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algebra m101 chapter 7, Lecture notes of Mathematics

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M101 COLLEGE AND ADVANCED ALGEBRA jvb
A LEARNING GUIDE
Presented to
the Instructional Materials Development
Committee Bohol Island State University Main
Campus Tagbilaran City
jvb
jvb
RENARIO G. HINAMPAS, JR.
TABLE OF CONTENTS
TITLE PAGE i TABLE OF CONTENTS ii
1 RADICALS AND RATIONAL EXPONENTS 1 1.1 Objectives . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . 1 1.2 Radicals . . . . . . . . . . . . . . . . . . . . . . . .
pf3
pf4
pf5

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M101 COLLEGE AND ADVANCED ALGEBRA jvb

A LEARNING GUIDE

Presented to

the Instructional Materials Development

Committee Bohol Island State University Main

Campus Tagbilaran City

jvb

jvb

RENARIO G. HINAMPAS, JR.

TABLE OF CONTENTS

TITLE PAGE i TABLE OF CONTENTS ii

1 RADICALS AND RATIONAL EXPONENTS 1 1.1 Objectives.........

..................... 1 1.2 Radicals........................

....... 1 1.3 Laws of Radicals......................... 2 1.

Operations of Radical Expressions................ 6 1.

Assessment Task 6........................ 10

BIBLIOGRAPHY 11

CHAPTER 1

COMPLEX NUMBERS

This chapter discusses the complex numbers and its operations.

Exam ples are given to further understand the concepts.

1.1 Objectives

Add, subtract, multiply and divide complex numbers.

1.2 Complex Numbers

Definition 1.2.1 The set of complex numbers is given by C = {x + yi : x, y

∈ R } such that i

2 = 1 with operations of addition (+) and multiplication ( · )

defined as follows:

(1) Addition: ( x 1 + y 1 i ) + ( x 2 + y 2 i ) = ( x 1 + x 2 ) + ( y 1 + y 2 ) i; (2)

Multiplication: ( x 1 + y 1 i ) · ( x 2 + y 2 i ) = ( x 1 x 2 − y 1 y 2 ) + ( x 1 y 2 + x 2 y 1 ) i. Example

1.2.2 Add (6 + 4 i ) + (5 7 i ).

Solutions:

(6 + 4 i ) + (5 7 i ) = (6 + 5) + (4 i − 7 i )

= 11 3 i.

Example 1.2.3 Subtract (2 + 5 i ) (8 + 6 i ).

Solutions:

(2 + 5 i ) (8 + 6 i ) = (2 8) + (5 i − 6 i )

= 6 − i.

Example 1.2.4 Multiply (8 + 4 i )(6 3 i ).

Example 1.2.9 Simplify

6+2 i

1 2 i.

Solutions:

6+2 i

1 2 i =

6+2 i

1 2

1+2 i

1+2 i

=

6+12 i +2 i +4 i 2

1 2 (2 i ) 2

6+14 i +4( 1)

1 4 i 2

6+14 i− 4

1 4 i 2

2+14 i

1 4( 1)

=

2+14 i

1+

=

2+14 i

5

Example 1.2.10 Simplify

4

2

√ − 9_._

Solutions

:

1+

√ − 4

2

√ − 9 =

1+

2

√ √

4( 1) 9( 1)

4 i 2

2

9 i 2

1+2 i

2 3 i

=

1+2 i

2 3

2+3 i

2+3 i

=

2+3 i +4 i +6 i 2

2 2 (3 i ) 2

2+7 i +6( 1) 4 9 i 2

2+7 i− 6

4 9 i 2

4+7 i 4 9( 1)

=

4+7 i 4+

=

4+7 i 13_._

4

1.3 Assessment Task 7

Direction: Simplify

(2_._ ) (3 + 2 i ) (5 + 8 i )

(4_._ ) (7 + 4 i )(9 −i )

2+7 i

1+ i

(6_._ ) (2 i− 5)

3

7 8 i

5

4+

√ − 16

BIBLIOGRAPHY

[1] W. L. Hart, College Algebra. D.C. Heath and Company, United State

of America, 1953

[2] R. Larson and R.P. Hostetler, Algebra and Trigonometry 8Ed.