = 0

Solutions

a) We factor the left-hand side of the equation x2+3x 10 = 0 and use Theorem 3.3.1 to solvefor x.

x2 + 3x 10 = 0(x+ 5)(x 2) = 0 factoring the left-hand side of the equation

x+ 5 = 0 or x 2 = 0 using Theorem 3.3.1x = 5 or x = 2

So, the solutions for our equation are x = 5 and x = 2.

60

b) Similarly, we factor the left-hand side of the equation 6x2 + 21x 12 = 0 and solve to have,

6x2 + 21x 12 = 0(3x+ 12)(2x 1) = 0 factoring the left-hand side of the equation

3x+ 12 = 0 or 2x 1 = 0 using Theorem 3.3.1x = 4 or x = 1

2

So, the solution for our equation are x = 4 and x = 12 .

3.3.2 Solving by Completing the Squares

The whole idea of completing the squares is to obtain a perfect square after factoring. Considerthe expression x2 + bx. To obtain a perfect square from this expression we need to add a thirdterm. For example for the expression x2 + 6x, we need to add the term 9, which is

(12 6

)2 toget x2 + 6x+ 9. Now, factoring this we get (x+ 3)2. Note that 3 is half of 6. This implies thatfor the expression x2 + bx we need to add

(b2

)2to get x2 + bx+

(b2

)2, and when we factor this

we get (x+ b2)2. This is the idea we use in completing the squares.

So, for the quadratic equation of the form ax2 + bx + c = 0, where a 6= 0, we first factor aout of the first two terms to get a(x2 + bax) + c = 0, and subtract c from both sides to geta(x2 + bax) = c. Now completing the squares on the expression inside the brackets on theleft-hand side we get,

a

(x2 +

b

ax

)= c

a

(x2 +

b

ax+

(b

2a

)2)= c+ b

2

4awhatever you do on the left, do it on the right as well.

a

(x+

b

2a

)2=

b2

4a c factoring inside the brackets on the left.

a

(x+

b

2a

)2=

b2 4ac4a(

x+b

2a

)2=

b2 4ac4a2

After obtaining the equation above, we then use the following fact to solve for x.

Theorem 3.3.2 The solutions of the equation x2 = c are x =c and x = c.

61

Lets therefore consider the following examples.

Example 3.3.3 Solve the following quadratic equations by completing the squares.

a) x2 + 6x+ 1 = 0

b) 2x2 + 4x 1 = 0

Solutions

a) We start by taking 1 to the right-hand side, and then proceed to complete the squares onthe left-hand side as follows.

x2 + 6x+ 1 = 0

x2 + 6x = 1x2 + 6x+ 9 = 1 + 9 adding 9 on both sides

(x+ 3)2 = 8 factorizing the left-hand side

x+ 3 = 8 taking the square root on both sides

x = 38

Therefore, the solutions to our equation are x = 3 +8 and x = 38.

b) As in a) above, we start by taking 1 to the right-hand side and then complete the squareson the left-hand side.

2x2 + 4x 1 = 02x2 + 4x = 1

x2 + 2x =12

dividing throughout by 2

x2 + 2x+ 1 =12+ 1 adding 1 on both sides

(x+ 1)2 =32

factoring the left-hand side

x+ 1 =

32

taking the square root on both sides

x = 1

32

Therefore, the solutions to our quadratic equation are x = 1 +

32 and x = 1

32 .

62

3.3.3 Solving by using The Quadratic Formula

For the quadratic equation ax2 + bx+ c = 0, the formula for obtaining its solutions, referred toas the quadratic formula, is obtained through the method of completing the squares. We willthen state it as a theorem, and prove it.

Theorem 3.3.3 (The Quadratic Formula) The solutions of the quadratic equation ax2 +bx+ c = 0, where a 6= 0, are

x1,2 =bb2 4ac

2a.

Proof: Consider the quadratic equation ax2+ bx+ c = 0, where a 6= 0. Completing the squareson this equation we get,

a

(x2 +

b

ax

)= c

a

(x2 +

b

ax+

(b

2a

)2)= c+ b

2

4awhatever you do on the left, do it on the right.

a

(x+

b

2a

)2=

b2

4a c factoring inside the brackets on the left.

a

(x+

b

2a

)2=

b2 4ac4a(

x+b

2a

)2=

b2 4ac4a2

x+b

2a=

b2 4ac4a2

taking the square root on both sides

x1,2 = b2a b2 4ac2a

x1,2 =bb2 4ac

2a.

As required.

If you are required to use the quadratic formula, you dont need to prove it again. Instead, justidentify your a, b and cs and substitute them in the formula to get your solutions. You howeverneed to understand the proof because you can definitely be asked to reproduce it during theevaluations. We will then try to use the quadratic formula in the following example.

63

Example 3.3.4 Use the quadratic formula to solve the following quadratic equations.

a) x2 + x 1 = 0

b) 2x2 + 4x 1 = 0

Solutions

a) Note that for our equation, a = 1, b = 1 and c = 1. So, we substitute these values into thequadratic formula to get,

x1,2 =bb2 4ac

2a

=112 4 1 (1)

2 1=

152

= 1252

So, the solutions are x = 12 +52 and x = 12

52 .

b) Here, a = 2, b = 4 and c = 1. So,

x1,2 =bb2 4ac

2a

=442 4 2 (1)

2 2=

4244

= 162

So, the solutions are x = 1 +62 and x = 1

62 .

Remark 3.3.4 From Theorem 3.3.3 it should be apparent now why and how the discriminant4 = b2 4ac determines whether the solutions for a given quadratic equation are real or not.

64

3.4 Simultaneous Nonlinear Equations

Before we conclude the Unit on equations, we will take a brief look at simultaneous nonlinearequations. In Section 3.2, we looked at ways of solving simultaneous linear equation, and some(or maybe all) of those ways can also be used to solve simultaneous nonlinear equations.

Simultaneous nonlinear equations is a set of equations where one or more of the equations arenot linear equations. For example,

x+ y = 2

xy = 1

are simultaneous nonlinear equations because the equation xy = 1 is not a linear equation.

Whether solutions for given simultaneous nonlinear equations exist or not, and the number ofequations, depends on the type of equations you are considering.

We will try to solve the simultaneous nonlinear equations above (and some others) using thesubstitution method in the following example.

Example 3.4.1 Solve the following simultaneous nonlinear equations.

a) x+ y = 2xy = 1

b) 3 + y = x2 + 2xy = x+ 1

Solutions

a) For these equations, we make y the subject of the formula in the first equation to have,y = 2 x. We then substitute this into the second equation for y, and obtain an equationin x only, which we will solve as follows;

xy = 1

x(2 x) = 1 substituting with 2 x for y2x x2 1 = 0x2 2x+ 1 = 0 dividing though by 1

(x 1)2 = 0x 1 = 0

x = 1

65

Substituting this value of x into the equation y = 2 x we get that y = 2 1 = 1.

b) For these equations, we use the second equation to substitute for y into the first equation,and then try to solve for x as follows;

y + 3 = x2 + 2x

(x+ 1) + 3 = x2 + 2x

x+ 4 = x2 + 2x

x2 + x 4 = 0 collecting the like terms

This is a quadratic equation in x, which we cant seem to factorize. So, we try the quadraticformula.

x1,2 =bb2 4ac

2a

=11 + 16

2

=117

2

So, we have x = 12 172 and x = 12 +

172 . Substituting these into the equation

y = x+1 we get, when x = 12 172 , y =

12

172 , and when x = 12 +

172 , y =

12 +

172 .

In the following example, we will try to solve simultaneous nonlinear equations using the addi-tion/subtraction method.

66

Example 3.4.2 Solve the following simultaneous equations.

a) y = 2x2 x+ 1y = x2 + 2x+ 5

b) 2x2 + 3y2 = 113x2 + 2y2 = 19

Solutions

a) Since y has the same coefficient in both equations, we subtract the second equation from thefirst one to get the following;

y = 2x2 x+ 1y = x2 + 2x+ 5

0 = x2 3x 4(x 4)(x+ 1) = 0

x 4 = 0 or x+ 1 = 0x = 4 or x = 1

Substituting these values into one of the equations (say the first one) we get, when x = 4,y = 2 42 4 + 1 = 29, and when x = 1, y = 2 (1)2 + 1 + 1 = 4.

b) Since in this case we dont have a variable with the same numerical coefficient in bothequations, we create it by multiplying through the first equation by 3 and thought thesecond equation by 2, and proceed to try and solve as follows.

6x2 + 9y2 = 33

6x2 + 4y2 = 38

5y2 = 5y2 = 1

Now, since the equation y2 = 1 has no real solutions (4 = 4), the simultaneousequations do not have real solutions as well.

This is where we leave it as far as equations are concerned. Solving equations is very importantin mathematics and its applications. If theres any type of equations which are not covered in

67

this Unit, its because it is believed that you should be able to work out their solutions usingone or more of the tools discussed in this Unit. From here onwards, it is assumed that you knowhow to solve equations.

Exercises 3.2

1. Determine the discriminant of the following quadratic equations, and then classify thesolutions of the equation as (i) two distinct real numbers, (ii) one real number (repeatedreal solutions), or (iii) no real solutions. You dont have to solve the equations.

a) x3 + 3x+ 3 = 0

b) x2 20x+ 100 = 0c) x2 + 3x 11 = 0

2. Solve the following equations by factoring.

a) 3x2 7x = 0b) x2 2x 15 = 0

3. Solve the following equations by completing the square.

a) 2x2 + 10x 3 = 0b) 2 + 10x 5x2 = 0c) x2 6x = 0

4. Solve the following equations by using the quadratic formula.

a) 2x2 + 4x 1 = 0b) x2 5x 24 = 0

5. Solve the following equations simultaneously.

y = x2

y = x+ 2

3.5 Summary

1. Whatever you do to the one side of the equal sign, you should do to the other side as well.

2. The existence of real solutions to the quadratic equation ax2 + bx + c = 0 is determinedby the discriminant 4 = b2 4ac.

68

3.6 Answers to Exercises in Unit 3

Exercise 3.1

1. a) x = 7b) x = 1

2

c) y =92

d) x = 6

2. x = 3 and y = 7.

3. x = 2 and y = 4.

Exercise 3.2

1. a) 4 = 3, So no real solutions.b) 4 = 0, So repeated real solutions.c) 4 = 53, So distinct real solutions.

2. a) x = 0 or x =73

b) x = 3 or x = 5

3. a) x = 52 1231 or x = 52 + 12

31

b) x = 1 1535 or x = 1 + 15

35

c) x = 0 or x = 6

4. a) x = 1 126 or x = 1 + 12

6

b) x = 6 or x = 1.

5. When x = 2, y = 4 and when x = 1, y = 1

69

Unit 4

Inequalities

Content

4.1 Linear Inequalities

4.2 Nonlinear Inequalities

Introduction

In this Unit, we will learn what each of the inequality signs mean, and we will look at methodsof solving inequalities. We will also learn how solutions to inequalities can be represented on anumber line.

Objectives

By the end of this Unit, you should be able to:

use the properties of inequalities.

solve linear inequalities.

solve nonlinear inequalities.

represent solutions of inequalities on a number line.

Activities and Study Times

All the activities and studying this Unit should not take you more than five (5) hours.

70

Its quite often that we hear people or ourselves making statements like: Life is not fair. Thisis because weve all observed that, not all