Maths in Focus Ext 1 Preliminary-4

Mathematics Extension 1 Preliminary Course

maths

Mathematics Extension 1 Preliminary Course

maths Margaret Grove

Text © 2010 Grove and Associates Pty Ltd Illustrations and design © 2010 McGraw-Hill Australia Pty Ltd Additional owners of copyright are acknowledged in on-page credits Every effort has been made to trace and acknowledge copyrighted material. The authors and publishers tender their apologies should any infringement have occurred. Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the institution (or the body that administers it) has sent a Statutory Educational notice to Copyright Agency Limited (CAL) and been granted a licence. For details of statutory educational and other copyright licences contact: Copyright Agency Limited, Level 15, 233 Castlereagh Street, Sydney NSW 2000. Telephone: (02) 9394 7600. Website: www.copyright.com.au Reproduction and communication for other purposes Apart from any fair dealing for the purposes of study, research, criticism or review, as permitted under the Act, no part of this publication may be reproduced, distributed or transmitted in any form or by any means, or stored in a database or retrieval system, without the written permission of McGraw-Hill Australia including, but not limited to, any network or other electronic storage. Enquiries should be made to the publisher via www.mcgraw-hill.com.au National Library of Australia Cataloguing-in-Publication Data Author: Grove, Margaret. Title: Maths in focus: mathematics extension preliminary course/Margaret Grove. Edition: 2nd ed. ISBN: 9780070278585 (pbk.) Target Audience: For secondary school age. Subjects: Mathematics–Problems, exercises, etc. Mathematics–Textbooks. Dewey Number: 510.76 Published in Australia by McGraw-Hill Australia Pty Ltd Level 2, 82 Waterloo Road, North Ryde NSW 2113 Publisher: Eiko Bron Managing Editor: Kathryn Fairfax Production Editor: Natalie Crouch Editorial Assistant: Ivy Chung Art Director: Astred Hicks Cover and Internal Design: Simon Rattray, Squirt Creative Cover Image: Corbis Proofreaders: Terence Townsend and Ron Buck CD-ROM Preparation: Nicole McKenzie Typeset in ITC Stone serif, 10/14 by diacriTech Printed in China on 80 gsm matt art by iBook

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Contents PREFACE

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ACKNOWLEDGEMENTS

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CREDITS

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FEATURES OF THIS BOOK

ix

SYLLABUS MATRIX

x

STUDY SKILLS

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Chapter 1: Basic Arithmetic

2

INTRODUCTION REAL NUMBERS DIRECTED NUMBERS FRACTIONS, DECIMALS AND PERCENTAGES POWERS AND ROOTS ABSOLUTE VALUE TEST YOURSELF 1 CHALLENGE EXERCISE 1 Chapter 2: Algebra and Surds INTRODUCTION SIMPLIFYING EXPRESSIONS BINOMIAL PRODUCTS FACTORISATION COMPLETING THE SQUARE ALGEBRAIC FRACTIONS SUBSTITUTION SURDS TEST YOURSELF 2 CHALLENGE EXERCISE 2 Chapter 3: Equations INTRODUCTION SIMPLE EQUATIONS SUBSTITUTION INEQUATIONS EQUATIONS AND INEQUATIONS INVOLVING ABSOLUTE VALUES EXPONENTIAL EQUATIONS QUADRATIC EQUATIONS FURTHER INEQUATIONS QUADRATIC INEQUATIONS SIMULTANEOUS EQUATIONS TEST YOURSELF 3 CHALLENGE EXERCISE 3

3 3 9 12 19 37 41 43 44 45 45 51 55 69 71 73 76 90 93 94 95 95 100 103 107 114 118 125 129 132 138 139

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Chapter 4: Geometry 1

140

INTRODUCTION NOTATION TYPES OF ANGLES PARALLEL LINES TYPES OF TRIANGLES CONGRUENT TRIANGLES SIMILAR TRIANGLES PYTHAGORAS’ THEOREM TYPES OF QUADRILATERALS POLYGONS AREAS TEST YOURSELF 4 CHALLENGE EXERCISE 4

141 141 142 149 153 159 163 171 177 184 188 195 197

Practice Assessment Task Set 1 Chapter 5: Functions and Graphs INTRODUCTION FUNCTIONS GRAPHING TECHNIQUES LINEAR FUNCTION QUADRATIC FUNCTION ABSOLUTE VALUE FUNCTION THE HYPERBOLA CIRCLES AND SEMI-CIRCLES OTHER GRAPHS LIMITS AND CONTINUITY FURTHER GRAPHS REGIONS TEST YOURSELF 5 CHALLENGE EXERCISE 5 Chapter 6: Trigonometry INTRODUCTION TRIGONOMETRIC RATIOS RIGHT-ANGLED TRIANGLE PROBLEMS APPLICATIONS EXACT RATIOS ANGLES OF ANY MAGNITUDE TRIGONOMETRIC EQUATIONS TRIGONOMETRIC IDENTITIES NON-RIGHT-ANGLED TRIANGLE RESULTS APPLICATIONS AREA TRIGONOMETRY IN THREE DIMENSIONS SUMS AND DIFFERENCES OF ANGLES FURTHER TRIGONOMETRIC EQUATIONS TEST YOURSELF 6 CHALLENGE EXERCISE 6

199 204 205 205 216 224 228 234 242 246 254 260 264 277 287 288 290 291 291 299 308 318 322 336 342 347 358 362 365 367 374 385 387

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Chapter 7: Linear Functions INTRODUCTION DISTANCE MIDPOINT GRADIENT EQUATION OF A STRAIGHT LINE PARALLEL AND PERPENDICULAR LINES INTERSECTION OF LINES PERPENDICULAR DISTANCE ANGLE BETWEEN TWO LINES RATIOS TEST YOURSELF 7 CHALLENGE EXERCISE 7 Chapter 8: Introduction to Calculus INTRODUCTION GRADIENT DIFFERENTIATION FROM FIRST PRINCIPLES SHORT METHODS OF DIFFERENTIATION TANGENTS AND NORMALS FURTHER DIFFERENTIATION AND INDICES COMPOSITE FUNCTION RULE PRODUCT RULE QUOTIENT RULE ANGLE BETWEEN 2 CURVES TEST YOURSELF 8 CHALLENGE EXERCISE 8 Practice Assessment Task Set 2 Chapter 9: Properties of the Circle INTRODUCTION PARTS OF A CIRCLE ARCS, ANGLES AND CHORDS CHORD PROPERTIES CONCYCLIC POINTS TANGENT PROPERTIES TEST YOURSELF 9 CHALLENGE EXERCISE 9 Chapter 10: The Quadratic Function INTRODUCTION GRAPH OF A QUADRATIC FUNCTION QUADRATIC INEQUALITIES THE DISCRIMINANT QUADRATIC IDENTITIES SUM AND PRODUCT OF ROOTS EQUATIONS REDUCIBLE TO QUADRATICS TEST YOURSELF 10 CHALLENGE EXERCISE 10

390 391 391 396 398 408 412 417 422 426 430 434 435 438 439 440 449 465 471 476 478 482 485 487 490 491 494 498 499 499 500 512 519 525 537 539 542 543 543 549 555 562 566 571 575 576

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Chapter 11: Locus and the Parabola INTRODUCTION LOCUS CIRCLE AS A LOCUS PARABOLA AS A LOCUS GENERAL PARABOLA TANGENTS AND NORMALS PARAMETRIC EQUATIONS OF THE PARABOLA CHORDS, TANGENTS AND NORMALS PROPERTIES OF THE PARABOLA LOCUS PROBLEMS TEST YOURSELF 11 CHALLENGE EXERCISE 11 Practice Assessment Task Set 3 Chapter 12: Polynomials 1 INTRODUCTION DEFINITION OF A POLYNOMIAL DIVISION OF POLYNOMIALS REMAINDER AND FACTOR THEOREMS GRAPH OF A POLYNOMIAL ROOTS AND COEFFICIENTS OF POLYNOMIAL EQUATIONS TEST YOURSELF 12 CHALLENGE EXERCISE 12 Chapter 13: Permutations and Combinations INTRODUCTION FUNDAMENTAL COUNTING PRINCIPLE PERMUTATIONS COMBINATIONS TEST YOURSELF 13 CHALLENGE EXERCISE 13 Practice Assessment Task Set 4 Answers

578 579 579 587 591 610 625 627 634 643 648 652 653 655 662 663 663 667 672 681 706 713 714 716 717 717 730 740 746 747 749 756

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PREFACE This book covers the Preliminary syllabus for Mathematics and Extension 1. The extension material is easy to see as it has green headings and there is green shading next to all extension question and answers. The syllabus is available through the NSW Board of Studies website on www.boardofstudies. nsw.edu.au. You can also access resources, study techniques, examination technique, sample and past examination papers through other websites such as www.math.nsw.edu.au and www.csu.edu. au. Searching the Internet generally will pick up many websites supporting the work in this course. Each chapter has comprehensive fully worked examples and explanations as well as ample sets of graded exercises. The theory follows a logical order, although some topics may be learned in any order. Each chapter contains Test Yourself and Challenge exercises, and there are several practice assessment tasks throughout the book. If you have trouble doing the Test Yourself exercises at the end of a chapter, you will need to go back into the chapter and revise it before trying them again. Don’t attempt to do the Challenge exercises until you are confident that you can do the Test Yourself exercises, as these are more difficult and are designed to test the more able students who understand the topic really well.

ACKNOWLEDGEMENTS Thanks go to my family, especially my husband Geoff, for supporting me in writing this book.

CREDITS Fairfax Photos: p 327 Istockphoto: p 101, p 171 Margaret Grove: p 37, p 163, p 206, p 246, p 260, p 291, p308 (bottom), p 310, p 311, p 313, p 316, p 391, p 499, p 543, p 591, p 717, p 719, p 726, p 729, p 730, p 739 Photolibrary: p 205 Shutterstock: p 74, p 164, p 229, p 308 (top), p 580

FEATURES OF THIS BOOK This second edition retains all the features of previous Maths in Focus books while adding in new improvements. The main feature of Maths in Focus is in its readability, its plentiful worked examples and straightforward language so that students can understand it and use it in self-paced learning. The logical progression of topics, the comprehensive fully worked examples and graded exercises are still major features. A wide variety of questions is maintained, with more comprehensive and more difficult questions included in each topic. At the end of each chapter is a consolidation set of exercises (Test yourself) in no particular order that will test whether the student has grasped the concepts contained in the chapter. There is also a challenge set for the more able students. The four practice assessment tasks provide a comprehensive variety of mixed questions from various chapters. These have been extended to contain questions in the form of sample examination questions, including short answer, free response and multiple-choice questions that students may encounter in assessments. The second edition also features a short summary of general study skills that students will find useful, both in the classroom and when doing assessment tasks and examinations. These study skills are also repeated in the HSC book.

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A syllabus matrix is included to show where each syllabus topic fits into the book. Topics are generally arranged in a logical order. For example, arithmetic and algebra are needed in most, if not all other topics, so these are treated at the beginning of the book. Some teachers like to introduce particular topics before others, e.g. linear functions before more general functions. However, part of the work on gradient requires some knowledge of trigonometry and the topic of angles of any magnitude in trigonometry needs some knowledge of functions. So the order of most chapters in the book have been carefully thought out. Some chapters, however, could be covered in a different order, such as geometry which is covered in Chapter 4, and quadratic functions and locus, which are near the end of the book.

SYLLABUS MATRIX This matrix shows how the syllabus is organised in the chapters of this book.

Mathematics (2 Unit) Basic arithmetic and algebra (1.1 – 1.4)

Chapter 1: Basic arithmetic Chapter 2: Algebra and surds Chapter 3: Equations

Real functions (4.1 – 4.4)

Chapter 5: Functions and graphs

Trigonometric ratios (5.1 – 5.5)

Chapter 6: Trigonometry

Linear functions (6.1 – 6.5, 6.7)

Chapter 7: Linear functions

The quadratic polynomial and the parabola (9.1 – 9.5)

Chapter 10: The quadratic function Chapter 11: Locus and the parabola

Plane geometry (2.1 – 2.4)

Chapter 4: Geometry 1

Tangent to a curve and derivative of a function (8.1 – 8.9)

Chapter 8: Introduction to calculus

Extension 1 Other inequalities (1.4E)

Chapter 3: Equations

Circle geometry (2.6 – 2.10E)

Chapter 9: Properties of the circle

Further trigonometry (5.6 – 5.9E)

Chapter 6: Trigonometry

Angles between two lines (6.6E)

Chapter 7: Linear functions

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Internal and external division of lines into given ratios (6.7E)

Chapter 7: Linear functions

Parametric representation (9.6E)

Chapter 11: Locus and the parabola

Permutations and combinations (18.1E)

Chapter 13: Permutations and combinations

Polynomials (16.1 – 16.3E)

Chapter 12: Polynomials 1

STUDY SKILLS You may have coasted through previous stages without needing to rely on regular study, but in this course many of the topics are new and you will need to systematically revise in order to build up your skills and to remember them. The Preliminary course introduces the basics of topics such as calculus that are then applied in the HSC course. You will struggle in the HSC if you don’t set yourself up to revise the preliminary topics as you learn new HSC topics. Your teachers will be able to help you build up and manage good study habits. Here are a few hints to get you started. There is no right or wrong way to learn. Different styles of learning suit different people. There is also no magical number of hours a week that you should study, as this will be different for every student. But just listening in class and taking notes is not enough, especially when learning material that is totally new. You wouldn’t go for your driver’s licence after just one trip in the car, or enter a dance competition after learning a dance routine once. These skills take a lot of practice. Studying mathematics is just the same. If a skill is not practised within the first 24 hours, up to 50% can be forgotten. If it is not practised within 72 hours, up to 85–90% can be forgotten! So it is really important that whatever your study timetable, new work must be looked at soon after it is presented to you. With a continual succession of new work to learn and retain, this is a challenge. But the good news is that you don’t have to study for hours on end!

In the classroom In order to remember, first you need to focus on what is being said and done. According to an ancient proverb:

‘I hear and I forget I see and I remember I do and I understand’

If you chat to friends and just take notes without really paying attention, you aren’t giving yourself a chance to remember anything and will have to study harder at home.

xii

If you have just had a fight with a friend, have been chatting about weekend activities or myriad other conversations outside the classroom, it helps if you can check these at the door and don’t keep chatting about them once the lesson starts. If you are unsure of something that the teacher has said, the chances are that others are also not sure. Asking questions and clarifying things will ultimately help you gain better results, especially in a subject like mathematics where much of the knowledge and skills depends on being able to understand the basics. Learning is all about knowing what you know and what you don’t know. Many students feel like they don’t know anything, but it’s surprising just how much they know already. Picking up the main concepts in class and not worrying too much about other less important parts can really help. The teacher can guide you on this. Here are some pointers to get the best out of classroom learning: ■ Take control and be responsible for your own learning ■ Clear your head of other issues in the classroom ■ Active, not passive, learning is more memorable ■ Ask questions if you don’t understand something ■ Listen for cues from the teacher ■ Look out for what are the main concepts Note taking varies from class to class, but there are some general guidelines that will help when you come to read over your notes later on at home: ■ Write legibly ■ Use different colours to highlight important points or formulae ■ Make notes in textbooks (using pencil if you don’t own the textbook) ■ Use highlighter pens to point out important points ■ Summarise the main points ■ If notes are scribbled, rewrite them at home

At home You are responsible for your own learning and nobody else can tell you how best to study. Some people need more revision time than others, some study better in the mornings while others do better at night, and some can work at home while others prefer a library. There are some general guidelines for studying at home: ■ Revise both new and older topics regularly ■ Have a realistic timetable and be flexible ■ Summarise the main points ■ Revise when you are fresh and energetic ■ Divide study time into smaller rather than longer chunks

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■ Study in a quiet environment ■ Have a balanced life and don’t forget to have fun! If you are given exercises out of a textbook to do for homework, consider asking the teacher if you can leave some of them till later and use these for revision. It is not necessary to do every exercise at one sitting, and you learn better if you can spread these over time. People use different learning styles to help them study. The more variety the better, and you will find some that help you more than others. Some people (around 35%) learn best visually, some (25%) learn best by hearing and others (40%) learn by doing. Here are some ideas to give you a variety of ways to study: ■ Summarise on cue cards or in a small notebook ■ Use colourful posters ■ Use mindmaps and diagrams ■ Discuss work with a group of friends ■ Read notes out aloud ■ Make up songs and rhymes ■ Do exercises regularly ■ Role play teaching someone else

Assessment tasks and exams Many of the assessment tasks for maths are closed book examinations. You will cope better in exams if you have practised doing sample exams under exam conditions. Regular revision will give you confidence and if you feel well prepared, this will help get rid of nerves in the exam. You will also cope better if you have had a reasonable night’s sleep before the exam. One of the biggest problems students have with exams is in timing. Make sure you don’t spend too much time on questions you’re unsure about, but work through and find questions you can do first. Divide the time up into smaller chunks for each question and allow some extra time to go back to questions you couldn’t do or finish. For example, in a 2 hour exam with 6 questions, allow around 15 minutes for each question. This will give an extra half hour at the end to tidy up and finish off questions. Here are some general guidelines for doing exams: ■ Read through and ensure you know how many questions there are ■ Divide your time between questions with extra time at the end ■ Don’t spend too much time on one question ■ Read each question carefully, underlining key words ■ Show all working out, including diagrams and formulae ■ Cross out mistakes with a single line so it can still be read ■ Write legibly

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And finally… Study involves knowing what you don’t know, and putting in a lot of time into concentrating on these areas. This is a positive way to learn. Rather than just saying, ‘I can’t do this’, say instead, ‘I can’t do this yet’, and use your teachers, friends, textbooks and other ways of finding out. With the parts of the course that you do know, make sure you can remember these easily under exam pressure by putting in lots of practice. Remember to look at new work ■ today ■ tomorrow ■ in a week ■ in a month Some people hardly ever find time to study while others give up their outside lives to devote their time to study. The ideal situation is to balance study with other aspects of your life, including going out with friends, working and keeping up with sport and other activities that you enjoy.

Good luck with your studies!

1 Basic Arithmetic TERMINOLOGY Absolute value: The distance of a number from zero on the number line. Hence it is the magnitude or value of a number without the sign Directed numbers: The set of integers or whole numbers f -3, -2, -1, 0, 1, 2, 3, f Exponent: Power or index of a number. For example 23 has a base number of 2 and an exponent of 3 Index: The power of a base number showing how many times this number is multiplied by itself e.g. 2 3 = 2 # 2 # 2. The index is 3

Indices: More than one index (plural) Recurring decimal: A repeating decimal that does not terminate e.g. 0.777777 … is a recurring decimal that can be written as a fraction. More than one digit can recur e.g. 0.14141414 ... Scientific notation: Sometimes called standard notation. A standard form to write very large or very small numbers as a product of a number between 1 and 10 and a power of 10 e.g. 765 000 000 is 7.65 # 10 8 in scientific notation

Chapter 1 Basic Arithmetic

INTRODUCTION THIS CHAPTER GIVES A review of basic arithmetic skills, including knowing the

correct order of operations, rounding off, and working with fractions, decimals and percentages. Work on significant figures, scientific notation and indices is also included, as are the concepts of absolute values. Basic calculator skills are also covered in this chapter.

Real Numbers Types of numbers Unreal or imaginary numbers Real numbers

Rational numbers

Irrational numbers

Integers

Integers are whole numbers that may be positive, negative or zero. e.g. - 4, 7, 0, -11 a Rational numbers can be written in the form of a fraction b • 3 where a and b are integers, b ! 0. e.g. 1 , 3.7, 0. 5, - 5 4 a Irrational numbers cannot be written in the form of a fraction (that b is, they are not rational) e.g. 2 , r

EXAMPLE Which of these numbers are rational and which are irrational? • 3 r 3 , 1. 3, , 9 , , - 2.65 4 5

Solution r are irrational as they cannot be written as fractions (r is irrational). 4 • 3 13 1 1. 3 = 1 , 9 = and - 2.65 = - 2 so they are all rational. 3 1 20 3 and

3

4

Maths In Focus Mathematics Extension 1 Preliminary Course

Order of operations 1. Brackets: do calculations inside grouping symbols first. (For example, a fraction line, square root sign or absolute value sign can act as a grouping symbol.) 2. Multiply or divide from left to right. 3. Add or subtract from left to right.

EXAMPLE Evaluate 40 - 3 ] 5 + 4 g .

Solution 40 - 3 (5 + 4) = 40 - 3# 9 = 40 - 27 = 13

BRACKETS KEYS Use ( and ) to open and close brackets. Always use them in pairs. For example, to evaluate 40 - 3 ] 5 + 4 g press 40 - 3 #

( 5 + 4 ) = = 13 5.67 - 3.49 correct to 1 decimal place 1.69 + 2.77

To evaluate press :

(

( 5.67 - 3.49 )

'

( 1.69 + 2.77 )

)

=

= 0.7 correct to 1 decimal place

PROBLEM What is wrong with this calculation? 19 - 4 1+2 Press 19 - 4 ' 1 + 2 = 19 - 4 '1 + 2

Evaluate

What is the correct answer?

17

Chapter 1 Basic Arithmetic

MEMORY KEYS Use STO to store a number in memory. There are several memories that you can use at the same time—any letter from A to F, or X, Y and M on the keypad. To store the number 50 in, say, A press 50 STO A To recall this number, press ALPHA A = To clear all memories press SHIFT CLR

X -1 KEY Use this key to find the reciprocal of x. For example, to evaluate 1 - 7.6 # 2.1 -1 = press ( (-) 7.6 # 2.1 ) x = - 0.063 (correct to 3 decimal places)

Rounding off Rounding off is often done in everyday life. A quick look at a newspaper will give plenty of examples. For example in the sports section, a newspaper may report that 50 000 fans attended a football match. An accurate number is not always necessary. There may have been exactly 49 976 people at the football game, but 50 000 gives an idea of the size of the crowd.

EXAMPLES 1. Round off 24 629 to the nearest thousand.

Solution This number is between 24 000 and 25 000, but it is closer to 25 000.

` 24 629 = 25 000 to the nearest thousand

CONTINUED

Different calculators use different keys so check the instructions for your calculator.

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Maths In Focus Mathematics Extension 1 Preliminary Course

2. Write 850 to the nearest hundred.

Solution This number is exactly halfway between 800 and 900. When a number is halfway, we round it off to the larger number. ` 850 = 900 to the nearest hundred

In this course you will need to round off decimals, especially when using trigonometry or logarithms. To round a number off to a certain number of decimal places, look at the next digit to the right. If this digit is 5 or more, add 1 to the digit before it and drop all the other digits after it. If the digit to the right is less than 5, leave the digit before it and drop all the digits to the right.

EXAMPLES 1. Round off 0.6825371 correct to 1 decimal place. Add 1 to the 6 as the 8 is greater than 5.

Solution 0.6825371 # ` 0.6825371 = 0.7 correct to 1 decimal place 2. Round off 0.6825371 correct to 2 decimal places.

Drop off the 2 and all digits to the right as 2 is smaller than 5.

Solution 0.6825371 # ` 0.6825371 = 0.68 correct to 2 decimal places 3. Evaluate 3.56 ' 2.1 correct to 2 decimal places.

Check this on your calculator. Add 1 to the 69 as 5 is too large to just drop off.

Solution 3.56 ' 2.1 = 1.69 # 5238095

= 1.70 correct to 2 decimal places

Chapter 1 Basic Arithmetic

FIX KEY Use MODE or SET UP to fix the number of decimal places (see the instructions for your calculator). This will cause all answers to have a fixed number of decimal places until the calculator is turned off or switched back to normal.

While using a fixed number of decimal places on the display, the calculator still keeps track internally of the full number of decimal places.

EXAMPLE Calculate 3.25 ' 1.72 # 5.97 + 7.32 correct to 2 decimal places.

Solution 3.25 ' 1.72 # 5.97 + 7.32 = 1.889534884 # 5.97 + 7.32 = 11.28052326 + 7.32 = 18.60052326 = 18.60 correct to 2 decimal places If the FIX key is set to 2 decimal places, then the display will show 2 decimal places at each step. 3.25 ' 1.72 # 5.97 + 7.32 = 1.89 # 5.97 + 7.32 = 11.28 + 7.32 = 18.60 If you then set the calculator back to normal, the display will show the full answer of 18.60052326.

The calculator does not round off at each step. If it did, the answer might not be as accurate. This is an important point, since some students round off each step in calculations and then wonder why they do not get the same answer as other students and the textbook.

1.1 Exercises 1.

State which numbers are rational and which are irrational. (a) 169 (b) 0.546 (c) -17 r (d) 3

•

(e) 0.34 (f)

218

(g) 2 2 1 (h) 27 (i) 17.4% 1 (j) 5

Don’t round off at each step of a series of calculations.

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Maths In Focus Mathematics Extension 1 Preliminary Course

2.

3.

Evaluate (a) 20 - 8 ' 4 (b) 3 # 7 - 2 # 5 (c) 4 # ] 27 ' 3 g ' 6 (d) 17 + 3 # - 2 (e) 1.9 - 2 # 3.1 14 ' 7 (f) -1 + 3 3 1 2 (g) 2 - # 5 5 3 3 1 1 4 8 (h) 5 6 5 5 ' 8 6 (i) 1 1 + 4 8 1 7 3 5 10 (j) 1 1 1 4 2

7.

A crowd of 10 739 spectators attended a tennis match. Write this figure to the nearest thousand.

8.

A school has 623 students. What is this to the nearest hundred?

9.

A bank made loans to the value of $7 635 718 last year. Round this off to the nearest million.

Evaluate correct to 2 decimal places. (a) 2.36 + 4.2 ' 0.3 (b) ] 2.36 + 4.2 g ' 0.3 (c) 12.7 # 3.95 ' 5.7 (d) 8.2 ' 0.4 + 4.1# 0.54 (e) ] 3.2 - 6.5 g # ] 1.3 + 2.7 g 1 (f) 4.7 + 1.3 1 (g) 4.51 + 3.28

13. Round off 32.569148 to the nearest unit.

0.9 + 1.4 (h) 5.2 - 3.6 5.33 + 2.87 (i) 1.23 - 3.15 (j) 4.

1.7 2 + 8.9 2 - 3.94 2

Round off 1289 to the nearest hundred.

5.

Write 947 to the nearest ten.

6.

Round off 3200 to the nearest thousand.

10. A company made a profit of $34 562 991.39 last year. Write this to the nearest hundred thousand. 11. The distance between two cities is 843.72 km. What is this to the nearest kilometre? 12. Write 0.72548 correct to 2 decimal places.

14. Round off 3.24819 to 3 decimal places. 15. Evaluate 2.45 # 1.72 correct to 2 decimal places. 16. Evaluate 8.7 ' 5 correct to 1 decimal place. 17. If pies are on special at 3 for $2.38, find the cost of each pie. 18. Evaluate 7.48 correct to 2 decimal places. 6.4 + 2.3 correct to 8 1 decimal place.

19. Evaluate

20. Find the length of each piece of material, to 1 decimal place, if 25 m of material is cut into 7 equal pieces.

Chapter 1 Basic Arithmetic

21. How much will 7.5 m 2 of tiles cost, at $37.59 per m2?

3.5 + 9.8 5.6 + 4.35 15.9 + 6.3 - 7.8 (d) 7.63 - 5.12 1 (e) 6.87 - 3.21

(c)

22. Divide 12.9 grams of salt into 7 equal portions, to 1 decimal place. 23. The cost of 9 peaches is $5.72. How much would 5 peaches cost?

9.91 - ] 9.68 - 5.47 g 5.39 2 correct to 1 decimal place.

25. Evaluate

24. Evaluate correct to 2 decimal places. (a) 17.3 - 4.33 # 2.16 (b) 8.72 # 5.68 - 4.9 # 3.98

DID YOU KNOW? In building, engineering and other industries where accurate measurements are used, the number of decimal places used indicates how accurate the measurements are. For example, if a 2.431 m length of timber is cut into 8 equal parts, according to the calculator each part should be 0.303875 m. However, a machine could not cut this accurately. A length of 2.431 m shows that the measurement of the timber is only accurate to the nearest mm (2.431 m is 2431 mm). The cut pieces can also only be accurate to the nearest mm (0.304 m or 304 mm). The error in measurement is related to rounding off, as the error is half the smallest measurement. In the above example, the measurement error is half a millimetre. The length of timber could be anywhere between 2430.5 mm and 2431.5 mm.

Directed Numbers Many students use the calculator with work on directed numbers (numbers that can be positive or negative). Directed numbers occur in algebra and other topics, where you will need to remember how to use them. A good understanding of directed numbers will make your algebra skills much better.

^ - h KEY Use this key to enter negative numbers. For example, press (-) 3

=

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Maths In Focus Mathematics Extension 1 Preliminary Course

Adding and subtracting

To add: move to the right along the number line To subtract: move to the left along the number line -4

-3

-2

-1

0

1

Subtract

2

3

4

Add

EXAMPLES You can also do these on a calculator, or you may have a different way of working these out.

Evaluate 1. - 4 + 3

Solution Start at - 4 and move 3 places to the right.

-4

-3

-2

-1

0

1

2

3

4

2

3

4

- 4 + 3 = -1 2. -1 - 2

Solution Start at -1 and move 2 places to the left.

-4

-3

-2

-1

0

1

-1 - 2 = -3

Multiplying and dividing To multiply or divide, follow these rules. This rule also works if there are two signs together without a number in between e.g. 2 - -3

Same signs = + + + =+ - - =+ Different signs = + - =- + =-

Chapter 1 Basic Arithmetic

11

EXAMPLES Evaluate 1. - 2 #7

Solution Different signs (- 2 and + 7) give a negative answer. - 2 # 7 = -14 2. -12 ' - 4

Solution Same signs (-12 and - 4) give a positive answer. -12 ' - 4 = 3 3. -1 - - 3

Solution The signs together are the same (both negative) so give a positive answer. -

-1 - 3 = -1 + 3 =2

1.2 Exercises Evaluate 1.

-2 + 3

11. 5 - 3 # 4

2.

-7 - 4

12. - 2 + 7 # - 3

3.

8 # -7

13. 4 - 3 # - 2

4.

7 - ]-3 g

14. -1 - -2

5.

28 ' -7

15. 7 + - 2

6.

- 4.9 + 3.7

16. 2 - ] -1 g

7.

- 2.14 - 5.37

17. - 2 + 15 ' 5

8.

4.8 # -7.4

18. - 2 # 6 # - 5

9.

1.7 - ] - 4.87 g

19. - 28 ' -7 # - 5

10. -

3 2 -1 5 3

20. ] - 3 g2

Start at -1 and move 3 places to the right.

12

Maths In Focus Mathematics Extension 1 Preliminary Course

Fractions, Decimals and Percentages Conversions You can do all these conversions on your calculator using the b a or S + D key. c

EXAMPLES 1. Write 0.45 as a fraction in its simplest form.

Solution 45 5 ' 5 100 9 = 20

0.45 =

3 means 3 ' 8. 8

2. Convert

3 to a decimal. 8

Solution 0.375 8 g 3.000 3 So = 0.375 8 3. Change 35.5% to a fraction.

Solution 35.5 2 # 100 2 71 = 200

35.5% =

4. Write 0.436 as a percentage.

Solution Multiply by 100% to change a fraction or decimal to a percentage.

0.436 = 0.436 #100% = 43.6% 5. Write 20 g as a fraction of 1 kg in its simplest form.

Solution 1 kg = 1000 g 20 g 20 g = 1000 g 1 kg 1 = 50

Chapter 1 Basic Arithmetic

13

6. Find the percentage of people who prefer to drink Lemon Fuzzy, if 24 out of every 30 people prefer it.

Solution 24 100% # = 80% 30 1

Sometimes decimals repeat, or recur. Example • 1 = 0.33333333 f = 0. 3 3 There are different methods that can be used to change a recurring decimal into a fraction. Here is one way of doing it. Later you will discover another method when studying series. (See HSC Course book, Chapter 8.)

EXAMPLES

A rational number is any number that can be written as a fraction.

•

1. Write 0. 4 as a rational number.

Solution Let n = 0.44444 f Then 10n = 4.44444 f (2) - (1): 9n = 4 4 n= 9

( 1) ( 2)

Check this on your calculator by dividing 4 by 9.

• •

2. Change 1.329 to a fraction.

Solution n = 1.3292929 f Let Then 100n = 132.9292929 f (2) - (1): 99n = 131.6 131.6 10 n= # 99 10 1316 = 990 163 =1 495

( 1) ( 2)

CONTINUED

Try multiplying n by 10. Why doesn’t this work?

14

Maths In Focus Mathematics Extension 1 Preliminary Course

Another method Let n = 1.3292929 f Then 10n = 13.2929292 f and 1000n = 1329.292929 f (2) - (1): 990n = 1316 1316 n= 990 163 =1 495

This method avoids decimals in the fraction at the end.

(1 ) (2 )

1.3 Exercises 1.

2.

3.

Write each decimal as a fraction in its lowest terms. (a) 0.64 (b) 0.051 (c) 5.05 (d) 11.8 Change each fraction into a decimal. 2 (a) 5 7 (b) 1 8 5 (c) 12 7 (d) 11 Convert each percentage to a fraction in its simplest form. (a) 2% (b) 37.5% (c) 0.1% (d) 109.7%

4.

Write each percentage as a decimal. (a) 27% (b) 109% (c) 0.3% (d) 6.23%

5.

Write each fraction as a percentage. 7 20 1 (b) 3 (a)

4 15 1 (d) 1000

(c) 2

6.

Write each decimal as a percentage. (a) 1.24 (b) 0.7 (c) 0.405 (d) 1.2794

7.

Write each percentage as a decimal and as a fraction. (a) 52% (b) 7% (c) 16.8% (d) 109% (e) 43.4% 1 (f) 12 % 4

8.

Write these fractions as recurring decimals. 5 (a) 6 7 (b) 99 13 (c) 99 1 (d) 6 2 (e) 3

Chapter 1 Basic Arithmetic

5 33 1 (g) 7 2 (h) 1 11

31 99 13 + 6 (e) 7+4 (d) 1 -

(f)

9.

Express as fractions in lowest terms. •

(a) 0. 8 (b) (c) (d) (e) (f) (g)

•

0. 2 • 1. 5 • 3. 7 • • 0. 67 • • 0. 54 • 0.15 •

(h) 0.216 • • (i) 0.2 19 • • (j) 1.074 10. Evaluate and express as a decimal. 5 (a) 3+6 (b) 8 - 3 ' 5 4+7 (c) 12 + 3

11. Evaluate and write as a fraction. (a) 7.5 ' ] 4.1 + 7.9 g 15.7 - 8.9 (b) 4.5 - 1.3 6.3 + 1.7 (c) 12.3 - 8.9 + 7.6 4 .3 (d) 11.5 - 9.7 64 (e) 8100 12. Angel scored 17 out of 23 in a class test. What was her score as a percentage, to the nearest unit? 13. A survey showed that 31 out of 40 people watched the news on Monday night. What percentage of people watched the news? 14. What percentage of 2 kg is 350 g? 15. Write 25 minutes as a percentage of an hour.

Investigation Explore patterns in recurring decimals by dividing numbers by 3, 6, 9, 11, and so on. Can you predict what the recurring decimal will be if a fraction has 3 in the denominator? What about 9 in the denominator? What about 11? Can you predict what fraction certain recurring decimals will be? What denominator would 1 digit recurring give? What denominator would you have for 2 digits recurring?

Operations with fractions, decimals and percentages You will need to know how to work with fractions without using a calculator, as they occur in other areas such as algebra, trigonometry and surds.

15

16

Maths In Focus Mathematics Extension 1 Preliminary Course

The examples on fractions show how to add, subtract, multiply or divide fractions both with and without the calculator. The decimal examples will help with some simple multiplying and the percentage examples will be useful in Chapter 8 of the HSC Course book when doing compound interest. Most students use their calculators for decimal calculations. However, it is important for you to know how to operate with decimals. Sometimes the calculator can give a wrong answer if the wrong key is pressed. If you can estimate the size of the answer, you can work out if it makes sense or not. You can also save time by doing simple calculations in your head.

DID YOU KNOW? Some countries use a comma for the decimal point—for example, 0,45 for 0.45. This is the reason that our large numbers now have spaces instead of commas between digits—for example, 15 000 rather than 15,000.

EXAMPLES 1. Evaluate 1

3 2 - . 5 4

Solution 1

3 3 2 7 - = 5 4 5 4 28 15 = 20 20 13 = 20

2. Evaluate 2

1 ' 3. 2

Solution 2

3 5 1 '3 = ' 2 2 1 5 1 = # 2 3 5 = 6

3. Evaluate 0.056 # 100. Move the decimal point 2 places to the right.

Solution 0.056 #100 = 5.6

Chapter 1 Basic Arithmetic

17

4. Evaluate 0.02 # 0.3. Multiply the numbers and count the number of decimal places in the question.

Solution 0.02 # 0.3 = 0.006 5. Evaluate

8.753 . 10

Solution

Move the decimal point 1 place to the left.

8.753 ' 10 = 0.8753 1 6. The price of a $75 tennis racquet increased by 5 %. Find the new 2 price.

Solution 1 5 % = 0.055 2 1 ` 5 % of $75 = 0.055#$75 2 = $4.13

1 or 105 % of $75 = 1.055#$75 2 = $79.13

So the price increases by $4.13 to $79.13. 7. The price of a book increased by 12%. If it now costs $18.00, what did it cost before the price rise?

Solution The new price is 112% (old price 100%, plus 12%) $18.00 ` 1% = 112 $18.00 100 100% = # 112 1 = $16.07 So the old price was $16.07.

1.4 Exercises 1.

Write 18 minutes as a fraction of 2 hours in its lowest terms.

2.

Write 350 mL as a fraction of 1 litre in its simplest form.

3.

Evaluate 3 1 (a) + 5 4

2 7 -2 5 10 3 2 (c) #1 5 4 3 (d) ' 4 7 3 2 (e) 1 ' 2 5 3 (b) 3

18

Maths In Focus Mathematics Extension 1 Preliminary Course

3 of $912.60. 5

4.

Find

5.

5 Find of 1 kg, in grams correct 7 to 1 decimal place.

6.

Trinh spends sleeping,

1 of her day 3

7 1 at work and 24 12

eating. What fraction of the day is left? 7.

I get $150.00 a week for a casual 1 job. If I spend on bus fares, 10 2 1 on lunches and on outings, 15 3 how much money is left over for savings?

8.

John grew by

9.

17 of his height 200 this year. If he was 165 cm tall last year, what is his height now, to the nearest cm? Evaluate (a) 8.9 + 3 (b) 9 - 3.7 (c) 1.9 #10 (d) 0.032 #100 (e) 0.7 # 5 (f) 0.8 # 0.3 (g) 0.02 # 0.009 (h) 5.72 #1000 8.74 (i) 100 (j) 3.76 # 0.1

10. Find 7% of $750. 11. Find 6.5% of 845 mL. 12. What is 12.5% of 9217 g? 13. Find 3.7% of $289.45. 14. If Kaye makes a profit of $5 by selling a bike for $85, find the profit as a percentage of the selling price.

15. Increase 350 g by 15%. 1 16. Decrease 45 m by 8 %. 2 17. The cost of a calculator is now $32. If it has increased by 3.5%, how much was the old cost? 18. A tree now measures 3.5 m, which is 8.3% more than its previous year’s height. How high was the tree then, to 1 decimal place?

19. This month there has been a 4.9% increase in stolen cars. If 546 cars were stolen last month, how many were stolen this month? 20. George’s computer cost $3500. If it has depreciated by 17.2%, what is the computer worth now?

Chapter 1 Basic Arithmetic

19

PROBLEM If both the hour hand and minute hand start at the same position at 12 o’clock, when is the first time, correct to a fraction of a minute, that the two hands will be together again?

Powers and Roots A power (or index) of a number shows how many times a number is multiplied by itself.

EXAMPLES 1. 4 3 = 4 # 4 # 4 = 64 2. 2 5 = 2 # 2 # 2 # 2 # 2 = 32

A root of a number is the inverse of the power.

EXAMPLES 36 = 6 since 6 2 = 36

1.

2.

3

8 = 2 since 2 3 = 8

3.

6

64 = 2 since 2 6 = 64

DID YOU KNOW? Many formulae use indices (powers and roots). For example the compound interest formula that you will study in Chapter 8 of the HSC n Course book is A = P ^ 1 + r h 4 Geometry uses formulae involving indices, such as V = rr 3. Do you know what this 3 formula is for? In Chapter 7, the formula for the distance between 2 points on a number plane is d=

2

(x 2 - x 1) + (y 2 - y 1)

2

See if you can find other formulae involving indices.

In 4 3 the 4 is called the base number and the 3 is called the index or power.

20

Maths In Focus Mathematics Extension 1 Preliminary Course

POWER AND ROOT KEYS Use the x 2 and x 3 keys for squares and cubes. y Use the x or ^ key to find powers of numbers.

key for square roots.

Use the

These laws work for any m and n, including fractions and negative numbers.

Use the

3

key for cube roots.

Use the

x

for other roots.

Index laws There are some general laws that simplify calculations with indices.

am # an = am + n

Proof a m # a n = (a # a #f# a) # (a # a #f# a) 14444244443 14444244443 m times n times =a # # f # a a 14444244443 m + n times = am + n

am ' an = am - n

Proof am an a # a #f# a (m times) = a # a #f# a (n times) a # a #f# a (m - n times) = 1 = am - n

am ' an =

(a m)n = a mn

Proof (a m) n = a m # a m # a m #f# a m = am + m + m + f + m = a mn

(n times) (n times)

Chapter 1 Basic Arithmetic

(ab) n = a n b n

Proof (ab) n = ab # ab # ab #f# ab (n times) = (a # a #f# a) # (b # b #f# b) 14444244443 14444244443 n times n times = an bn

a n an c m = n b b

Proof a n a a a a c m = # # #f# b b b b b a # a # a #f # a = b # b # b #f # b an = n b

(n times) (n times) (n times)

EXAMPLES Simplify 1. m 9 # m 7 ' m 2

Solution m9 #m7 ' m2 = m9 + 7 - 2 = m 14 2. (2y 4)3

Solution (2y 4) 3 = 2 3 (y 4) 3 = 23 y4 # 3 = 8y 12

CONTINUED

21

22

Maths In Focus Mathematics Extension 1 Preliminary Course

3.

(y 6) 3 # y - 4 y5

Solution (y 6) 3 # y - 4 y5

= = =

y 18 # y - 4 y5 y 18 + (- 4) y5 y

14

y5 = y9

1.5 Exercises 1.

Evaluate without using a calculator. (a) 5 3 # 2 2 (b) 3 4 + 8 2 1 3 (c) c m 4 (d) (e)

2.

3.

3 4

(h) (i) (j) (k)

5

x2 p y9 w6 # w7 (m) w3 2 p #(p 3) 4 (n) p9 6 x ' x7 (o) x2 2 a # ( b 2) 6 (p) a4 # b9 (x 2) - 3 #(y 3) 2 (q) x -1 # y 4 (l) f

27 16

Evaluate correct to 1 decimal place. (a) 3.7 2 (b) 1.06 1.5 (c) 2.3 - 0.2 (d) 3 19 (e) 3 34.8 - 1.2 # 43.1 1 (f) 3 0.99 + 5.61 Simplify (a) a 6 # a 9 # a 2 (b) y 3 # y - 8 # y 5 (c) a -1 # a -3 1

1

(d) w 2 # w 2 (e) x 6 ' x (f) p 3 ' p - 7 y 11 (g) 5 y

(x 7) 3 (2x 5) 2 (3y - 2) 4 a3 #a5 ' a7

4.

Simplify (a) x 5 # x 9 (b) a -1 # a - 6 m7 (c) m3 (d) k 13 # k 6 ' k 9 (e) a - 5 # a 4 # a - 7 2

3

(f) x 5 # x 5 m5 # n4 (g) 4 m # n2

Chapter 1 Basic Arithmetic

1

1

p2 # p2

(h)

10. (a) Simplify

2

p (i) (3x 11) 2 (x 4) 6 (j) x3

5.

1

2 6 11. Evaluate (a ) when a = c m . 3 12. Evaluate b=

(2m 7) 3

m4 xy 3 #(xy 2) 4 (f) xy 8 4 (2k ) (g) (6k 3) 3 y 12 7 (h) _ 2y 5 i # 8

y=

Evaluate a3b2 when a = 2 and 3 b= . 4

7.

If x = of

8.

2 1 and y = , find the value 3 9

x3 y2 xy 5

.

1 1 1 , b = and c = , 4 2 3 a2 b3 evaluate 4 as a fraction. c

If a =

11

9.

(a) Simplify

8

a b . a8 b7

(b) Hence evaluate a=

a 11 b 8 when a8 b7

5 2 and b = as a fraction. 5 8

x5 y5

when x =

1 and 3

14. Evaluate

k-5 1 when k = . 3 k-9

15. Evaluate

a4 b6 3 when a = and 3 2 2 4 a (b )

b=

6.

x4 y7

2 . 9

-3

p

a3 b6 1 when a = and 2 b4

2 . 3

13. Evaluate

a6 # a4 o a 11 3 5xy 9 x8 # y3

p5 q8 r4

4 3

(d) (7a5b)2

(j) f

.

as a p4 q6 r2 7 2 fraction when p = , q = and 8 3 3 r= . 4

a 8 (b) c m b 4a 3 (c) d 4 n b

(i) e

p4 q6 r2

(b) Hence evaluate

Simplify (a) (pq 3) 5

(e)

p5 q8 r4

1 . 9

a6 # b3 as a fraction a5 # b2 3 1 when a = and b = . 4 9

16. Evaluate

a2 b7 as a fraction in a3 b 2 4 index form when a = c m and 5 5 3 b=c m. 8

17. Evaluate

18. Evaluate

(a 3) 2 b 4 c

as a fraction a (b 2) 4 c 3 6 1 7 when a = , b = and c = . 7 3 9

23

24

Maths In Focus Mathematics Extension 1 Preliminary Course

Negative and zero indices

Class Investigation Explore zero and negative indices by looking at these questions. For example simplify x 3 ' x 5 using (i) index laws and (ii) cancelling. (i) x 3 ' x 5 = x - 2 by index laws 3 x# x# x (ii) x = 5 x x# x# x# x # x 1 = 2 x 1 So x - 2 = 2 x Now simplify these questions by (i) index laws and (ii) cancelling. (a) x 2 ' x 3 (b) x 2 ' x 4 (c) x 2 ' x 5 (d) x 3 ' x 6 (e) x 3 ' x 3 (f) x 2 ' x 2 (g) x ' x 2 (h) x 5 ' x 6 (i) x 4 ' x 7 (j) x ' x 3 Use your results to complete: x0 = x-n =

x0 = 1

Proof xn ' xn = xn - n = x0 xn xn ' xn = n x =1 `

x0 = 1

Chapter 1 Basic Arithmetic

x-n =

1 xn

Proof x0 ' xn = x0 - n = x-n x0 x0 ' xn = n x 1 = n x 1 ` x-n = n x

EXAMPLES 0

1. Simplify e

ab 5 c o . abc 4

Solution 0

e

ab 5 c o =1 abc 4

2. Evaluate 2 - 3 .

Solution 1 23 1 = 8

2-3 =

3. Write in index form. 1 x2 3 (b) 5 x 1 (c) 5x 1 (d) x +1 (a)

CONTINUED

25

26

Maths In Focus Mathematics Extension 1 Preliminary Course

Solution 1 = x-2 x2 3 (b) 5 = 3# 15 x x -5 = 3x 1 1 1 = #x (c) 5x 5 1 -1 = x 5 1 1 = (d) x +1 (x + 1) 1 = ] x + 1 g-1 (a)

4. Write a−3 without the negative index.

Solution a-3 =

1 a3

1.6 Exercises 1.

Evaluate as a fraction or whole number. (a) 3 - 3 (b) 4 - 1 (c) 7 - 3 (d) 10 - 4 (e) 2 - 8 (f) 60 (g) 2 - 5 (h) 3 - 4 (i) 7 - 1 (j) 9 - 2 (k) 2 - 6 (l) 3 - 2 (m) 40 (n) 6 - 2 (o) 5 - 3 (p) 10 - 5 (q) 2 - 7 (r) 2 0 (s) 8 - 2 (t) 4 - 3

2.

Evaluate (a) 2 0 1 -4 (b) c m 2 2 -1 (c) c m 3 5 -2 (d) c m 6 x + 2y 0 p (e) f 3x - y 1 -3 (f) c m 5 3 -1 (g) c m 4 1 -2 (h) c m 7 2 -3 (i) c m 3 1 -5 (j) c m 2 3 -1 (k) c m 7

Chapter 1 Basic Arithmetic

8 0 (l) c m 9 6 -2 (m)c m 7 9 -2 (n) c m 10 6 0 (o) c m 11 1 -2 (p) c - m 4 2 -3 (q) c - m 5 2 -1 (r) c - 3 m 7 3 0 (s) c - m 8 1 -2 (t) c - 1 m 4 3.

Change into index form. (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k)

1 m3 1 x 1 p7 1 d9 1 k5 1 x2 2 x4 3 y2 1 2z 6 3 5t 8 2 7x

5 2m 6 2 (m) 7 3y (l)

1 (3x + 4) 2 1 (o) ( a + b) 8 1 (p) x-2 (n)

1 (5p + 1) 3 2 (r) (4t - 9) 5 1 (s) 4 (x + 1) 11 5 (t) 9 ( a + 3 b) 7 (q)

4.

Write without negative indices. (a) t - 5 (b) x - 6 (c) y - 3 (d) n - 8 (e) w - 10 (f) 2x -1 (g) 3m - 4 (h) 5x - 7 (i) ]2xg- 3 (j) ] 4n g-1 (k) ] x + 1 g- 6 (l) ^ 8y + z h-1 (m) ]k - 3g- 2 (n) ^ 3x + 2y h- 9 1 -5 (o) b x l 1 -10 (p) c y m 2 -1 (q) d n p 1 -2 m a+b x + y -1 (s) e x - y o (r) c

(t) e

2w - z - 7 o 3x + y

27

28

Maths In Focus Mathematics Extension 1 Preliminary Course

Fractional indices

Class Investigation Explore fractional indices by looking at these questions. 1 2

For example simplify (i) ` x 2 j and (ii) ^ x h . 1 2

(i) ` x 2 j = x 1 =x 2 (ii) ^ x h = x

2

^ by index laws h

1 2

So ` x 2 j = ^ x h = x 2

1

`

x2 =

x

Now simplify these questions. 1

(a) ^ x 2 h 2 x2

(b)

1 3

(c) ` x 3 j

1

(d) ^ x 3 h 3 3 (e) ^ 3 x h

(f)

3

x3 1 4

4 (g) ` x j

1

(h) ^ x 4 h 4 4 (i) ^ 4 x h

(j)

4

x4

Use your results to complete: 1

xn =

1 n

a =n a

Proof 1 n

`an j = a ^ n a hn = a 1 n

` a =n a

^ by index laws h

Chapter 1 Basic Arithmetic

EXAMPLES 1. Evaluate (a) 49

1 2 1

(b) 27 3

Solution 1 2

(a) 49 = 49 =7 1 3

(b) 27 = 3 27 =3 2. Write

3x - 2 in index form.

Solution 1

3x - 2 = (3x - 2) 2 1

3. Write (a + b) 7 without fractional indices.

Solution 1

( a + b) 7 = 7 a + b

Putting the fractional and negative indices together gives this rule.

a

1 -n

=

n

1 a

Here are some further rules.

m n

a = n am = (n a ) m

Proof m

1 m

m n

1 n

n n a = `a j m = ^n a h

a = ^ am h = n am

29

30

Maths In Focus Mathematics Extension 1 Preliminary Course

a -n b n c m = bal b

Proof a -n 1 c m = b a n c m b 1 = n a bn an bn bn =1# n a bn = n a b n = bal =1'

EXAMPLES 1. Evaluate 4

(a) 8 3 (b) 125

-

1 3

2 -3 (c) c m 3

Solution 4

(a) 8 3 = (3 8 ) 4 (or 3 8 4 ) = 24 = 16 (b) 125

-

1 3

=

1 1

125 3 1 =3 125 1 = 5

Chapter 1 Basic Arithmetic

-3

(c) c 2 m 3

3 3 =c m 2 27 = 8 3 =3 8

2. Write in index form. x5

(a) (b)

1 (4x - 1) 2 2

3

Solution 5

x5 = x 2 1

(a) (b)

(4x - 1) 2

3

2

=

1 2

(4x 2 - 1) 3 -

= (4x 2 - 1)

3. Write r

-

3 5

2 3

without the negative and fractional indices.

Solution r

-

3 5

= =

1 3

r5 1 5

r3

DID YOU KNOW? Nicole Oresme (1323–82) was the first mathematician to use fractional indices. John Wallis (1616–1703) was the first person to explain the significance of zero, negative and fractional indices. He also introduced the symbol 3 for infinity. Do an Internet search on these mathematicians and find out more about their work and backgrounds. You could use keywords such as indices and infinity as well as their names to find this information.

31

32

Maths In Focus Mathematics Extension 1 Preliminary Course

1.7 Exercises 1.

Evaluate (a) 81

3.

1 2

Write without fractional indices. 1

(a) y 3

1

2

(b) 27 3

(b) y 3

1

(c) 16 2

(c) x

1

-

1 2 1

(d) (2x + 5) 2

(d) 8 3 1

(e) (3x - 1)

(e) 49 2 1

-

1 2

1

(f) (6q + r) 3

(f) 1000 3 1

(g) (x + 7)

(g) 16 4

-

2 5

1

(h) 64 2 (i) 64 (j) 1

4.

1 3

(a)

1 7

(k) 81 (l) 32

Write in index form. (b)

t 5

x3

(c)

1 4

(d) (e)

1 5

3

1

(m) 0 8

(f)

(n) 125

1 3

(g)

1 1 1

(r) 9 (s) 8

(i)

(t) 64 2.

(x - 2) 2 1 (j) 2 y+7 5 (k) 3 x+4 2 (l) 3 y2 - 1 3 (m) 5 4 (x 2 + 2) 3

3 2 -

1 3 -

2 3

Evaluate correct to 2 decimal places. 1

(a) 23 4 (b) 4 45.8 (c) (d) (e)

7

5

8

5 .9 # 3 .7 8.79 - 1.4

4

(f)

1.24 + 4.3 2 1 12.9 3 .6 - 1 .4 1 .5 + 3 .7

(3x + 1) 5 1

(h)

(q) 256 4

9-x 4s + 1 1 2t + 3 1 (5x - y) 3

(o) 343 3 (p) 128 7

y

5.

3

Write in index form and simplify. (a) x x x (b) x x (c) 3 x x2 (d) 3 x (e) x 4 x

Chapter 1 Basic Arithmetic

6.

Expand and simplify, and write in index form.

7.

(a) ( x + x) 2 (b) (3 a + 3 b ) (3 a - 3 b ) 1 2 (c) f p + p p 1 2 ) x x ( x 2 - 3x + 1 )

(d) ( x + (e)

x3

33

Write without fractional or negative indices. (a) (a - 2b) (b) (y - 3)

-

-

1 3

2 3

-

4 7

-

2 9

(c) 4 (6a + 1) ( x + y) (d) 3

-

5 4

6 (3 x + 8 ) (e) 7

Scientific notation (standard form) Very large or very small numbers are usually written in scientific notation to make them easier to read. What could be done to make the figures in the box below easier to read?

DID YOU KNOW? The Bay of Fundy, Canada, has the largest tidal changes in the world. About 100 000 000 000 tons of water are moved with each tide change. The dinosaurs dwelt on Earth for 185 000 000 years until they died out 65 000 000 years ago. The width of one plant cell is about 0.000 06 m. In 2005, the total storage capacity of dams in Australia was 83 853 000 000 000 litres and households in Australia used 2 108 000 000 000 litres of water.

A number in scientific notation is written as a number between 1 and 10 multiplied by a power of 10.

EXAMPLES 1. Write 320 000 000 in scientific notation.

Solution 320 000 000 = 3.2 #10 8

Write the number between 1 and 10 and count the decimal places moved.

2. Write 7.1#10 -5 as a decimal number.

Solution 7.1#10

-5

= 7.1 ' 10 = 0.000 071 5

Count 5 places to the left.

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Maths In Focus Mathematics Extension 1 Preliminary Course

SCIENTIFIC NOTATION KEY Use the EXP or #10 x key to put numbers in scientific notation. For example, to evaluate 3.1#10 4 ' 2.5 #10 - 2, press 3.1 EXP 4 ' 2.5 EXP (-) 2 = = 1 240 000

DID YOU KNOW? Engineering notation is similar to scientific notation, except the powers of 10 are always multiples of 3. For example, 3.5 # 10

3

15.4 # 10

-6

SIGNIFICANT FIGURES The concept of significant figures is related to rounding off. When we look at very large (or very small) numbers, some of the smaller digits are not significant. For example, in a football crowd of 49 976, the 6 people are not really significant in terms of a crowd of about 50 000! Even the 76 people are not significant. When a company makes a profit of $5 012 342.87, the amount of 87 cents is not exactly a significant sum! Nor is the sum of $342.87. To round off to a certain number of significant figures, we count from the first non-zero digit. In any number, non-zero digits are always significant. Zeros are not significant, except between two non-zero digits or at the end of a decimal number. Even though zeros may not be significant, they are still necessary. For example 31, 310, 3100, 31 000 and 310 000 all have 2 significant figures but are very different numbers! Scientific notation uses the significant figures in a number.

EXAMPLES 12 000 = 1.2 #10 4 0.000 043 5 = 4.35#10 - 5 0.020 7 = 2.07 #10 - 2

(2 significant figures) (3 significant figures) (3 significant figures)

When rounding off to significant figures, use the usual rules for rounding off.

Chapter 1 Basic Arithmetic

35

EXAMPLES 1. Round off 4 592 170 to 3 significant figures.

Solution 4 592 170 = 4 590 000 to 3 significant figures 2. Round off 0.248 391 to 2 significant figures.

Solution 0.248 391 = 0.25 to 2 significant figures 3. Round off 1.396 794 to 3 significant figures.

Solution 1.396 794 = 1.40 to 3 significant figures

1.8 Exercises 1.

Write in scientific notation. (a) 3 800 (b) 1 230 000 (c) 61 900 (d) 12 000 000 (e) 8 670 000 000 (f) 416 000 (g) 900 (h) 13 760 (i) 20 000 000 (j) 80 000

3.

Write as a decimal number. (a) 3.6 #10 4 (b) 2.78 #10 7 (c) 9.25#10 3 (d) 6.33#10 6 (e) 4 #10 5 (f) 7.23#10 - 2 (g) 9.7 #10 - 5 (h) 3.8 # 10 - 8 (i) 7 #10 - 6 (j) 5#10 - 4

2.

Write in scientific notation. (a) 0.057 (b) 0.000 055 (c) 0.004 (d) 0.000 62 (e) 0.000 002 (f) 0.000 000 08 (g) 0.000 007 6 (h) 0.23 (i) 0.008 5 (j) 0.000 000 000 07

4.

Round these numbers to 2 significant figures. (a) 235 980 (b) 9 234 605 (c) 10 742 (d) 0.364 258 (e) 1.293 542 (f) 8.973 498 011 (g) 15.694 (h) 322.78 (i) 2904.686 (j) 9.0741

Remember to put the 0’s in!

36

Maths In Focus Mathematics Extension 1 Preliminary Course

5.

Evaluate correct to 3 significant figures. (a) 14.6 # 0.453 (b) 4.8 ' 7 (c) 4.47 + 2.59 #1.46 1 (d) 3.47 - 2.7

6.

Evaluate 4.5#10 4 # 2.9 #10 5, giving your answer in scientific notation.

7.

Calculate

8.72 #10 - 3 and write 1.34 #10 7 your answer in standard form correct to 3 significant figures.

Investigation A logarithm is an index. It is a way of finding the power (or index) to which a base number is raised. For example, when solving 3 x = 9, the solution is x = 2. The 3 is called the base number and the x is the index or power. You will learn about logarithms in the HSC course. The a is called the base number and the x is the index or power.

If a x = y then log a y = x

1. The expression log7 49 means the power of 7 that gives 49. The solution is 2 since 7 2 = 49. 2. The expression log2 16 means the power of 2 that gives 16. The solution is 4 since 2 4 = 16. Can you evaluate these logarithms? 1. log3 27 2. log5 25 3. log10 10 000 4. log2 64 5. log4 4 6. log7 7 7. log3 1 8. log4 2 1 9. log 3 3 1 10. log 2 4

Chapter 1 Basic Arithmetic

37

Absolute Value Negative numbers are used in maths and science, to show opposite directions. For example, temperatures can be positive or negative.

But sometimes it is not appropriate to use negative numbers. For example, solving c 2 = 9 gives two solutions, c = !3. However when solving c 2 = 9, using Pythagoras’ theorem, we only use the positive answer, c = 3, as this gives the length of the side of a triangle. The negative answer doesn’t make sense. We don’t use negative numbers in other situations, such as speed. In science we would talk about a vehicle travelling at –60k/h going in a negative direction, but we would not commonly use this when talking about the speed of our cars!

Absolute value definitions We write the absolute value of x as x

x =)

We can also define x as the distance of x from 0 on the number line. We will use this in Chapter 3.

x when x $ 0 - x when x 1 0

EXAMPLES 1. Evaluate 4 .

Solution 4 = 4 since 4 $ 0

CONTINUED

38

Maths In Focus Mathematics Extension 1 Preliminary Course

2. Evaluate - 3 .

Solution -3 = - ] - 3 g since - 3 1 0 =3

The absolute value has some properties shown below.

Properties of absolute value

| ab | = | a |#| b |

e.g. | 2 # - 3 | = | 2 |#| - 3 | = 6

|a | = a

e.g. | - 3 | 2 = ] - 3 g2 = 9

2

2

a2 = | a | |- a | = | a | |a - b | = | b - a | | a + b |#| a | + | b |

e.g. 5 2 = | 5 | = 5 e.g. | -7 | = | 7 | = 7 e.g. | 2 - 3 | = | 3 - 2 | = 1 e.g. | 2 + 3 | = | 2 | + | 3 | but | - 3 + 4 | 1 | - 3 | + | 4 |

EXAMPLES 1. Evaluate 2 - -1 + - 3 2.

Solution 2 - -1 + - 3 2 = 2 - 1 + 3 2 =2 -1 + 9 = 10 2. Show that a + b # a + b when a = - 2 and b = 3.

Solution LHS means Left Hand Side.

LHS = a + b = -2 + 3 = 1 =1

Chapter 1 Basic Arithmetic

RHS means Right Hand Side.

RHS = a + b = -2 + 3 = 2+3 =5 Since 11 5 a+b # a + b 3. Write expressions for 2x - 4 without the absolute value signs.

Solution 2x - 4 = 2x - 4 when 2x - 4 $ 0 i.e. 2x $ 4 x$2 2x - 4 = - ] 2x - 4 g when 2x - 4 1 0 = - 2x + 4 i.e.

2x 1 4 x12

Class Discussion Are these statements true? If so, are there some values for which the expression is undefined (values of x or y that the expression cannot have)?

2.

x =1 x 2x = 2x

3.

2x = 2 x

4.

x + y = x+y

5.

2 x = x2

6. 7.

3 x = x3 x +1 = x +1

1.

3x - 2 =1 3x - 2 x 9. =1 x2 10. x $ 0 8.

Discuss absolute value and its definition in relation to these statements.

39

40

Maths In Focus Mathematics Extension 1 Preliminary Course

1.9 Exercises 1.

2.

3.

Evaluate (a) 7 (b) - 5 (c) - 6 (d) 0 (e) 2 (f) -11 (g) - 2 3 (h) 3 - 8 2 (i) - 5 (j) - 5 3 Evaluate (a) 3 + - 2 (b) - 3 - 4 (c) - 5 + 3 (d) 2 #-7 (e) - 3 + -1 2 (f) 5 - - 2 # 6 (g) - 2 + 5# -1 (h) 3 - 4 (i) 2 - 3 - 3 - 4 (j) 5 - 7 + 4 - 2

(i) (j)

Show that a + b # a + b when (a) a = 2 and b = 4 (b) a = -1 and b = - 2 (c) a = - 2 and b = 3 (d) a = - 4 and b = 5 (e) a = -7 and b = - 3.

6.

Show that x 2 = x when (a) x = 5 (b) x = - 2 (c) x = - 3 (d) x = 4 (e) x = - 9.

7.

Use the definition of absolute value to write each expression without the absolute value signs (a) x + 5 (b) b - 3 (c) a + 4 (d) 2y - 6 (e) 3x + 9 (f) 4 - x (g) 2k + 1 (h) 5x - 2 (i) a + b (j) p - q

8.

Find values of x for which x = 3.

9.

n Simplify n where n ! 0.

a = 5 and b = 2 a = -1 and b = 2 a = - 2 and b = - 3 a = 4 and b = 7 a = -1 and b = - 2.

Write an expression for (a) a when a 2 0 (b) (c) (d) (e) (f) (g)

a when a 1 0 a when a = 0 3a when a 2 0 3a when a 1 0 3a when a = 0 a + 1 when a 2 -1

x - 2 when x 2 2 x - 2 when x 1 2.

5.

Evaluate a - b if (a) (b) (c) (d) (e)

4.

(h) a + 1 when a 1 -1

x-2 and state which x-2 value x cannot be.

10. Simplify

Chapter 1 Basic Arithmetic

Test Yourself 1 1.

2.

Convert (a) 0.45 to a fraction (b) 14% to a decimal 5 (c) to a decimal 8 (d) 78.5% to a fraction (e) 0.012 to a percentage 11 (f) to a percentage 15

6.

(b) (c) (d) (e)

(a) 7 - 2 (b) 5 -1

3.

-

7.

1 2

(b) 36 2 (c) - 5 2 - 2 3 (d) 4 - 3 as fraction

4.5 2 + 7.6 2

(b) 4.3 0.3 2 (c) 3 5.7 (d)

(e) 6 4.

5.

2

(e) 8 3 (f) - 2 - 1

1.3#10 9 3.8 #10 6 -

(g) 49

2 3

-

1 2

as a fraction

1 4

Evaluate (a) |-3 | -| 2 | (b) | 4 - 5 | (c) 7 + 4 # 8 (d) [(3 + 2)#(5 - 1) - 4] ' 8 (e) - 4 + 3 - 9 (f) - 2 - -1 (g) - 24 ' - 6

(h) 16 (i) ] -3 g0 (j) 4 - 7 2 - -2 - 3 8.

(a) x 5 # x 7 ' x 3 (b) (5y 3) 2 (a 5) 4 b 7 (c) a9 b 3 2x 6 n (d) d 3 0

ab 4 o a5 b6

Simplify (a) a 14 ' a 9 6 (b) _ x 5 y 3 i (c) p 6 # p 5 ' p 2 4 (d) ^ 2b 9h (2x 7) 3 y 2 (e) x 10 y

Simplify

(e) e

Evaluate (a) - 4 1

Evaluate correct to 3 significant figures. (a)

3 7 5 8 6 2 #3 7 3 3 9' 4 2 1 +2 5 10 5 15# 6

(a) 1

Evaluate as a fraction.

(c) 9

Evaluate

9.

Write in index form. (a)

n 1 (b) 5 x 1 (c) x+y (d)

4

x +1

41

42

Maths In Focus Mathematics Extension 1 Preliminary Course

(e)

7

(c) If he spends 3 hours watching TV, what fraction of the day is this? (d) What percentage of the day does he spend sleeping?

a+b

2 (f) x 1 (g) 2x 3 (h)

3

x4

(i)

7

(5x + 3) 9 1

4

m3

(j)

17. Rachel scored 56 out of 80 for a maths test. What percentage did she score?

10. Write without fractional or negative indices. (a) a - 5 1

(b) n 4 1

(c) (x + 1) 2 (d) (x - y) -1 (e) (4t - 7) - 4 1

(f) (a + b) 5 (g) x

3

(i) (2x + 3) -

18. Evaluate 2118, and write your answer in scientific notation correct to 1 decimal place. 19. Write in index form. (a) x 1 (b) y x+3 1 (d) (2x - 3) 11

1 3

(h) b 4

(j) x

16. The price of a car increased by 12%. If the car cost $34 500 previously, what is its new price?

4 3

3 2

(c)

6

(e)

3

y7

20. Write in scientific notation. (a) 0.000 013

11. Show that a + b # a + b when a = 5 and b = - 3. 9 2 12. Evaluate a b when a = and b = 1 . 25 3 2 4

3 1 4 13. If a = c m and b = , evaluate ab 3 as a 4 3 fraction. 14. Increase 650 mL by 6%. 1 of his 24-hour day 3 1 sleeping and at work. 4 (a) How many hours does Johan spend at work? (b) What fraction of his day is spent at work or sleeping?

15. Johan spends

(b) 123 000 000 000 21. Convert to a fraction. • (a) 0. 7 • • (b) 0.124 22. Write without the negative index. (a) x - 3 (b) (2a + 5)- 1 a -5 (c) c m b 23. The number of people attending a football match increased by 4% from last week. If there were 15 080 people at the match this week, how many attended last week? 24. Show that | a + b | # a + b when a = - 2 and b = - 5.

Chapter 1 Basic Arithmetic

Challenge Exercise 1 3 2 2 7 + 3 m ' c4 - 1 m. 4 5 3 8

1.

Simplify c 8

2.

3 5 149 7 Simplify + + . 5 12 180 30

3.

4.

Arrange in increasing order of size: 51%, • 51 0.502, 0. 5, . 99 1 1 of his day sleeping, 3 12 1 of the day eating and of the day 20 watching TV. What percentage of the day is left? Mark spends

-

2 3

5.

Write 64

6.

Express 3.2 ' 0.014 in scientific notation correct to 3 significant figures.

7.

as a rational number.

11. Show that 2 (2 k - 1) + 2 k + 1 = 2 (2 k + 1 - 1) . 12. Find the value of

3 2 2 4 1 3 a = c m , b = c - m and c = c m . 5 5 3 13. Which of the following are rational • 3 numbers: 3 , - 0.34, 2, 3r, 1. 5, 0, ? 7 14. The percentage of salt in 1 L of water is 10%. If 500 mL of water is added to this mixture, what percentage of salt is there now? 15. Simplify

25

1 out of 20 for a maths 2 1 test, 19 out of 23 for English and 55 2 out of 70 for physics. Find his average score as a percentage, to the nearest whole percentage.

Vinh scored 17

• • •

a in index form if b3 c2

|x + 1 | x2 - 1

for x ! !1.

4.3 1.3 - 2.9 correct to 2.4 3 + 3.31 2 2 decimal places.

16. Evaluate 6

17. Write 15 g as a percentage of 2.5 kg. 18. Evaluate 2.3 1.8 + 5.7 #10 - 2 correct to 3 significant figures. - 3.4 #10 - 3 + 1.7 #10 - 2 and (6.9 #10 5) 3 express your answer in scientific notation correct to 3 significant figures.

8.

Write 1.3274 as a rational number.

19. Evaluate

9.

The distance from the Earth to the moon is 3.84 #10 5 km. How long would it take a rocket travelling at 2.13#10 4 km h to reach the moon, to the nearest hour?

20. Prove | a + b | # | a | + | b | for all real a, b.

8.3# 4.1 correct to 0.2 + 5.4 ' 1.3 3 significant figures.

10. Evaluate 3

43

2 Algebra and Surds TERMINOLOGY Binomial: A mathematical expression consisting of two terms such as x + 3 or 3x - 1 Binomial product: The product of two binomial expressions such as (x + 3) (2x - 4) Expression: A mathematical statement involving numbers, pronumerals and symbols e.g. 2x - 3 Factorise: The process of writing an expression as a product of its factors. It is the reverse operation of expanding brackets i.e. take out the highest common factor in an expression and place the rest in brackets e.g. 2y - 8 = 2 (y - 4) Pronumeral: A letter or symbol that stands for a number

Rationalising the denominator: A process for replacing a surd in the denominator by a rational number without altering its value Surd: From ‘absurd’. The root of a number that has an irrational value e.g. 3 . It cannot be expressed as a rational number Term: An element of an expression containing pronumerals and/or numbers separated by an operation such as + , - , # or ' e.g. 2x, - 3 Trinomial: An expression with three terms such as 3x 2 - 2x + 1

Chapter 2 Algebra and Surds

45

INTRODUCTION THIS CHAPTER REVIEWS ALGEBRA skills, including simplifying expressions, removing grouping symbols, factorising, completing the square and simplifying algebraic fractions. Operations with surds, including rationalising the denominator, are also studied in this chapter.

DID YOU KNOW? One of the earliest mathematicians to use algebra was Diophantus of Alexandria. It is not known when he lived, but it is thought this may have been around 250 AD. In Baghdad around 700–800 AD a mathematician named Mohammed Un-Musa Al-Khowarezmi wrote books on algebra and Hindu numerals. One of his books was named Al-Jabr wa’l Migabaloh, and the word algebra comes from the first word in this title.

Simplifying Expressions Addition and subtraction

EXAMPLES Simplify 1. 7x - x

Solution

Here x is called a pronumeral.

7x - x = 7x - 1 x = 6x 2. 4x 2 - 3x 2 + 6x 2

Solution 4x 2 - 3x 2 + 6x 2 = x 2 + 6 x 2 = 7x 2

CONTINUED

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Maths In Focus Mathematics Extension 1 Preliminary Course

3. x 3 - 3x - 5x + 4 Only add or subtract ‘like’ terms. These have the same pronumeral (for example, 3x and 5x).

Solution x 3 - 3 x - 5x + 4 = x 3 - 8 x + 4 4. 3a - 4b - 5a - b

Solution 3a - 4b - 5a - b = 3a - 5a - 4b - b = - 2a - 5b

2.1 Exercises Simplify 1.

2x + 5x

16. 7b + b - 3b

2.

9a - 6a

17. 3b - 5b + 4b + 9b

3.

5z - 4z

18. - 5x + 3x - x - 7x

4.

5a + a

19. 6x - 5y - y

5.

4b - b

20. 8a + b - 4b - 7a

6.

2r - 5r

21. xy + 2y + 3xy

7.

- 4y + 3y

22. 2ab 2 - 5ab 2 - 3ab 2

8.

- 2x - 3x

23. m 2 - 5m - m + 12

9.

2a - 2a

24. p 2 - 7p + 5p - 6

10. - 4k + 7k

25. 3x + 7y + 5x - 4y

11. 3t + 4t + 2t

26. ab + 2b - 3ab + 8b

12. 8w - w + 3w

27. ab + bc - ab - ac + bc

13. 4m - 3m - 2m

28. a 5 - 7x 3 + a 5 - 2x 3 + 1

14. x + 3x - 5x

29. x 3 - 3xy 2 + 4x 2 y - x 2 y + xy 2 + 2y 3

15. 8h - h - 7h

30. 3x 3 - 4x 2 - 3x + 5x 2 - 4x - 6

Chapter 2 Algebra and Surds

47

Multiplication EXAMPLES Simplify 1. - 5x # 3y # 2x

Solution - 5x # 3y # 2x = - 30xyx = - 30x 2 y 2. - 3x 3 y 2 # - 4xy 5

Solution

Use index laws to simplify this question.

- 3x 3 y 2 # - 4xy 5 = 12x 4 y 7

2.2 Exercises Simplify 1.

5 # 2b

5 11. ^ 2x 2h

2.

2x # 4y

12. 2ab 3 # 3a

3.

5p # 2p

13. 5a 2 b # - 2ab

4.

- 3z # 2w

14. 7pq 2 # 3p 2 q 2

5.

- 5a # - 3b

15. 5ab # a 2 b 2

6.

x # 2y # 7z

16. 4h 3 # - 2h 7

7.

8ab # 6c

17. k 3 p # p 2

8.

4d # 3d

4 18. ^ - 3t 3 h

9.

3a # 4a # a

19. 7m 6 # - 2m 5

10. ^ - 3y h3

20. - 2x 2 # 3x 3 y # - 4xy 2

48

Maths In Focus Mathematics Extension 1 Preliminary Course

Division Use cancelling or index laws to simplify divisions.

EXAMPLES Simplify 1. 6v 2 y ' 2vy

Solution By cancelling, 6v 2 y ' 2vy = =

6v 2 y 2vy 63 # v # v1 # y1 21 # v # y1

= 3v Using index laws, 6v 2 y ' 2vy = 3v 2 - 1 y 1 - 1 = 3v 1 y 0 = 3v 2.

5a 3 b 15ab 2

Solution 5a 3 b = 1 a3 -1 b1- 2 3 15ab 2 = 1 a 2 b -1 3 a2 = 3b

2.3 Exercises Simplify 1.

30x ' 5

2.

2y ' y 2

3.

8a 2

4.

8a 2 a

5.

8a 2 2a

6.

xy 2x

7.

12p 3 ' 4p 2

8.

3a 2 b 2 6ab

9.

20x 15xy

10.

- 9x 7 3x 4

Chapter 2 Algebra and Surds

11. -15ab ' - 5b 12.

2ab 6a 2 b 3

13.

- 8p 4pqs

16.

7pq 3

17. 5a 9 b 4 c - 2 ' 20a 5 b -3 c -1 2 ^ a -5 h b 4 2

18.

14. 14cd 2 ' 21c 3 d 3 15.

42p 5 q 4

-1

4a - 9 ^ b 2 h

19. - 5x 4 y 7 z ' 15xy 8 z - 2

2xy 2 z 3

20. - 9 ^ a 4 b -1 h ' -18a -1 b 3 3

4x 3 y 2 z

Removing grouping symbols The distributive law of numbers is given by

a ] b + c g = ab + ac

EXAMPLE 7 # (9 + 11) = 7 # 20 = 140 Using the distributive law, 7 # (9 + 11) = 7 # 9 + 7 # 11 = 63 + 77 = 140

This rule is used in algebra to help remove grouping symbols.

EXAMPLES Expand and simplify. 1. 2 ] a + 3 g

Solution 2 (a + 3) = 2 # a + 2 # 3 = 2a + 6

CONTINUED

49

50

Maths In Focus Mathematics Extension 1 Preliminary Course

2. - ] 2x - 5 g

Solution -(2x - 5) = -1 (2x - 5) = -1 # 2x - 1 # - 5 = - 2x + 5 3. 5a 2]4 + 3ab - c g

Solution 5a 2 (4 + 3ab - c) = 5a 2 # 4 + 5a 2 # 3ab - 5a 2 # c = 20a 2 + 15a 3 b - 5a 2 c 4. 5 - 2 ^ y + 3 h

Solution 5 - 2 (y + 3 ) = 5 - 2 # y - 2 # 3 = 5 - 2y - 6 = - 2y - 1 5. 2 ] b - 5 g - ] b + 1 g

Solution 2 (b - 5) - (b + 1) = 2 # b + 2 # - 5 - 1 # b -1 # 1 = 2b - 10 - b - 1 = b - 11

2.4 Exercises Expand and simplify 1.

2]x - 4 g

7.

ab ] 2a + b g

2.

3 ] 2h + 3 g

8.

5n ] n - 4 g

3.

-5 ] a - 2 g

9.

3x 2 y _ xy + 2y 2 i

4.

x ^ 2y + 3 h

10. 3 + 4 ] k + 1 g

5.

x]x - 2 g

11. 2 ] t - 7 g - 3

6.

2a ] 3a - 8b g

12. y ^ 4y + 3 h + 8y

Chapter 2 Algebra and Surds

13. 9 - 5 ] b + 3 g

20. 2ab ] 3 - a g - b ] 4a - 1 g

14. 3 - ] 2x - 5 g

21. 5x - ] x - 2 g - 3

15. 5] 3 - 2m g + 7 ] m - 2 g

22. 8 - 4 ^ 2y + 1 h + y

16. 2 ] h + 4 g + 3 ] 2h - 9 g

23. ] a + b g - ] a - b g

17. 3 ] 2d - 3 g - ] 5d - 3 g

24. 2 ] 3t - 4 g - ] t + 1 g + 3

18. a ] 2a + 1 g - ^ a 2 + 3a - 4 h

25. 4 + 3 ] a + 5 g - ] a - 7 g

51

19. x ] 3x - 4 g - 5 ] x + 1 g

Binomial Products A binomial expression consists of two numbers, for example x + 3. A set of two binomial expressions multiplied together is called a binomial product. Example: ] x + 3 g ] x - 2 g. Each term in the first bracket is multiplied by each term in the second bracket.

] a + b g ^ x + y h = ax + ay + bx + by

Proof ]a + bg]c + d g = a ]c + d g + b ]c + d g = ac + ad + bc + bd

EXAMPLES Expand and simplify 1. ^ p + 3h^ q - 4h

Solution ^ p + 3 h ^ q - 4 h = pq - 4p + 3q - 12 2. ]a + 5g2

Solution ] a + 5 g2 = (a + 5)(a + 5) = a 2 + 5a + 5a + 25 = a 2 + 10a + 25

Can you see a quick way of doing this?

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Maths In Focus Mathematics Extension 1 Preliminary Course

The rule below is not a binomial product (one expression is a trinomial), but it works the same way.

] a + b g ^ x + y + z h = ax + ay + az + bx + by + bz

EXAMPLE Expand and simplify ] x + 4 g ^ 2x - 3y - 1 h .

Solution (x + 4) (2x - 3y - 1) = 2x 2 - 3xy - x + 8x - 12y - 4 = 2x 2 - 3xy + 7x - 12y - 4

2.5 Exercises Expand and simplify 1.

]a + 5g]a + 2g

17. ]a + 2bg]a - 2bg

2.

]x + 3g]x - 1g

18. ^ 3x - 4y h^ 3x + 4y h

3.

^ 2y - 3h^ y + 5h

19. ]x + 3g]x - 3g

4.

]m - 4g]m - 2g

20. ^ y - 6h^ y + 6h

5.

]x + 4g]x + 3g

21. ] 3a + 1 g ] 3a - 1 g

6.

^ y + 2h^ y - 5h

22. ]2z - 7g]2z + 7g

7.

]2x - 3g]x + 2g

23. ]x + 9g^ x - 2y + 2h

8.

]h - 7g]h - 3g

24. ] b - 3 g ] 2a + 2b - 1 g

9.

]x + 5g]x - 5g

25. ]x + 2g^ x 2 - 2x + 4h

10. ] 5a - 4 g ] 3a - 1 g

26. ]a - 3g^ a 2 + 3a + 9h

11. ^ 2y + 3h^ 4y - 3h

27. ]a + 9g2

12. ]x - 4g^ y + 7h

28. ]k - 4g2

13. ^ x 2 + 3h]x - 2g

29. ]x + 2g2

14. ]n + 2g]n - 2g

30. ^ y - 7h2

15. ]2x + 3g]2x - 3g

31. ]2x + 3g2

16. ^ 4 - 7y h^ 4 + 7y h

32. ]2t - 1g2

Chapter 2 Algebra and Surds

33. ]3a + 4bg2

37. ] a + b g2

34. ^ x - 5y h2

38. ] a - b g2

35. ]2a + bg2

39. ] a + b g ^ a 2 - ab + b 2 h

36. ] a - b g ] a + b g

40. ] a - b g ^ a 2 + ab + b 2 h

Some binomial products have special results and can be simplified quickly using their special properties. Binomial products involving perfect squares and the difference of two squares occur in many topics in mathematics. Their expansions are given below.

Difference of 2 squares ] a + b g ] a - b g = a2 - b2

Proof (a + b) (a - b) = a 2 - ab + ab - b 2 = a2 - b2

Perfect squares ] a + b g2 = a 2 + 2ab + b 2

Proof ] a + b g2 = (a + b) (a + b) = a 2 + ab + ab + b 2 = a 2 + 2ab + b 2

]a - bg2 = a 2 - 2ab + b 2

Proof ] a - b g2 = (a - b) (a - b) = a 2 - ab - ab + b 2 = a 2 - 2ab + b 2

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EXAMPLES Expand and simplify 1. ]2x - 3g2

Solution ] 2x - 3 g2 = ] 2x g2 - 2 (2x) 3 + 3 2 = 4x 2 - 12x + 9 2. ^ 3y - 4h^ 3y + 4h

Solution (3y - 4) (3y + 4) = ^ 3y h2 - 4 2 = 9y 2 - 16

2.6 Exercises Expand and simplify 1.

]t + 4g2

16. ^ p + 1 h ^ p - 1 h

2.

]z - 6g2

17. ]r + 6g]r - 6g

3.

] x - 1 g2

18. ] x - 10 g ] x + 10 g

4.

^ y + 8h2

19. ]2a + 3g]2a - 3g

5.

^ q + 3h2

20. ^ x - 5y h^ x + 5y h

6.

]k - 7g2

21. ] 4a + 1 g ] 4a - 1 g

7.

] n + 1 g2

22. ]7 - 3xg]7 + 3xg

8.

]2b + 5g2

23. ^ x 2 + 2h^ x 2 - 2h

9.

]3 - xg2

2 24. ^ x 2 + 5h

10. ^ 3y - 1 h2

25. ]3ab - 4cg]3ab + 4c g

11. ^ x + y h2

2 2 26. b x + x l

12. ] 3a - b g2 13. ]4d + 5eg2

1 1 27. b a - a lb a + a l

14. ]t + 4g]t - 4g

28. _ x + 6 y - 2 @ i _ x - 6 y - 2 @ i

15. ] x - 3 g ] x + 3 g

29. 6]a + bg + c @2

Chapter 2 Algebra and Surds

30. 7 ] x + 1 g - y A

36. ] x - 4 g3

2

55

Expand (x - 4) (x - 4) 2 .

31. ] a + 3 g2 - ] a - 3 g2

1 2 1 2 37. b x - x l - b x l + 2

32. 16 - ]z - 4g]z + 4g

38. _ x 2 + y 2 i - 4x 2 y 2

33. 2x + ]3x + 1g2 - 4

39. ]2a + 5g3

34. ^ x + y h2 - x ^ 2 - y h

40. ] 2x - 1 g ] 2x + 1 g ] x + 2 g2

2

35. ] 4n - 3 g ] 4n + 3 g - 2n 2 + 5

PROBLEM Find values of all pronumerals that make this true. a b d f e i i i h i i c c

c e b g b

#

Try c = 9.

Factorisation Simple factors Factors are numbers that exactly divide or go into an equal or larger number, without leaving a remainder.

EXAMPLES The numbers 1, 2, 3, 4, 6, 8, 12 and 24 are all the factors of 24. Factors of 5x are 1, 5, x and 5x.

To factorise an expression, we use the distributive law.

ax + bx = x ] a + b g

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Maths In Focus Mathematics Extension 1 Preliminary Course

EXAMPLES Factorise 1. 3x + 12

Solution Divide each term by 3 to find the terms inside the brackets.

The highest common factor is 3. 3x + 12 = 3 ] x + 4 g 2. y 2 - 2y

Solution Check answers by expanding brackets.

The highest common factor is y. y 2 - 2y = y ^ y - 2 h 3. x 3 - 2x 2

Solution x and x2 are both common factors. We take out the highest common factor which is x2. x 3 - 2x 2 = x 2 ] x - 2 g 4. 5] x + 3 g + 2y ] x + 3 g

Solution The highest common factor is x + 3. 5 ] x + 3 g + 2y ] x + 3 g = ] x + 3 g ^ 5 + 2 y h 5. 8a 3 b 2 - 2ab 3

Solution There are several common factors here. The highest common factor is 2ab2. 8a 3 b 2 - 2ab 3 = 2ab 2 ^ 4a 2 - bh

Chapter 2 Algebra and Surds

2.7 Exercises Factorise 1.

2y + 6

19. x ] m + 5 g + 7 ] m + 5 g

2.

5x - 10

20. 2 ^ y - 1 h - y ^ y - 1 h

3.

3m - 9

21. 4^ 7 + y h - 3x ^ 7 + y h

4.

8x + 2

22. 6x ]a - 2g + 5]a - 2g

5.

24 - 18y

23. x ] 2t + 1 g - y ] 2t + 1 g

6.

x 2 + 2x

7.

m 2 - 3m

24. a ] 3x - 2 g + 2b ] 3x - 2 g - 3c ] 3x - 2 g

8.

2y 2 + 4y

9.

15a - 3a 2

25. 6x 3 + 9x 2 26. 3pq 5 - 6q 3 27. 15a 4 b 3 + 3ab

10. ab 2 + ab

28. 4x 3 - 24x 2

11. 4x 2 y - 2xy

29. 35m 3 n 4 - 25m 2 n

12. 3mn 3 + 9mn

30. 24a 2 b 5 + 16ab 2

13. 8x 2 z - 2xz 2 14. 6ab + 3a - 2a

31. 2rr 2 + 2rrh

2

32. ]x - 3g2 + 5]x - 3g

15. 5x 2 - 2x + xy

33. y 2 ]x + 4g + 2]x + 4g

16. 3q 5 - 2q 2

34. a ] a + 1 g - ] a + 1 g2

17. 5b 3 + 15b 2

35. 4ab ^ a 2 + 1 h - 3 ^ a 2 + 1 h

18. 6a 2 b 3 - 3a 3 b 2

Grouping in pairs If an expression has 4 terms, it may be factorised in pairs.

ax + bx + ay + by = x(a + b) + y (a + b) = ( a + b) ( x + y)

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EXAMPLES Factorise 1. x 2 - 2x + 3x - 6

Solution x 2 - 2x + 3x - 6 = x (x - 2) + 3 (x - 2) = (x - 2) (x + 3) 2. 2x - 4 + 6y - 3xy

Solution 2x - 4 + 6y - 3xy = 2 (x - 2) + 3y (2 - x) = 2 ( x - 2) - 3y ( x - 2 ) = (x - 2) (2 - 3y) or 2x - 4 + 6y - 3xy = 2 (x - 2) - 3y (- 2 + x) = 2 ( x - 2) - 3y ( x - 2 ) = (x - 2) (2 - 3y)

2.8 Exercises Factorise 1.

2x + 8 + bx + 4b

12. m - 2 + 4y - 2my

2.

ay - 3a + by - 3b

13. 2x 2 + 10xy - 3xy - 15y 2

3.

x 2 + 5x + 2x + 10

14. a 2 b + ab 3 - 4a - 4b 2

4.

m 2 - 2m + 3m - 6

15. 5x - x 2 - 3x + 15

5.

ad - ac + bd - bc

16. x 4 + 7x 3 - 4x - 28

6.

x 3 + x 2 + 3x + 3

17. 7x - 21 - xy + 3y

7.

5ab - 3b + 10a - 6

18. 4d + 12 - de - 3e

8.

2xy - x 2 + 2y 2 - xy

19. 3x - 12 + xy - 4y

9.

ay + a + y + 1

20. 2a + 6 - ab - 3b

10. x 2 + 5x - x - 5

21. x 3 - 3x 2 + 6x - 18

11. y + 3 + ay + 3a

22. pq - 3p + q 2 - 3q

Chapter 2 Algebra and Surds

23. 3x 3 - 6x 2 - 5x + 10

27. 4x 3 - 6x 2 + 8x - 12

24. 4a - 12b + ac - 3bc

28. 3a 2 + 9a + 6ab + 18b

25. xy + 7x - 4y - 28

29. 5y - 15 + 10xy - 30x

26. x 4 - 4x 3 - 5x + 20

30. rr 2 + 2rr - 3r - 6

59

Trinomials A trinomial is an expression with three terms, for example x 2 - 4x + 3. Factorising a trinomial usually gives a binomial product. x 2 + ] a + b g x + ab = ] x + a g ] x + b g

Proof x 2 + (a + b) x + ab = x 2 + ax + bx + ab = x(x + a) + b(x + a) = (x + a) (x + b)

EXAMPLES Factorise 1. m 2 - 5m + 6

Solution a + b = - 5 and ab = + 6 -2 +6 ' -3 -5 Numbers with sum - 5 and product + 6 are - 2 and - 3. ` m 2 - 5m + 6 = [m + ] - 2 g] [m + ] - 3 g] = ]m - 2g]m - 3g

Guess and check by trying - 2 and - 3 or -1 and - 6.

2. y 2 + y - 2

Solution a + b = + 1 and ab = - 2 +2 -2 ' -1 +1 Two numbers with sum + 1 and product - 2 are + 2 and -1. ` y2 + y - 2 = ^ y + 2 h ^ y - 1 h

Guess and check by trying 2 and -1 or - 2 and 1.

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Maths In Focus Mathematics Extension 1 Preliminary Course

2.9 Exercises Factorise 1.

x 2 + 4x + 3

14. a 2 - 4a + 4

2.

y 2 + 7y + 12

15. x 2 + 14x - 32

3.

m 2 + 2m + 1

16. y 2 - 5y - 36

4.

t 2 + 8t + 16

17. n 2 - 10n + 24

5.

z2 + z - 6

18. x 2 - 10x + 25

6.

x 2 - 5x - 6

19. p 2 + 8p - 9

7.

v 2 - 8v + 15

20. k 2 - 7k + 10

8.

t 2 - 6t + 9

21. x 2 + x - 12

9.

x 2 + 9x - 10

22. m 2 - 6m - 7

10. y 2 - 10y + 21

23. q 2 + 12q + 20

11. m 2 - 9m + 18

24. d 2 - 4d - 5

12. y 2 + 9y - 36

25. l 2 - 11l + 18

13. x 2 - 5x - 24

The result x 2 + ] a + b g x + ab = ] x + a g ] x + b g only works when the coefficient of x 2 (the number in front of x 2) is 1. When the coefficient of x 2 is not 1, for example in the expression 5x 2 - 2x + 4, we need to use a different method to factorise the trinomial. There are different ways of factorising these trinomials. One method is the cross method. Another is called the PSF method. Or you can simply guess and check.

EXAMPLES Factorise 1. 5y 2 - 13y + 6

Solution—guess and check For 5y2, one bracket will have 5y and the other y: ^ 5y h ^ y h . Now look at the constant (term without y in it): + 6.

Chapter 2 Algebra and Surds

The two numbers inside the brackets must multiply to give + 6. To get a positive answer, they must both have the same signs. But there is a negative sign in front of 13y so the numbers cannot be both positive. They must both be negative. ^ 5y - h ^ y - h To get a product of 6, the numbers must be 2 and 3 or 1 and 6. Guess 2 and 3 and check: ^ 5y - 2 h ^ y - 3 h = 5y 2 - 15y - 2y + 6 = 5y 2 - 17y + 6 This is not correct. Notice that we are mainly interested in checking the middle two terms, -15y and - 2y. Try 2 and 3 the other way around: ^ 5y - 3 h ^ y - 2 h . Checking the middle terms: -10y - 3y = -13y This is correct, so the answer is ^ 5y - 3 h ^ y - 2 h . Note: If this did not check out, do the same with 1 and 6.

Solution—cross method Factors of 5y 2 are 5y and y. Factors of 6 are -1 and - 6 or - 2 and - 3. Possible combinations that give a middle term of -13y are 5y

-2

5y

-3

5y

-1

5y

-6

y

-3

y

-2

y

-6

y

-1

By guessing and checking, we choose the correct combination. -3 5y # - 2 = -10y 5y y

-2

y # - 3 = - 3y -13y

` 5y 2 - 13y + 6 = ^ 5y - 3 h ^ y - 2 h

Solution—PSF method P: Product of first and last terms S: Sum or middle term F: Factors of P that give S - 3y 30y 2 ) -10y -13y

30y 2 -13y - 3y, -10y

` 5y 2 - 13y + 6 = 5y 2 - 3y - 10y + 6 = y ^ 5y - 3 h - 2 ^ 5 y - 3 h = ^ 5y - 3 h ^ y - 2 h

CONTINUED

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2. 4y 2 + 4y - 3

Solution—guess and check For 4y2, both brackets will have 2y or one bracket will have 4y and the other y. Try 2y in each bracket: ^ 2y h ^ 2y h . Now look at the constant: - 3. The two numbers inside the brackets must multiply to give - 3. To get a negative answer, they must have different signs. ^ 2y - h ^ 2y + h To get a product of 3, the numbers must be 1 and 3. Guess and check: ^ 2y - 3 h ^ 2 y + 1 h Checking the middle terms: 2y - 6y = - 4y This is almost correct, as the sign is wrong but the coefficient is right (the number in front of y). Swap the signs around: ^ 2y - 1 h ^ 2 y + 3 h = 4y 2 + 6 y - 2 y - 3 = 4y 2 + 4y - 3 This is correct, so the answer is ^ 2y - 1 h ^ 2y + 3 h .

Solution—cross method Factors of 4y 2 are 4y and y or 2y and 2y. Factors of 3 are -1 and 3 or - 3 and 1. Trying combinations of these factors gives 3 2y 2y # - 1 = - 2 y 2y

-1

2y # 3 =

6y 4y

` 4y 2 + 4y - 3 = ^ 2 y + 3 h ^ 2 y - 1 h

Solution—PSF method P: Product of first and last terms -12y 2 S: Sum or middle term 4y F: Factors of P that give S + 6y, - 2y 2 + 6y -12y ) -2y + 4y ` 4y 2 + 4y - 3 = 4 y 2 + 6 y - 2 y - 3 = 2y ^ 2y + 3 h - 1 ^ 2 y + 3 h = ^ 2y + 3 h ^ 2y - 1 h

Chapter 2 Algebra and Surds

2.10

Exercises

Factorise 1.

2a 2 + 11a + 5

16. 4n 2 - 11n + 6

2.

5y 2 + 7y + 2

17. 8t 2 + 18t - 5

3.

3x 2 + 10x + 7

18. 12q 2 + 23q + 10

4.

3x 2 + 8x + 4

19. 8r 2 + 22r - 6

5.

2b 2 - 5b + 3

20. 4x 2 - 4x - 15

6.

7x 2 - 9x + 2

21. 6y 2 - 13y + 2

7.

3y 2 + 5y - 2

22. 6p 2 - 5p - 6

8.

2x 2 + 11x + 12

23. 8x 2 + 31x + 21

9.

5p 2 + 13p - 6

24. 12b 2 - 43b + 36

10. 6x 2 + 13x + 5

25. 6x 2 - 53x - 9

11. 2y 2 - 11y - 6

26. 9x 2 + 30x + 25

12. 10x 2 + 3x - 1

27. 16y 2 + 24y + 9

13. 8t 2 - 14t + 3

28. 25k 2 - 20k + 4

14. 6x 2 - x - 12

29. 36a 2 - 12a + 1

15. 6y 2 + 47y - 8

30. 49m 2 + 84m + 36

Perfect squares You have looked at some special binomial products, including ]a + bg2 = a 2 + 2ab + b 2 and ]a - bg2 = a 2 - 2ab + b 2 . When factorising, use these results the other way around.

a 2 + 2ab + b 2 = ] a + b g2 a 2 - 2ab + b 2 = ] a - b g2

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EXAMPLES In a perfect square, the constant term is always a square number.

Factorise 1. x 2 - 8x + 16

Solution x 2 - 8x + 16 = x 2 - 2 (4) x + 4 2 = ] x - 4 g2 2. 4a 2 + 20a + 25

Solution 4a 2 + 20a + 25 = ] 2a g2 + 2 (2a) (5) + 5 2 = ] 2a + 5 g2

2.11

Exercises

Factorise 1.

y 2 - 2y + 1

12. 16k 2 - 24k + 9

2.

x 2 + 6x + 9

13. 25x 2 + 10x + 1

3.

m 2 + 10m + 25

14. 81a 2 - 36a + 4

4.

t 2 - 4t + 4

15. 49m 2 + 84m + 36

5.

x 2 - 12x + 36

16. t 2 + t +

6.

4x 2 + 12x + 9

7.

16b 2 - 8b + 1

8.

9a 2 + 12a + 4

4x 4 + 3 9 6y 1 18. 9y 2 + + 5 25

9.

25x 2 - 40x + 16

19. x 2 + 2 +

10. 49y 2 + 14y + 1 11. 9y 2 - 30y + 25

1 4

17. x 2 -

1 x2

20. 25k 2 - 20 +

4 k2

Chapter 2 Algebra and Surds

Difference of 2 squares A special case of binomial products is ] a + b g ] a - b g = a 2 - b 2. a2 - b2 = ] a + b g ] a - b g

EXAMPLES Factorise 1. d 2 - 36

Solution d 2 - 36 = d 2 - 6 2 = ]d + 6 g]d - 6 g 2. 9b 2 - 1

Solution 9b 2 - 1 = ] 3b g2 - 1 2 = ( 3 b + 1) ( 3 b - 1 ) 3. (a + 3) 2 - (b - 1) 2

Solution ] a + 3 g2 - ] b - 1 g2 = [(a + 3) + (b - 1)] [(a + 3) - (b - 1)] = (a + 3 + b - 1) ( a + 3 - b + 1)

= ( a + b + 2 ) (a - b + 4 )

2.12

Exercises

Factorise 1.

a2 - 4

7.

1 - 4z 2

2.

x2 - 9

8.

25t 2 - 1

3.

y2 - 1

9.

9t 2 - 4

4.

x 2 - 25

10. 9 - 16x 2

5.

4x 2 - 49

11. x 2 - 4y 2

6.

16y 2 - 9

12. 36x 2 - y 2

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13. 4a 2 - 9b 2

20.

14. x 2 - 100y 2 15. 4a - 81b 2

21. ] x + 2 g2 - ^ 2y + 1 h2

2

22. x 4 - 1

16. ]x + 2g2 - y 2 17. ] a - 1 g - ] b - 2 g 2

2

18. z - ] 1 + w g 2

19. x 2 -

y2 -1 9

2

1 4

23. 9x 6 - 4y 2 24. x 4 - 16y 4 25. a 8 - 1

Sums and differences of 2 cubes

a 3 + b 3 = ] a + b g ^ a 2 - ab + b 2 h

Proof (a + b) (a 2 - ab + b 2) = a 3 - a 2 b + ab 2 + a 2 b - ab 2 + b 3 = a3 + b3 a 3 - b 3 = ] a - b g ^ a 2 + ab + b 2 h

Proof (a - b) (a 2 + ab + b 2) = a 3 + a 2 b + ab 2 - a 2 b - ab 2 - b 3 = a3 - b3

EXAMPLES Factorise 1. 8x 3 + 1

Solution 8x 3 + 1 = ] 2x g3 + 1 3 = (2x + 1) [] 2x g2 - (2x) (1) + 1 2] = (2x + 1 ) (4 x 2 - 2 x + 1 )

Chapter 2 Algebra and Surds

2. 27a 3 - 64b 3

Solution 27a 3 - 64b 3 = ] 3a g3 - ] 4b g3 = (3a - 4b) [] 3a g2 + (3a) (4b) + ] 4b g2] = (3a - 4b) (9a 2 + 12ab + 16b 2)

2.13

Exercises

Factorise 1.

b3 - 8

2.

x 3 + 27

3.

t3 + 1

4.

a 3 - 64

14. ] x + 1 g3 - y 3

5.

1 - x3

15. 125x 3 y 3 + 216z 3

6.

8 + 27y 3

16. ]a - 2g3 - ]a + 1g3

7.

y 3 + 8z 3

8.

x 3 - 125y 3

9.

8x 3 + 27y 3

10. a 3 b 3 - 1 11. 1000 + 8t 3

12.

x3 - 27 8

13.

1000 1 + 3 3 a b

17. 1 -

x3 27

18. y 3 + ]3 + xg3 19. ] x + 1 g3 + ^ y - 2 h3 20. 8]a + 3g3 - b 3

Mixed factors Sometimes more than one method of factorising is needed to completely factorise an expression.

EXAMPLE Factorise 5x 2 - 45.

Solution 5x 2 - 45 = 5 (x 2 - 9) = 5 (x + 3) (x - 3)

(using simple factors) (the difference of two squares)

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2.14

Exercises

Factorise 1.

2x 2 - 18

16. x 3 - 3x 2 - 10x

2.

3p 2 - 3p - 36

17. x 3 - 3x 2 - 9x + 27

3.

5y 3 - 5

18. 4x 2 y 3 - y

4.

4a 3 b + 8a 2 b 2 - 4ab 2 - 2a 2 b

19. 24 - 3b 3

5.

5a 2 - 10a + 5

20. 18x 2 + 33x - 30

6.

- 2x 2 + 11x - 12

21. 3x 2 - 6x + 3

7.

3z 3 + 27z 2 + 60z

22. x 3 + 2x 2 - 25x - 50

8.

9ab - 4a 3 b 3

23. z 3 + 6z 2 + 9z

9.

x3 - x

24. 4x 4 - 13x 2 + 9

10. 6x 2 + 8x - 8

25. 2x 5 + 2x 2 y 3 - 8x 3 - 8y 3

11. 3m - 15 - 5n + mn

26. 4a 3 - 36a

12. ] x - 3 g2 - ] x + 4 g2

27. 40x - 5x 4

13. y 2 ^ y + 5 h - 16 ^ y + 5 h

28. a 4 - 13a 2 + 36

14. x 4 - x 3 + 8x - 8

29. 4k 3 + 40k 2 + 100k

15. x 6 - 1

30. 3x 3 + 9x 2 - 3x - 9

DID YOU KNOW? Long division can be used to find factors of an expression. For example, x - 1 is a factor of x 3 + 4x - 5. We can find the other factor by dividing x 3 + 4x - 5 by x - 1. x2 + x + 5 x - 1 x3 + 4x - 5

g

x3

-

x2 x 2 + 4x x2

You will study this in Chapter 12.

-

x 5x - 5 5x - 5

0 So the other factor of x 3 + 4x - 5 is x 2 + x + 5 ` x 3 + 4x - 5 = (x - 1) (x 2 + x + 5)

Chapter 2 Algebra and Surds

69

Completing the Square Factorising a perfect square uses the results a 2 ! 2ab + b 2 = ] a ! b g2

EXAMPLES 1. Complete the square on x 2 + 6x.

Solution Using a 2 + 2ab + b 2: a=x 2ab = 6x Substituting a = x: 2xb = 6x b=3

Notice that 3 is half of 6.

To complete the square: a 2 + 2ab + b 2 = ] a + b g2 2 x + 2x ] 3 g + 3 2 = ] x + 3 g2 x 2 + 6x + 9 = ] x + 3 g2 2. Complete the square on n 2 - 10n.

Solution Using a 2 - 2ab + b 2: a=n 2ab = 10x Substituting a = n: 2nb = 10n b=5

Notice that 5 is half of 10.

To complete the square: a 2 - 2ab + b 2 = ] a - b g2 n 2 - 2n ] 5 g + 5 2 = ] n - 5 g2 n 2 - 10n + 25 = ] n - 5 g2

To complete the square on a 2 + pa, divide p by 2 and square it. p 2 p 2 a 2 + pa + d n = d a + n 2 2

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Maths In Focus Mathematics Extension 1 Preliminary Course

EXAMPLES 1. Complete the square on x 2 + 12x.

Solution Divide 12 by 2 and square it: x 2 + 12x + c

12 2 m = x 2 + 12x + 6 2 2 = x 2 + 12x + 36 = ]x + 6g2

2. Complete the square on y 2 - 2y.

Solution Divide 2 by 2 and square it: 2 2 y 2 - 2y + c m = y 2 - 2 y + 1 2 2 = y 2 - 2y + 1 = ^ y - 1 h2

2.15

Exercises

Complete the square on 1.

x 2 + 4x

12. y 2 + 3y

2.

b 2 - 6b

13. x 2 - 7x

3.

x 2 - 10x

14. a 2 + a

4.

y 2 + 8y

15. x 2 + 9x

5.

m 2 - 14m

16. y 2 -

6.

q 2 + 18q

5y 2

7.

x 2 + 2x

17. k 2 -

11k 2

8.

t 2 - 16t

18. x 2 + 6xy

9.

x 2 - 20x

19. a 2 - 4ab

10. w 2 + 44w 11. x 2 - 32x

20. p 2 - 8pq

Chapter 2 Algebra and Surds

71

Algebraic Fractions Simplifying fractions EXAMPLES Simplify 4x + 2 2

1.

Solution 2 ] 2x + 1 g 4x + 2 = 2 2 = 2x + 1

Factorise first, then cancel.

2x 2 - 3x - 2 x3 - 8

2.

Solution ] 2x + 1 g ] x - 2 g 2x 2 - 3x - 2 = 3 ] x - 2 g ^ x 2 + 2x + 4 h x -8 2x + 1 = 2 x + 2x + 4

2.16

Exercises

Simplify 1.

5a + 10 5

9.

b3 - 1 b2 - 1

2.

6t - 3 3

10.

3.

8y + 2 6

2p 2 + 7p - 15 6p - 9

11.

a2 - 1 a + 2a - 3

4. 5. 6.

8 4d - 2 2

x 5x 2 - 2x y-4

12. 13.

y - 8y + 16

2

3 ]x - 2g + y ]x - 2g x3 - 8 x 3 + 3x 2 - 9x - 27 x 2 + 6x + 9

2

7.

2ab - 4a 2 a 2 - 3a

8.

s2 + s - 2 s 2 + 5s + 6

14.

15.

2p 2 - 3p - 2 8p 3 + 1 ay - ax + by - bx 2ay - by - 2ax + bx

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Maths In Focus Mathematics Extension 1 Preliminary Course

Operations with algebraic fractions EXAMPLES Simplify 1.

x+3 x-1 5 4

Solution Do algebraic fractions the same way as ordinary fractions.

4 ]x - 1 g - 5 ]x + 3 g x -1 x +3 = 5 4 20 4x - 4 - 5x - 15 = 20 - x - 19 = 20

2.

2a 2 b + 10ab a 2 - 25 ' 3 4b + 12 b + 27

Solution 2a 2 b + 10ab a 2 - 25 2a 2 b + 10ab 4b + 12 ' = # 2 4b + 12 b 3 + 27 b 3 + 27 a - 25 2ab ] a + 5 g 4 ]b + 3 g = # 2 ] a + 5 g]a - 5 g ] b + 3 g ^ b - 3b + 9 h 8ab = ] a - 5 g ^ b 2 - 3b + 9 h

3.

2 1 + x-5 x+2

Solution 2 ]x + 2g + 1 ]x - 5g 2 1 + = x-5 x+2 ]x - 5g]x + 2g 2x + 4 + x - 5 = ]x - 5g]x + 2g 3x - 1 = ]x - 5g]x + 2g

Chapter 2 Algebra and Surds

2.17 1.

2.

Exercises

Simplify x 3x (a) + 4 2 y + 1 2y (b) + 5 3 a+2 a (c) 4 3 p-3 p+2 (d) + 6 2 x-5 x-1 (e) 2 3 4.

Simplify 3 b 2 + 2b # (a) b + 2 6a - 3

1 1 + x+1 x-3

(g)

3 2 x 2 + x -4

(h)

1 1 + a 2 + 2a + 1 a + 1

(i)

5 2 1 + y+2 y+3 y-1

(j)

2 7 x 2 - 16 x 2 - x - 12

2

Simplify (a)

y2 - 9 3x 2 x 2 - 2x - 8 # # 4y - 12 6x - 24 y 3 + 27

q3 + 1 (b) 2 # q + 2q + 1 p + 2

(b)

2 a 2 - 5a 3a - 15 y - y - 2 ' # 5ay y 2 - 4y + 4 y2 - 4

3ab 2 12ab - 6a (c) ' 2 5xy x y + 2xy 2

(c)

3 x 2 + 3x 2x + 8 + 2 # x-3 4x - 16 x -9

(d)

5b b2 b ' 2 2b + 6 b 1 + b +b-6

(e)

x 2 - 8x + 15 x 2 - 9 x 2 + 5x + 6 ' # 2 2x - 10 5x + 10x 10x 2

p2 - 4

(d)

ax - ay + bx - by x2 - y2

#

x3 + y3 ab 2 + a 2 b

x 2 - 6x + 9 x 2 - 5x + 6 (e) ' x 2 - 25 x 2 + 4x - 5 3.

(f)

5.

Simplify 2 3 (a) x + x

Simplify (a)

1 2 4 + x 2 - 7x + 10 x 2 - 2x - 15 x 2 + x - 6

1 2 x-1 x

(b)

3 5 2 + 2 2 x x x -4

(c) 1 +

3 a+b

(c)

3 2 + p 2 + pq pq - q 2

(d) x -

x2 x+2

(d)

a b 1 + a + b a - b a2 - b2

(b)

(e) p - q +

1 p+q

2

x+y y x (e) x - y + y - x - 2 y - x2

Substitution Algebra is used in writing general formulae or rules. For example, the formula A = lb is used to find the area of a rectangle with length l and breadth b. We can substitute any values for l and b to find the area of different rectangles.

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EXAMPLES 1. P = 2l + 2b is the formula for finding the perimeter of a rectangle with length l and breadth b. Find P when l = 1.3 and b = 3.2.

Solution P = 2 l + 2b = 2 ] 1 . 3 g + 2 ] 3 .2 g = 2 .6 + 6 . 4 =9 2. V = rr 2 h is the formula for finding the volume of a cylinder with radius r and height h. Find V (correct to 1 decimal place) when r = 2.1 and h = 8.7.

Solution V = rr 2 h = r ] 2.1 g2 (8.7) = 120.5 correct to 1 decimal place

9C + 32 is the formula for changing degrees Celsius ] °C g into 5 degrees Fahrenheit ] °F g find F when C = 25. 3. If F =

Solution 9C + 32 5 9 ] 25 g = + 32 5 225 = + 32 5 225 + 160 = 5 385 = 5 = 77 This means that 25°C is the same as 77°F. F=

Chapter 2 Algebra and Surds

2.18 1.

Exercises

Given a = 3.1 and b = - 2.3 find, correct to 1 decimal place. (a) ab (b) 3b (c) 5a 2 (d) ab 3 (e) ]a + bg2 (f)

a-b

(g) - b 2 2.

T = a + ] n - 1 g d is the formula for finding the term of an arithmetic series. Find T when a = - 4, n = 18 and d = 3.

3.

Given y = mx + b, the equation of a straight line, find y if m = 3, x = - 2 and b = - 1.

4.

If h = 100t - 5t 2 is the height of a particle at time t, find h when t = 5.

5.

Given vertical velocity v = - gt, find v when g = 9.8 and t = 20.

6.

If y = 2 x + 3 is the equation of a function, find y when x = 1.3, correct to 1 decimal place.

7.

S = 2r r ] r + h g is the formula for the surface area of a cylinder. Find S when r = 5 and h = 7, correct to the nearest whole number.

8.

A = rr 2 is the area of a circle with radius r. Find A when r = 9.5, correct to 3 significant figures.

9.

n-1

Given u n = ar is the nth term of a geometric series, find u n if a = 5, r = - 2 and n = 4.

10. Given V = 1 lbh is the volume 3 formula for a rectangular pyramid, find V if l = 4.7, b = 5.1 and h = 6.5. 11. The gradient of a straight line is y2 - y1 given by m = x - x . Find m 2 1 if x 1 = 3, x 2 = -1, y 1 = - 2 and y 2 = 5. 12. If A = 1 h ] a + b g gives the area 2 of a trapezium, find A when h = 7, a = 2.5 and b = 3.9. 13. Find V if V = 4 rr 3 is the volume 3 formula for a sphere with radius r and r = 7.6, to 1 decimal place.

14. The velocity of an object at a certain time t is given by the formula v = u + at. Find v when u = 1 , a = 3 and t = 5 . 4 5 6 a 15. Given S = , find S if a = 5 1-r and r = 2 . S is the sum to infinity 3 of a geometric series. 16. c = a 2 + b 2 , according to Pythagoras’ theorem. Find the value of c if a = 6 and b = 8. 17. Given y = 16 - x 2 is the equation of a semicircle, find the exact value of y when x = 2.

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18. Find the value of E in the energy equation E = mc 2 if m = 8.3 and c = 1.7. 19. A = P c 1 +

r n m is the formula 100

for finding compound interest. Find A when P = 200, r = 12 and n = 5, correct to 2 decimal places.

20. If S =

a ^rn - 1h is the sum of r -1

a geometric series, find S if a = 3, r = 2 and n = 5. 21. Find the value of

a3 b2 if c2

2 3 1 4 a = c 3 m , b = c 2 m and c = c m . 4 3 2

Surds An irrational number is a number that cannot be written as a ratio or fraction (rational). Surds are special types of irrational numbers, such as 2, 3 and 5 . Some surds give rational values: for example, 9 = 3. Others, like 2 , do not have an exact decimal value. If a question involving surds asks for an exact answer, then leave it as a surd rather than giving a decimal approximation.

Simplifying surds

Class Investigations 1. Is there an exact decimal equivalent for 2 ? 2. Can you draw a line of length exactly 2 ? 3. Do these calculations give the same results? (a) 9 # 4 and 9 # 4 (b)

4

and

4 9

(c)

9 9 + 4 and

9 +

4

(d)

9 - 4 and

9 -

4

Here are some basic properties of surds.

a# b =

ab

a' b =

a

^ x h2 =

b

=

x2 = x

a b

Chapter 2 Algebra and Surds

77

EXAMPLES 1. Express in simplest surd form

45 .

45 also equals 3 # 15 but this will not simplify. We look for a number that is a perfect square.

Solution 45 = 9 # 5 = 9 # 5 =3# 5 =3 5 2. Simplify 3 40 .

Solution

Find a factor of 40 that is a perfect square.

3 40 = 3 4 # 10 = 3 # 4 # 10 = 3 # 2 # 10 = 6 10 3. Write 5 2 as a single surd.

Solution 5 2 = =

2.19 1.

25 # 2 50

Exercises

Express these surds in simplest surd form.

(k)

112

(l)

300

(a)

12

(b)

63

(c)

24

(d)

50

(e)

72

(f)

200

(g)

48

(h)

75

(i)

32

(a) 2 27

(j)

54

(b) 5 80

(m) 128

2.

(n)

243

(o)

245

(p)

108

(q)

99

(r)

125

Simplify

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Maths In Focus Mathematics Extension 1 Preliminary Course

(c) 4 98

(g) 3 13

(d) 2 28

(h) 7 2

(e) 8 20

(i) 11 3

(f) 4 56

(j) 12 7

(g) 8 405

4.

(h) 15 8

(a)

(i) 7 40

x =3 5

(b) 2 3 =

x

(c) 3 7 =

x

Write as a single surd.

(d) 5 2 =

x

(a) 3 2

(e) 2 11 =

(b) 2 5

(f)

(c) 4 11

(g) 4 19 =

(d) 8 2

(h)

(e) 5 3

(i) 5 31 =

(f) 4 10

(j)

(j) 8 45 3.

Evaluate x if

x

x =7 3 x

x = 6 23 x

x = 8 15

Addition and subtraction Calculations with surds are similar to calculations in algebra. We can only add or subtract ‘like terms’ with algebraic expressions. This is the same with surds.

EXAMPLES 1. Simplify 3 2 + 4 2 .

Solution 3 2+4 2 =7 2 2. Simplify

3 - 12 .

Solution First, change into ‘like’ surds. 3 - 12 = 3 - 4 # 3 = 3 -2 3 =- 3 3. Simplify 2 2 - 2 + 3 .

Solution 2 2- 2+ 3=

2+ 3

Chapter 2 Algebra and Surds

2.20

79

Exercises

Simplify 1.

5 +2 5

14.

50 -

32

2.

3 2 -2 2

15.

28 +

63

3.

3 +5 3

16. 2 8 -

18

4.

7 3 -4 3

5.

5 -4 5

6.

4 6 -

17. 3 54 + 2 24 18.

90 - 5 40 - 2 10

19. 4 48 + 3 147 + 5 12

6

7.

2 -8 2

20. 3 2 + 8 - 12

8.

5 +4 5 +3 5

21.

63 - 28 - 50

9.

2 -2 2 -3 2

22.

12 - 45 - 48 - 5

10.

5 +

45

23.

150 + 45 + 24

11.

8 -

2

24.

32 - 243 - 50 + 147

12.

3 +

48

25.

80 - 3 245 + 2 50

13.

12 -

27

Multiplication and division To get a b # c d = ac bd , multiply surds with surds and rationals with rationals.

a # b = ab a b # c d = ac bd a# a =

a b

=

a2 = a

a b

EXAMPLES Simplify 1. 2 2 #- 5 7

Solution 2 2 #- 5 7 = -10 14

CONTINUED

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Maths In Focus Mathematics Extension 1 Preliminary Course

2. 4 2 # 5 18

Solution 4 2 # 5 18 = 20 36 = 20 # 6 = 120

3.

2 14 4 2

Solution 2 14 4 2

=

2 2 # 7 2

=

4.

7

4 2

3 10 15 2

Solution 3 10 15 2

=

3# 5 # 2 15 2

5 = 5

5. d

2

10 n 3

Solution 2 ^ 10 h 10 n = 3 ^ 3 h2 10 = 3 =31 3

2

d

Chapter 2 Algebra and Surds

2.21

Exercises

Simplify 1.

7 #

2.

3# 5

3.

2 #3 3

4.

5 7 #2 2

5.

-3 3 #2 2

6.

5 3 #2 3

7.

- 4 5 # 3 11

8.

2 7# 7

9.

2 3 # 5 12

3

10.

6# 2

11.

8 #2 6

23.

5 8 10 2

24.

16 2

25.

10 30

26.

2 2

27. 28.

2 12

5 10

6 20 4 2 8 10 3 3 15

29.

2

30.

3 15

8

12. 3 2 # 5 14 13.

10 # 2 2

14. 2 6 #-7 6 15. ^ 2 h

2

2 16. ^ 2 7 h

17.

31.

5 12

32.

15 18

33.

15

5 8

10 10

3# 5# 2

18. 2 3 # 7 #- 5 19.

6 10

2 # 6 #3 3

2 6 2n 3

35. d

5n 7

20. 2 5 # - 3 2 # - 5 5 21.

4 12

22.

12 18

2 2

2

34. d

2

3 6

Expanding brackets The same rules for expanding brackets and binomial products that you use in algebra also apply to surds.

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Simplifying surds by removing grouping symbols uses these general rules.

a^ b + ch=

ab + ac

Proof a^ b + ch = =

a# b + ab + ac

a# c

Binomial product:

^ a + b h^ c + d h =

ac +

ad +

bc +

bd

Proof ^ a + b h^ c + d h = a # c + a # d + b # c + b # d = ac + ad + bc + bd Perfect squares:

^ a + b h2 = a + 2 ab + b

Proof ^ a + b h2 = ^ a + b h ^ a + b h = a 2 + ab + ab + b 2 = a + 2 ab + b ^ a - b h2 = a - 2 ab + b

Proof ^ a - b h2 = ^ a - b h ^ a - b h = a 2 - ab - ab + b 2 = a - 2 ab + b Difference of two squares:

^ a + b h^ a - b h = a - b

Proof ^ a + b h ^ a - b h = a 2 - ab + ab - b 2 =a-b

Chapter 2 Algebra and Surds

83

EXAMPLES Expand and simplify 1. 2 ^ 5 + 2 h

Solution 2( 5 +

2) = = =

2# 5 + 10 + 4 10 + 2

2# 2

2. 3 7 ^ 2 3 - 3 2 h

Solution 3 7 (2 3 - 3 2 ) = 3 7 # 2 3 - 3 7 # 3 2 = 6 21 - 9 14 3. ^ 2 + 3 5 h ^ 3 -

2h

Solution ( 2 + 3 5)( 3 -

2) = =

2# 3 - 2# 2 +3 5# 3 -3 5# 2 6 - 2 + 3 15 - 3 10

4. ^ 5 + 2 3 h ^ 5 - 2 3 h

Solution ( 5 + 2 3 ) ( 5 - 2 3 ) = 5 # 5 - 5 #2 3 + 2 3 # 5 - 2 3 #2 3 = 5 - 2 15 + 2 15 - 4#3 = 5 - 12 = -7 Another way to do this question is by using the difference of two squares. 2 2 ( 5 + 2 3)( 5 - 2 3) = ^ 5 h - ^2 3 h = 5 - 4#3 = -7

Notice that using the difference of two squares gives a rational answer.

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Maths In Focus Mathematics Extension 1 Preliminary Course

2.22 1.

Exercises (m)^ 2 11 + 5 2 h^ 2 11 - 5 2 h

Expand and simplify (a)

2^ 5 + 3h

(b)

3 ^2 2 - 5 h

(n) ^ 5 + 2 h

2

2 (o) ^ 2 2 - 3 h

(c) 4 3 ^ 3 + 2 5 h (d)

2 (p) ^ 3 2 + 7 h

7 ^5 2 - 2 3 h

2 (q) ^ 2 3 + 3 5 h

(e) - 3 ^ 2 - 4 6 h (f)

2 (r) ^ 7 - 2 5 h

3 ^ 5 11 + 3 7 h

2 (s) ^ 2 8 - 3 5 h

(g) - 3 2 ^ 2 + 4 3 h (h)

5^ 5 - 5 3h

(i)

3 ^ 12 + 10 h

2 (t) ^ 3 5 + 2 2 h

3.

If a = 3 2 , simplify (a) a2 (b) 2a3 (c) (2a)3 (d) ]a + 1g2 (e) ] a + 3 g ] a – 3 g

4.

Evaluate a and b if 2 (a) ^ 2 5 + 1h = a + b

(j) 2 3 ^ 18 + 3 h (k) - 4 2 ^ 2 - 3 6 h (l) - 7 5 ^ - 3 20 + 2 3 h (m) 10 3 ^ 2 - 2 12 h (n) - 2 ^ 5 + 2 h (o) 2 3 ^ 2 - 12 h 2.

(b) ^ 2 2 - 5 h ^ 2 - 3 5 h = a + b 10

Expand and simplify (a) ^ 2 + 3h^ 5 + 3 3 h (b) ^ 5 - 2 h^ 2 - 7 h

5.

Expand and simplify (a) ^ a + 3 - 2 h ^ a + 3 + 2 h 2 (b) _ p - 1 - p i

6.

Evaluate k if ^ 2 7 - 3 h ^ 2 7 + 3 h = k.

7.

Simplify _ 2 x + y i _ x - 3 y i .

8.

If ^ 2 3 - 5 h = a - b , evaluate a and b.

9.

Evaluate a and b if ^ 7 2 - 3 h2 = a + b 2 .

(c) ^ 2 + 5 3 h^ 2 5 - 3 2 h (d) ^ 3 10 - 2 5 h^ 4 2 + 6 6 h (e) ^ 2 5 - 7 2 h^ 5 - 3 2 h (f) ^ 5 + 6 2 h^ 3 5 - 3 h (g) ^ 7 + 3 h^ 7 - 3 h (h) ^ 2 - 3 h^ 2 + 3 h (i) ^ 6 + 3 2 h^ 6 - 3 2 h (j) ^ 3 5 + 2 h^ 3 5 - 2 h (k) ^ 8 - 5 h^ 8 + 5 h (l) ^ 2 + 9 3 h^ 2 - 9 3 h

2

10. A rectangle has sides 5 + 1 and 2 5 - 1. Find its exact area.

Rationalising the denominator Rationalising the denominator of a fractional surd means writing it with a rational number (not a surd) in the denominator. For example, after 3 5 3 rationalising the denominator, becomes . 5 5

Chapter 2 Algebra and Surds

85

DID YOU KNOW? A major reason for rationalising the denominator used to be to make it easier to evaluate the fraction (before calculators were available). It is easier to divide by a rational number than an irrational one; for example, 3 = 3 ' 2.236 5 3

5 5

This is hard to do without a calculator.

This is easier to calculate.

= 3 # 2.236 ' 5

Squaring a surd in the denominator will rationalise it since ^ x h = x. 2

Multiplying by

b a b a # = b b b

b

b is the same as multiplying by 1.

Proof b a b a # = b b b2 a b = b

EXAMPLES 1. Rationalise the denominator of

Solution

3 . 5

5 3 5 3 # = 5 5 5 2. Rationalise the denominator of

Solution

2 5 3

. Don’t multiply by 5

2 5 3

#

3 3

=

2 3

5 9 2 3 = 5# 3 2 3 = 15

3

as it takes 5 3 longer to simplify.

86

Maths In Focus Mathematics Extension 1 Preliminary Course

When there is a binomial denominator, we use the difference of two squares to rationalise it, as the result is always a rational number.

To rationalise the denominator of

a+ b c+ d

, multiply by

Proof a+ b c+ d

^ a + b h^ c - d h c- d ^ c + d h^ c - d h ^ a + b h^ c - d h = ^ c h2 - ^ d h2 ^ a + b h^ c - d h = c-d c- d

#

=

EXAMPLES 1. Write with a rational denominator 5 2 -3 Multiply by the conjugate surd 2 + 3.

.

Solution 5 2 -3

2 +3

#

2 +3

=

5 ^ 2 + 3h

^ 2 h2 - 3 2 10 + 3 5 = 2-9 10 + 3 5 = -7 10 + 3 5 =7

2. Write with a rational denominator 2 3+ 5 3+4 2

.

Solution 2 3 +

5

3 +4 2

#

3 -4 2 3 -4 2

=

^2 3 + 5 h^ 3 - 4 2 h

^ 3 h2 - ^ 4 2 h2 2 # 3 - 8 6 + 15 - 4 10 = 3 - 16 # 2

c- d c- d

Chapter 2 Algebra and Surds

6 - 8 6 + 15 - 4 10 - 29 - 6 + 8 6 - 15 + 4 10 = 29 =

3. Evaluate a and b if

3 3 3- 2

= a + b.

Solution 3 3 3- 2

#

3+ 2 3+ 2

=

3 3^ 3 + 2h

^ 3 - 2 h^ 3 + 2 h 3 9+3 6 = ^ 3 h2 - ^ 2 h2 3#3+3 6 3-2 9+3 6 = 1 =9+3 6 =

=9+ 9# 6 = 9 + 54 So a = 9 and b = 54. 4. Evaluate as a fraction with rational denominator 2 + 3+2

5 3-2

.

Solution 2 + 3+2

5 3 -2

=

2^ 3 - 2h + 5 ^ 3 + 2h

^ 3 + 2h ^ 3 - 2h 2 3 - 4 + 15 + 2 5 = ^ 3 h2 - 2 2 2 3 - 4 + 15 + 2 5 3-4 2 3 - 4 + 15 + 2 5 = -1 = - 2 3 + 4 - 15 - 2 5 =

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Maths In Focus Mathematics Extension 1 Preliminary Course

2.23 1.

Express with rational denominator (a) (b) (c) (d) (e)

2.

Exercises 3.

1 7

(a)

3

(b)

2 2 2 3

(c)

5 6 7

(d)

5 2 1+

2 3

6 -5

(g)

5 +2 2

8+3 2

(j)

4 3 -2 2

(f)

1 5 +

2

2 -

7

2 +

3

2 +3

4 5 (j)

7 5 (k)

4 3 +

(l) 2 4.

3 2 -7 5 +2 6 3 -4 3 +4 3 3 3 +

(b) (c)

2 +5 2 2

2 5 +3 2

3 2 +

+

3 3 2 -

#

3

6 -

3

2 3 2 +3 5 6 +2 2 +7 4+

2 3 +

3 -2

3

6 +

1 3

+

2

-

2

2

3 -

2

(d) (e)

2 5 3 4 2

2

2 -1

+

+

5 -

3

5

2

3

3 5 3 2 4-

3

2+

3

3 +1

Find a and b if (a)

2 3

-

1 where z = 1 + z2

(h) (i)

1 2 -1

1 where t = t

3 2 +4

2 7

Express with rational denominator

(e)

3

(g)

5

(i)

(d)

2 -

2

3 2 -4

(c)

2

(f) z 2 -

(h)

(b)

1 + 2 +1

(e) t +

(f)

(a)

Express as a single fraction with rational denominator

=

a b

=

a 6 b

2 =a+b 5 5 +1 2 7 7 -4 2 +3 2 -1

=a+b 7 =a+

b

2 -

2 6 -1

Chapter 2 Algebra and Surds

5.

2 -1

Show that

2 +1

+

4 is 2

7.

If x =

(b) x 2 +

8.

1 x2

2

+

1 5 -

2

-

as a single fraction with 3 rational denominator.

3 + 2, simplify

1 (a) x + x

2 5 +

5 +1

rational. 6.

Write

Show that

8 2 + is 3+2 2 2

rational. 2

1 (c) b x + x l

9.

1 If 2 + x = 3 , where x ! 0, find x as a surd with rational denominator.

10. Rationalise the denominator of b +2 ]b ! 4 g b -2

89

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Maths In Focus Mathematics Extension 1 Preliminary Course

Test Yourself 2 1.

2.

3.

4.

Simplify (a) 5y - 7y 3a + 12 (b) 3 (c) - 2k 3 # 3k 2 y x (d) + 5 3 (e) 4a - 3b - a - 5b (f) 8 + 32 (g) 3 5 - 20 + 45 Factorise (a) x 2 - 36 (b) a 2 + 2a - 3 (c) 4ab 2 - 8ab (d) 5y - 15 + xy - 3x (e) 4n - 2p + 6 (f) 8 - x 3 Expand and simplify (a) b + 3 ] b - 2 g (b) ] 2x - 1 g ] x + 3 g (c) 5 ] m + 3 g - ] m - 2 g (d) ]4x - 3g2 (e) ^ p - 5h^ p + 5h (f) 7 - 2 ] a + 4 g - 5a (g) 3 ^ 2 2 - 5 h (h) ^ 3 + 7 h^ 3 - 2h Simplify 4a - 12 10b (a) # 3 5b 3 a - 27 (b)

5.

5m + 10 m2 - 4 ' 2 m - m - 2 3m + 3

The volume of a cube is V = s 3. Evaluate V when s = 5.4.

6.

(a) Expand and simplify ^ 2 5 + 3 h ^ 2 5 - 3 h. (b) Rationalise the denominator of 3 3 . 2 5+ 3

7.

Simplify

8.

If a = 4, b = - 3 and c = - 2, find the value of (a) ab 2 (b) a - bc (c) a (d) ]bcg3 (e) c ] 2a + 3b g

9.

Simplify 3 12 (a) 6 15 (b)

3 1 2 + - 2 . x-2 x+3 x +x-6

4 32 2 2

10. The formula for the distance an object falls is given by d = 5t 2 . Find d when t = 1.5. 11. Rationalise the denominator of 2 (a) 5 3 (b)

1+ 3 2

12. Expand and simplify (a) ^ 3 2 - 4h^ 3 - 2 h 2 (b) ^ 7 + 2h 13. Factorise fully (a) 3x 2 - 27 (b) 6x 2 - 12x - 18 (c) 5y 3 + 40

Chapter 2 Algebra and Surds

14. Simplify 3x 4 y (a) 9xy 5 (b)

5 15x - 5

15. Simplify 2 (a) ^ 3 11 h 3 (b) ^ 2 3 h 16. Expand and simplify (a) ] a + b g ] a - b g (b) ] a + b g 2 (c) ] a - b g 2 17. Factorise (a) a 2 - 2ab + b 2 (b) a 3 - b 3 1 18. If x = 3 + 1, simplify x + x and give your answer with a rational denominator. 19. Simplify 4 3 (a) a + b (b)

x-3 x-2 5 2

20. Simplify

2 3 , writing 5+2 2 2-1

your answer with a rational denominator. 21. Simplify (a) 3 8 (b) - 2 2 # 4 3 (c) 108 - 48 (d)

23. Rationalise the denominator of 3 (a) 7 (b)

2

5 3 2 (c) 5 -1 (d) (e)

2 2 3 2+ 3 5+ 2 4 5-3 3

24. Simplify 3x x-2 (a) 5 2 a+2 2a - 3 (b) + 7 3 1 2 (c) 2 1 x + x -1 4 1 (d) 2 + k + 2k - 3 k + 3 (e)

3 2+ 5

-

5 3- 2

25. Evaluate n if (a) 108 - 12 = (b)

112 + 7 =

n n

8 6

(c) 2 8 + 200 =

2 18

(d) 4 147 + 3 75 = n 180 (e) 2 245 + = n 2

(e) 5a # - 3b # - 2a (f)

22. Expand and simplify (a) 2 2 ^ 3 + 2 h (b) ^ 5 7 - 3 5 h^ 2 2 - 3 h (c) ^ 3 + 2 h^ 3 - 2 h (d) ^ 4 3 - 5 h^ 4 3 + 5 h 2 (e) ^ 3 7 - 2 h

2m 3 n 6m 2 n 5

(g) 3x - 2y - x - y

n

91

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Maths In Focus Mathematics Extension 1 Preliminary Course

26. Evaluate x 2 +

1+2 3 1 if x = 2 x 1-2 3

27. Rationalise the denominator of

3

2 7 (there may be more than one answer). 21 (a) 28 2 21 (b) 28 21 (c) 14 21 (d) 7 x-3 x +1 . 5 4 -]x + 7 g 20 x+7 20 x + 17 20 - ] x + 17 g 20

28. Simplify (a) (b) (c) (d)

(a) (b) (c) (d)

32. Simplify 5ab - 2a 2 - 7ab - 3a 2 . (a) 2ab + a 2 (b) - 2ab - 5a 2 (c) - 13a 3 b (d) - 2ab + 5a 2 33. Simplify (a) (b) (c)

29. Factorise x 3 - 4x 2 - x + 4 (there may be more than one answer). (a) ^ x 2 - 1 h ] x - 4 g (b) ^ x 2 + 1 h ] x - 4 g (c) x 2 ] x - 4 g (d) ] x - 4 g ] x + 1 g ] x - 1 g 30. Simplify 3 2 + 2 98 . (a) 5 2 (b) 5 10 (c) 17 2 (d) 10 2

3 2 1 + . x-2 x+2 x2 - 4 x+5 ]x + 2g]x - 2g x+1 ]x + 2g]x - 2g x+9 ]x + 2g]x - 2g x-3 ]x + 2g]x - 2g

31. Simplify

(d)

80 . 27

4 5 3 3 4 5 9 3 8 5 9 3 8 5 3 3

34. Expand and simplify ^ 3x - 2y h2 . (a) 3x 2 - 12xy - 2y 2 (b) 9x 2 - 12xy - 4y 2 (c) 3x 2 - 6xy + 2y 2 (d) 9x 2 - 12xy + 4y 2 35. Complete the square on a 2 - 16a. (a) a 2 - 16a + 16 = ^ a - 4 h2 (b) a 2 - 16a + 64 = ^ a - 8 h2 (c) a 2 - 16a + 8 = ^ a - 4 h2 (d) a 2 - 16a + 4 = ^ a - 2 h2

Chapter 2 Algebra and Surds

Challenge Exercise 2 1.

2.

Expand and simplify (a) 4ab ] a - 2b g - 2a 2 ] b - 3a g (b) _ y 2 - 2 i_ y 2 + 2 i (c) ] 2x - 5 g3 Find the value of x + y with rational denominator if x = 3 + 1 and 1 y= . 2 5-3 2 3

11. Simplify

2x + y x-y 3x + 2y . + - 2 x-3 x+3 x +x-6

12. (a) Expand ^ 2x - 1 h3. 6x 2 + 5x - 4 (b) Simplify . 8x 3 - 12x 2 + 6x - 1 13. Expand and simplify ] x - 1 g ^ x - 3 h2. 14. Simplify and express with rational 2 +

5

5 3

3.

Simplify

4.

b Complete the square on x 2 + a x.

15. Complete the square on x 2 + 2 x. 3

5.

Factorise (a) (x + 4)2 + 5 (x + 4) (b) x 4 - x 2 y - 6y 2 (c) 125x 3 + 343 (d) a 2 b - 2a 2 - 4b + 8

16. If x =

6. 7.

8.

9.

7 6 - 54

.

denominator

Simplify

d=

4x 2 - 16x + 12

| ax 1 + by 1 + c |

.

Simplify

10. Factorise

^a + 1h a3 + 1

.

a2 4 - 2. 2 x b

.

lx 1 + kx 2

17. Find the exact value with rational 1 denominator of 2x 2 - 3x + x if x = 2 5 . 18. Find the exact value of 1+2 3 1 (a) x 2 + 2 if x = x 1-2 3 (b) a and b if

is the formula for

a2 + b2 the perpendicular distance from a point to a line. Find the exact value of d with a rational denominator if a = 2, b = -1, c = 3, x 1 = - 4 and y 1 = 5. 3

2 -1

, find the value of x when k+l k = 3, l = - 2, x 1 = 5 and x 2 = 4.

Complete the square on 4x 2 + 12x. 2xy + 2x - 6 - 6y

3 +4

-

3 -4 2+3 3

=a+b 3

19. A = 1 r 2 i is the area of a sector of a 2 circle. Find the value of i when A = 12 and r = 4. 20. If V = rr 2 h is the volume of a cylinder, find the exact value of r when V = 9 and h = 16. 21. If s = u + 1 at 2, find the exact value of s 2 when u = 2, a = 3 and t = 2 3 .

93

3 Equations TERMINOLOGY Absolute value: the distance of a number from zero on a number line.

pronumeral that is solved to find values that make the statement true e.g. 2x - 3 2 4

Equation: A mathematical statement that has a pronumeral or unknown number and an equal sign. An equation can be solved to find the value of the unknown number e.g. 2x - 3 = 5

Quadratic equation: An equation involving x 2 as the highest power of x that may have two, one or no solutions

Exponential equation: Equation where the unknown pronumeral is the power or index e.g. 2 x = 8 Inequation: A mathematical statement involving an inequality sign, 1, 2, # or $ that has an unknown

Simultaneous equations: Two or more independent equations that can be solved together to produce a solution that makes each equation true at the same time. The number of equations required is the same as the number of unknowns

Chapter 3 Equations

95

INTRODUCTION EQUATIONS ARE FOUND IN most branches of mathematics. They are also

important in many other fields, such as science, economics, statistics and engineering. In this chapter you will revise basic equations and inequations. Equations involving absolute values, exponential equations, quadratic equations and simultaneous equations are also covered here.

DID YOU KNOW? Algebra was known in ancient civilisations. Many equations were known in Babylonia, although general solutions were difficult because symbols were not used in those times. Diophantus, around 250 AD, first used algebraic notation and symbols (e.g. the minus sign). He wrote a treatise on algebra in his Arithmetica, comprising 13 books. Only six of these books survived. About 400 AD, Hypatia of Alexandria wrote a commentary on them. Hypatia was the daughter of Theon, a mathematician who ensured that she had the best education. She was the first female mathematician on record, and was a philosopher and teacher. She was murdered for her philosophical views by a fanatical Christian sect. In 1799 Carl Friedrich Gauss proved the Fundamental Theorem of Algebra: that every algebraic equation has a solution.

PROBLEM The age of Diophantus at his death can be calculated from this epitaph: Diophantus passed one-sixth of his life in childhood, one-twelfth in youth, and one-seventh more as a bachelor; five years after his marriage a son was born who died four years before his father at half his father’s final age. How old was Diophantus?

Simple Equations Here are the four rules for changing numbers or pronumerals from one side of an equation to the other.

• • • •

ch3.indd 95

If a number is added, subtract it from both sides If a number is subtracted, add it to both sides If a number is multiplied, divide both sides by the number If a number is divided, multiply both sides by the number

Do the opposite operation to take a number to the other side of an equation.

8/11/09 10:59:40 AM

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EXAMPLES Solve 1. 3x + 5 = 17

Solution 3x + 5 = 17 3x + 5 - 5 = 17 - 5 3x = 12 3x 12 = 3 3 x=4 You can check the solution by substituting the value into the equation. LHS = 3x + 5 = 3 ( 4) + 5 = 12 + 5 = 17 = RHS Since LHS = RHS, x = 4 is the correct solution. 2. 4y - 3 = 8y + 21

Solution 4y - 3 4 y - 4y - 3 -3 - 3 - 21 - 24

`

= 8y + 21 = 8y - 4y + 21 = 4y + 21 = 4y + 21 - 21 = 4y 4y - 24 = 4 4 -6 = y y = -6

3. 2 ] 3x + 7 g = 6 - ] x - 1 g Check these solutions by substituting them into the equation.

Solution 2 (3 x + 7 ) = 6 - ( x - 1 ) 6x + 14 = 6 - x + 1 =7-x 6x + x + 14 = 7 - x + x 7x + 14 = 7

Chapter 3 Equations

7x + 14 - 14 7x 7x 7 x

= 7 - 14 = -7 -7 7 = -1 =

3.1 Exercises Solve 1.

t + 4 = -1

2.

z + 1.7 = -3.9

3.

y - 3 = -2

4.

w - 2 .6 = 4 .1

18. 3x + 5 = 17

5.

5 = x -7

19. 4a + 7 = - 21

6.

1.5x = 6

20. 7y - 1 = 20

7.

5y = 1 3

8.

b =5 7

9.

-2 =

10.

r 2 = 6 3

16.

x -3 =7 2

17.

m + 7 = 11 5

21. 8b - 4 = - 36 22. 3 (x + 2) = 15 23. -2 (3a + 1) = 8

n 8

11. 2y + 1 = 19 12. 33 = 4k + 9 13. 7d - 2 = 12 14. -2 = 5x - 27 y 15. +4=9 3

24. 7t + 4 = 3t - 12 25. x - 3 = 6x - 9 26. 2 (a - 2) = 4 - 3a 27. 5b + 2 = - 3(b - 1) 28. 3 (t + 7) = 2 (2t - 9) 29. 2 + 5( p - 1) = 5p - ( p - 2) 30. 3.7x + 1.2 = 5.4x - 6.3

A S TA R T L I N G FA C T ! Half full = half empty ` full = empty

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Equations involving fractions There are different ways to solve this type of equation. One way is to multiply both sides of the equation by the common denominator of the fractions.

EXAMPLES Solve m 1 1. -4= 3 2

Solution Multiply by the common denominator, 6.

m 1 -4 = 3 2 m m - 6 (4) = 6 c 1 m 2 3 2m - 24 = 3 2m - 24 + 24 = 3 + 24 2m = 27 2m 27 = 2 2 27 m= 2 = 13 1 2 6c

2.

x+1 x + =5 4 3

Solution The common denominator of 3 and 4 is 12.

x +1 x + =5 4 3 x +1 x m + 12 c m = 12 (5) 12 c 4 3 4 (x + 1) + 3x = 60 4x + 4 + 3x = 60 7x + 4 = 60 7x + 4 - 4 = 60 - 4 7x = 56 7x 56 = 7 7 x=8

Chapter 3 Equations

3.

99

y +1 y-2 5 = 5 3 6

Solution y +1 y-2 5 = 5 3 6 y +1 y-2 o - 30 e o = 30 c 5 m 30 e 5 3 6 6 (y + 1) - 10 (y - 2) = 25 6y + 6 - 10y + 20 = 25 - 4y + 26 = 25 - 4y + 26 - 26 = 25 - 26 - 4y = -1 - 4y -1 = -4 -4 y=1 4 When there is a fraction on either side of the equation, multiplying by the common denominator is the same as cross multiplying.

EXAMPLES 5 8 1. Solve x = (x ! 0 ) 3

Solution 5 8 x =3 8x = 15 8x 15 = 8 8 7 x=1 8 2. Solve

3 8 ^n ! 0h = 5 2n

Solution 3 8 = 5 2n 16n = 15 16n 15 = 16 16 15 n= 16

The common denominator of 5, 3 and 6 is 30.

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3.2 Exercises Solve 1.

b 2 = 5 3

14.

3 x x - = 5 2 10

2.

7 1 x = 5 (x ! 0 )

15.

x+4 x + =1 3 2

3.

9 4 y = 10 (y ! 0)

16.

p-3 2p + =2 2 3

4.

5x 11 = 4 7

17.

t +3 t -1 + =4 7 3

5.

9 4 = ( k ! 0) 5 2k

18.

x+5 x+2 =1 5 9

6.

x -4=8 3

19.

q-1 q-2 =2 4 3

7.

3 5t = 4 4

20.

x+3 x +7 +2= 5 2

8.

5+x 2 = 7 7

21.

3b 1 b - = 4 5 2

9.

y 3 =5 2

22.

a 3 5 + = 4 3 8

10.

x 2 - =7 9 3

23.

3 5 =x x+2

^ x ! 0, -2 h

11.

w-3 =5 2

24.

1 1 = y +1 3y - 1

c y ! -1,

12.

2t t - =2 5 3

25.

2 1 + = 0 ^ t ! 3, - 4 h t-3 t+4

13.

x 1 + =4 4 2

1 m 3

Substitution Sometimes substituting values into a formula involves solving an equation.

Investigation Body mass index (BMI) is a formula that is used to measure body fatness and is used by health professionals to screen for weight categories that may lead to health problems.

Chapter 3 Equations

This is not the only measure that is used when looking for health problems, however. For example, there are other factors in cardiac (heart) disease. Research these to find out what other things doctors look for. The BMI is used in a different way with children and teens, and is taken in relation to the child’s age. w The formula for BMI is BMI = 2 where w is weight in kg and h is height h in metres. For adults over 20, a BMI under 18.5 means that the person is underweight and over 25 is overweight. Over 30 is obese. The BMI may not always be reliable in measuring body fat. Can you think of some reasons? Is it important where the body fat is stored? Does it make a difference if it is on the hips or the stomach? Research these questions and find out more about BMI generally.

EXAMPLES 1. The formula for the surface area of a rectangular prism is given by S = 2 (lb + bh + lh) . Find the value of b when S = 180, l = 9 and h = 6.

Solution S = 2 (lb + bh + lh) 180 = 2 (9b + 6b + 9 # 6) = 2 (15b + 54) = 30b + 108 180 - 108 = 30b + 108 - 108 72 = 30b 30b 72 = 30 30 2. 4 = b

Another way of doing this would be to change the subject of the formula first.

CONTINUED

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2. The volume of a cylinder is given by V = rr 2 h. Evaluate the radius r, correct to 2 decimal places, when V = 350 and h = 6.5.

Solution V = rr 2 h 350 = rr 2 (6.5) r r 2 ( 6 .5 ) 350 = 6 .5 r 6.5r 350 = r2 6 .5 r 350 = r2 6 .5 r 350 =r 6 .5 r 4.14 = r

3.3 Exercises 1.

Given that v = u + at is the formula for the velocity of a particle at time t, find the value of t when u = 17.3, v = 100.6 and a = 9.8.

7.

The area of a rhombus is given by the formula A = 1 xy where x and 2 y are its diagonals. Find the value of x correct to 2 decimal places when y = 7.8 and A = 25.1.

2.

The sum of an arithmetic series is n given by S = (a + l ) . Find l if 2 a = 3, n = 26 and S = 1625.

8.

The simple interest formula is Pr n I= . Find n if r = 14.5, 100 P = 150 and I = 326.25.

3.

The formula for finding the area of a triangle is A = 1 bh. Find b 2 when A = 36 and h = 9.

9.

The gradient of a straight y2 - y1 line is given by m = x - x . 2 1

4.

The area of a trapezium is given by A = 1 h (a + b) . Find 2 the value of a when A = 120, h = 5 and b = 7.

5.

Find the value of y when x = 3, given the straight line equation 5x - 2y - 7 = 0.

6.

The area of a circle is given by A = rr 2 . Find r correct to 3 significant figures if A = 140.

Find y 1 when m = - 5 , 6 y 2 = 7, x 2 = - 3 and x 1 = 1. 10. The surface area of a cylinder is given by the formula S = 2rr ] r + h g . Evaluate h correct to 1 decimal place if S = 232 and r = 4.5.

Chapter 3 Equations

11. The formula for body mass index w is BMI = 2 . Evaluate h (a) the BMI when w = 65 and h = 1.6 (b) w when BMI = 21.5 and h = 1.8 (c) h when BMI = 19.7 and w = 73.8.

16. If the surface area of a sphere is S = 4rr 2, evaluate r to 3 significant figures when S = 56.3.

12. A formula for depreciation is D = P ] 1 - r g n . Find r if D = 12 000, P = 15 000 and n = 3.

18. If y =

13. The x-value of the midpoint is x1 + x2 given by x = . Find x1 2 when x = - 2 and x 2 = 5.

19. Given y = 2x + 5 , evaluate x when y = 4.

14. Given the height of a particle at time t is h = 5t 2, evaluate t when h = 23.

15. If y = x 2 + 1, evaluate x when y = 5.

17. The area of a sector of a circle 1 is A = r 2 i. Evaluate r when 2 A = 24.6 and i = 0.45. 2 , find the value of x x3 - 1 when y = 3.

20. The volume of a sphere is 4 V = rr 3. Evaluate r to 1 decimal 3 place when V = 150.

Inequations

• • • •

2 means greater than 1 means less than $ means greater than or equal to # means less than or equal to

In order to solve inequations, we need to see what effect one operation applied to both sides has on the inequality sign.

If a 2 b then a + c 2 b + c for all c

For example, 3 2 2 and 3 + 1 2 2 + 1 are both true.

If a 2 b then a - c 2 b - c for all c

For example, 3 2 2 and 3 - 1 2 2 - 1 are both true.

103

There are two solutions to this question.

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Maths In Focus Mathematics Extension 1 Preliminary Course

If a 2 b then ac 2 bc for all c 2 0

For example, 3 2 2 and 3 # 2 2 2 # 2 are both true.

If a 2 b then ac 1 bc for all c 1 0

For example, 3 2 2 but 3 # -2 1 2 # -2.

If a 2 b then a ' c 2 b ' c for all c 2 0

For example, 6 2 4 and 6 ' 2 2 4 ' 2 are both true.

If a 2 b then a ' c 1 b ' c for all c 1 0

For example, 6 2 4 but 6 ' -2 1 4 ' -2.

1 1 If a 2 b then a 1 for all positive numbers a and b b

For example, 3 2 2 but

1 1 1 . 3 2

The inequality sign reverses when: • multiplying by a negative • dividing by a negative • taking the reciprocal of both sides

On the number plane, we graph inequalities using arrows and circles (open for greater than and less than and closed in for greater than or equal to and less than or equal to) 1 2 # $

Chapter 3 Equations

105

EXAMPLES Solve and show the solutions on a number line 1. 5x + 7 $ 17

Solution 5x + 7 $ 17 5x + 7 - 7 $ 17 - 7 5x $ 10 5x 10 $ 5 5 x$2 -4

-3

-2

-1

0

1

2

3

4

2. 3t - 2 2 5t + 4

Solution 3t - 2 2 5t + 3t - 3t - 2 2 5t -2 2 2t + - 2 - 4 2 2t + -6 2 2t 2t -6 2 2 2 -3 2 t

4 3t + 4 4 4-4

or 3t - 2 3t - 5t - 2 -2t - 2 - 2t - 2 + 2 -2t -2t -2 t -4

2 5t + 4 2 5t - 5t + 4 24 24+2 26 6 2 -2 1 -3 -3

-2

Remember to change the inequality sign when dividing by -2.

-1

0

1

2

3

4

CONTINUED

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Maths In Focus Mathematics Extension 1 Preliminary Course

3. Solve 1 1 2z + 7 # 11.

Solution Method 1: Separate into two separate questions. 1 1 2z + 7 (i) 1 - 7 1 2z + 7 - 7 - 6 1 2z -6 2z 1 2 2 -3 1 z (ii)

2z + 7 # 11 2z + 7 - 7 # 11 - 7 2z # 4 2z 4 # 2 2 z #2

Putting these together gives the solution -3 1 z # 2. Method 2: Do as a single question. 1 1 2z + 7 # 11 1 - 7 1 2z + 7 - 7 # 11 - 7 -6 1 2z # 4 -6 2z 4 # 1 2 2 2 -3 1 z # 2

Solving this inequation as a single question is quicker than splitting it into two parts. Notice that the circle is not filled in for 1 and filled in for #.

-4

-3

-2

-1

0

1

2

3

4

3.4 Exercises 1.

Solve and plot the solution on a number line (a) x + 4 2 7 (b) y - 3 # 1

2.

Solve (a) 5t 2 35 (b) 3x - 7 $ 2 (c) 2 (p + 5) 2 8 (d) 4 - (x - 1) # 7 (e) 3y + 5 2 2y - 4 (f) 2a - 6 # 5a - 3 (g) 3 + 4y $ - 2 (1 - y)

(h) 2x + 9 1 1 - 4 (x + 1) a (i) # - 3 2 2y (j) 8 2 3 b (k) + 5 1 - 4 2 x (l) - 4 2 6 3 x 1 (m) + # 1 4 5 (n)

m 2 -3 2 4 3

Chapter 3 Equations

2b 1 - $6 5 2 r-3 (p) # -6 2 z+1 (q) +223 9 w 2w + 5 (r) + 14 6 3 (o)

(s)

x+1 x-2 $7 2 3

(t)

t+3 t+2 #2 7 2

(u)

q-2 3q 12+ 4 3

3.

(v)

2x x -1 2 2 3 2 9

(w)

2b - 5 b+6 +3# 8 12

Solve and plot the solutions on a number line (a) 3 1 x + 2 1 9 (b) -4 # 2p 1 10 (c) 2 1 3x - 1 1 11 (d) -6 # 5y + 9 # 34 (e) -2 1 3 (2y - 1) 1 7

PROBLEM Find a solution for this sum. Is it a unique solution? CR OS S +RO A DS DANGE R

Equations and Inequations Involving Absolute Values On a number line, x means the distance of x from zero in either direction.

EXAMPLES Plot on a number line and evaluate x 1. x = 2

Solution x = 2 means the distance of x from zero is 2 (in either direction). 2

-4

-3

-2

-1

2

0

1

2

3

4

x = !2

CONTINUED

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2. x # 2

Solution x # 2 means the distance of x from zero is less than or equal to 2 (in either direction). 2

-4 The solution of | x | 1 2 would be - 2 1 x 1 2.

-3

2

-1

-2

1

0

2

3

4

Notice that there is one region on the number line. We can write this as the single statement - 2 # x # 2. 3. x 2 2

Solution x 2 2 means the distance of x from zero is greater than 2 (in either direction). 2

-4 The solution of | x | $ 2 would be x # - 2, x $ 2.

-3

-1

-2

2

0

1

2

3

4

There are two regions on the number line, so we write two separate inequalities x 1 - 2, x 2 2.

x = a means x = ! a x 1 a means -a 1 x 1 a x 2 a means x 2 a, x 1 -a

Class Discussion What does a - b mean as a distance along the number line? Select different values of a and b to help with this discussion.

We use absolute value as a distance on a number line to solve equations and inequations involving absolute values.

Chapter 3 Equations

109

EXAMPLES Solve 1. x + 4 = 7

Solution This means that the distance from x + 4 to zero is 7 in either direction. So x + 4 = ! 7. x+4 =7 x+4=7 or x + 4 = -7 x+4-4=7-4 x + 4 - 4 = -7 - 4 x=3 x = -11 2. 2y - 1 1 5

Solution This means that the distance from 2y - 1 to zero is less than 5 in either direction. So it means - 5 1 2y - 1 1 5. - 5 1 2y - 1 1 5 - 5 + 1 1 2y - 1 + 1 1 5 + 1 2y 6 -4 1 1 2 2 2 -2 1 y 1 3

You could solve these as two separate inequations.

3. 5b - 7 $ 3

Solution 5b - 7 $ 3 means that the distance from 5b - 7 to zero is greater than or equal to 3 in either direction. 5b - 7 # - 3

5b - 7 $ 3

5b - 7 + 7 # -3 + 7 5b # 4 5b 4 # 5 5 4 b # 5 4 So b # , b $ 2. 5

5b - 7 + 7 $ 3 + 7 5b $ 10 5b 10 $ 5 5 b$2

These must be solved and written as two separate inequations.

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While it is always a good habit to check solutions to equations and inequations by substituting in values, in these next examples it is essential to check, as some of the solutions are impossible!

EXAMPLES Solve 1. 2x + 1 = 3x - 2

Solution 2x + 1 = 3x - 2 means that 2x + 1 is at a distance of 3x - 2 from zero. 2x + 1 = ! ] 3x - 2 g This question is impossible if 3x - 2 is negative. Can you see why? If 2x + 1 is equal to a negative number, this is impossible as the absolute value is always positive. Case (i) 2x + 1 = 3x - 2 2 x - 2 x + 1 = 3x - 2 x - 2 1=x-2 1+2=x-2+2 3=x Check solution is possible: Substitute x = 3 into 2x + 1 = 3x - 2. LHS = 2 # 3 + 1 = 7 =7 RHS = 3 # 3 - 2 =9-2 =7 Since LHS = RHS, x = 3 is a solution. Case (ii) 2 x + 1 = - ( 3x - 2 ) = - 3x + 2 2 x + 3x + 1 = - 3 x + 3x + 2 5x + 1 = 2 5x + 1 - 1 = 2 - 1 5x = 1 5x 1 = 5 5 1 x= 5

Chapter 3 Equations

Check: 1 Substitute x = into 2x + 1 = 3x - 2. 5 1 LHS = 2 # + 1 5 2 = 1 5 2 =1 5 1 RHS = 3 # - 2 5 3 = -2 5 2 = -1 5 1 Since LHS ! RHS, x = is not a solution. 5 So the only solution is x = 3.

It is often easier to solve these harder equations graphically. You will do this in Chapter 5.

2. 2x - 3 + x + 1 = 9

Solution In this question it is difficult to use distances on the number line, so we use the definition of absolute value. 2x - 3 2x - 3 = ' - (2 x - 3) +1 x + 1 = ' -(xx + 1)

when 2x - 3 $ 0 when 2x - 3 1 0 when x + 1 $ 0 when x + 1 1 0

This gives 4 cases: (i) (2x - 3) + (x + 1) = 9 (ii) (2x - 3) - (x + 1) = 9 (iii) -(2x - 3) + (x + 1) = 9 (iv) -(2x - 3) - (x + 1) = 9 Case (i) ( 2x - 3 ) + ( x + 1 ) = 9 2x - 3 + x + 1 = 9 3x - 2 = 9 3x - 2 + 2 = 9 + 2 3x = 11 3x 11 = 3 3 2 x=3 3 Check by substituting x = 3

111

2 into 2x - 3 + x + 1 = 9. 3 CONTINUED

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2 2 -3 + 3 +1 3 3 1 2 = 4 + 4 3 3 1 2 =4 +4 3 3 =9 = RHS 2 So x = 3 is a solution. 3 Case (ii) ( 2 x - 3 ) - (x + 1 ) = 9 2x - 3 - x - 1 = 9 x-4=9 x-4+4=9+4 x = 13 Check by substituting x = 13 into 2x - 3 + x + 1 = 9. LHS = 2 # 13 - 3 + 13 + 1 = 23 + 14 = 23 + 14 = 37 ! RHS So x = 13 is not a solution. Case (iii) -(2x - 3) + (x + 1) = 9 - 2x + 3 + x + 1 = 9 -x + 4 = 9 -x + 4 - 4 = 9 - 4 -x = 5 -x 5 = -1 -1 x = -5 LHS = 2 # 3

Check by substituting x = - 5 into 2x - 3 + x + 1 = 9. LHS = 2 # - 5 - 3 + - 5 + 1 = - 13 + - 4 = 13 + 4 = 17 ! RHS So x = - 5 is not a solution. Case (iv) - (2x - 3) - (x + 1) = 9 - 2x + 3 - x - 1 = 9 - 3x + 2 = 9 - 3x + 2 - 2 = 9 - 2 - 3x = 7

Chapter 3 Equations

113

- 3x 7 = -3 -3 1 3 1 Check by substituting x = - 2 into 2x - 3 + x + 1 = 9. 3 1 1 LHS = 2 # - 2 - 3 + - 2 + 1 3 3 2 1 = -7 + -1 3 3 2 1 = 7 +1 3 3 =9 = RHS 1 So x = - 2 is a solution. 3 2 1 So solutions are x = 3 , - 2 . 3 3 x = -2

While you should always check solutions, you can see that there are some cases where this is really important.

You will learn how to solve equations involving absolute values graphically in Chapter 5. With graphical solutions it is easy to see how many solutions there are.

3.5 Exercises 1.

Solve

3.

Solve (a) x + 2 = 5x - 3 (b) 2a - 1 = a + 2 (c) b - 3 = 2b - 4 (d) 3k - 2 = k - 4 (e) 6y + 23 = y - 7 (f) 4x + 3 = 5x - 4 (g) 2m - 5 = m (h) 3d + 1 = d + 6 (i) 5 - y = 4y + 1 (j) 2t - 7 = 3 - t

4.

Solve

(a) x = 5 (b) y = 8 (c) a 1 4 (d) k $ 1 (e) x 2 6 (f) p # 10 (g) x = 0 (h) a 2 14 (i) y 1 12 (j) b $ 20 2.

Solve

(a) x + 3 = 3x - 1

(a) x + 2 = 7

(b) 2y - 5 = y - 2 (c) 3a + 1 = 2a - 9

(b) n - 1 = 3

(d) 2x + 5 + x = 17

(c) 2a 2 4

(e) 3d - 2 + d + 4 = 18

(d) x - 5 # 1 (e) 9 = 2x + 3 (f) 7x - 1 = 34 (g) 4y + 3 1 11 (h) 2x - 3 $ 15 x (i) =4 3 a (j) -3 #2 2

5.

(a) Solve 4t - 3 + t - 1 = 11. (b) By plotting the solutions on a number line and looking at values in between the solutions, solve 4t - 3 + t - 1 1 11.

Remember to check solutions in questions 3, 4 and 5.

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Exponential Equations An exponential equation involves an unknown index or power e.g. 2 x = 8. We can also solve other equations involving indices. In order to solve these, you need to understand their relationship. For example, squares and square roots are the reverse of each other (we call them inverse operations). Similarly cubes and cube roots are inverses, and this extends to all indices. To solve equations, use inverse operations: For squares, take the square root For cubes, take the cube root For square roots, take the square For cube roots, take the cube You have previously used these rules when substituting into formulae involving squares and cubes.

EXAMPLES Solve 1. x 2 = 9 There are two possible solutions for x – one positive and one negative since 3 2 = 9 and (- 3) 2 = 9.

Solution x2 = 9 x2 = ! 9 ` x= !3 2. 5n 3 = 40

Solution

There is only one answer for this question since 2 3 = 8 but (- 2) 3 = -8.

5n 3 = 40 5n 3 40 = 5 5 3 n =8 3

n3 = 3 8 n=2

Chapter 3 Equations

2

3. a 3 = 4

Solution 2 3

3 2

3 2

2 3

We use the fact that ` a j = ` a j = a. 2

a3 = 4 2 3

3 2

3

`a j = 4 2 3

`

a= 42 3 a = ^ 4h = 23 =8

Investigation Investigate equations of the type x n = k where k is a constant, for example, x n = 9. Look at these questions: 1. 2. 3. 4. 5. 6.

What is the solution when n = 0? What is the solution when n = 1? How many solutions are there when n = 2? How many solutions are there when n = 3? How many solutions are there when n is even? How many solutions are there when n is odd?

In other types of equations, the pronumeral (or unknown variable) is in the index. We call these exponential equations, and we use the fact that if the base numbers are equal, then the powers (or indices or exponents) must be equal.

EXAMPLES Solve 1. 3 x = 81

Solution 3 x = 81 Equating indices: 3x = 34 `x=4 CONTINUED

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2. 5 2k - 1 = 25

Solution 5 2k - 1 = 25 5 2k - 1 = 5 2 ` 2k - 1 = 2 2k - 1 + 1 = 2 + 1 2k = 3 3 2k = 2 2 1 k=1 2

We can check this solution 1 by substituting k = 1 into 2 2k -1 the equation 5 = 25.

3. 8 n = 4

Solution It is hard to write 8 as a power of 4 or 4 as a power of 8, but both can be written as powers of 2. 8n = 4 (2 ) = 2 2 2 3n = 2 2 ` 3n = 2 3n 2 = 3 3 2 n= 3 3 n

3.6 Exercises 1.

Solve (a) x 3 = 27 (b) y 2 = 64 (c) n 4 = 16 (d) x 2 = 20 (give the exact answer) (e) p 3 = 1000 (f) 2x 2 = 50 (g) 6y 4 = 486 (h) w 3 + 7 = 15 (i) 6n 2 - 4 = 92 (j) 3q 3 + 20 = - 4

2.

Solve and give the answer correct to 2 decimal places. (a) p 2 = 45 (b) x 3 = 100 (c) n 5 = 240 (d) 2x 2 = 70 (e) 4y 3 + 7 = 34 d4 (f) = 14 3 k2 (g) -3=7 2 x3 - 1 (h) =2 5 (i) 2y 2 - 9 = 20 (j) 7y 3 + 9 = 200

Chapter 3 Equations

3.

Solve

6.

Solve (a) 2 n = 16 (b) 3 y = 243 (c) 2 m = 512 (d) 10 x = 100 000 (e) 6 m = 1 (f) 4 x = 64 (g) 4 x + 3 = 19 (h) 5 (3 x ) = 45 (i) 4 x = 4 6k (j) = 18 2

7.

Solve (a) 3 2x = 81 (b) 2 5x - 1 = 16 (c) 4 x + 3 = 4 (d) 3 n - 2 = 1 (e) 7 2x + 1 = 7 (f) 3 x - 3 = 27 (g) 5 3y + 2 = 125 (h) 7 3x - 4 = 49 (i) 2 4x = 256 (j) 9 3a + 1 = 9

8.

Solve (a) 4 m = 2 (b) 27 x = 3 (c) 125 x = 5

2 3

(a) n = 9 3

(b) t 4 = 8 2

(c) x 5 = 4 4

(d) t 3 = 16 3

(e) p 5 = 27 3

(f) 2m 4 = 250 2

(g) b 3 + 3 = 39 4

(h) 5y 3 = 405 2

(i) 3a 7 - 2 = 10 3 4

(j) 4.

5.

t =9 3

Solve (all pronumerals ! 0) (a) x - 1 = 5 (b) a - 3 = 8 (c) y - 5 = 32 (d) x - 2 + 1 = 50 (e) 2n - 1 = 3 1 (f) a - 3 = 8 1 -2 (g) x = 4 1 (h) b - 1 = 9 1 (i) x - 2 = 2 4 16 (j) b - 4 = 81

1 k m =7 49 1 k m = 100 (e) c 1000 (f) 16 n = 8 (g) 25 x = 125 (h) 64 n = 16 (d) c

Solve (all pronumerals ! 0) (a) x

-

1 3

-

3 2

-

1 4

-

3 4

(b) x (c) a

(d) k

(e) 3x

-

3 2

=8 =

8 125

=3 = 125 2 3

= 12

1 8 2 1 3 (g) y = 4 2 4 (h) n 5 = 9 (f) x

(i) b

-

(j) m

5 3

2 3

1 3k (i) c m = 2 4 (j) 8 x - 1 = 4

=

= =

1 32 36 49

9.

Solve (a) 2 4x + 1 = 8 x (b) 3 5x = 9 x - 2 (c) 7 2k + 3 = 7 k - 1 (d) 4 3n = 8 n + 3 (e) 6 x - 5 = 216 x (f) 16 2x - 1 = 4 x - 4 (g) 27 x + 3 = 3 x 1 x 1 2x + 3 m (h) c m = c 2 64

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3 x 27 2x - 3 m (i) c m = c 4 64 1 x-9 m (j) ] 5 g- x = c 25 10. Solve (a) 4 m =

2

9 k+3 m (b) c = 25 1 (c) = 4 2x - 5 2

3 5

(d) 3 k = 3 3 (e) c

3 1 3n + 1 m = 27 81

5 -n 2 3n + 1 (f) c m =c m 5 2 1 (g) 32 - x = 16 (h) 9 2b + 5 = 3 b 3 (i) 81 x + 1 =

3x

1 3m - 5 (j) 25 - m = c m 5

PUZZLE Test your logical thinking and that of your friends. 1. How many months have 28 days? 2. If I have 128 sheep and take away all but 10, how many do I have left? 3. A bottle and its cork cost $1.10 to make. If the bottle costs $1 more than the cork, how much does each cost? 4. What do you get if you add 1 to 15 four times? 5. On what day of the week does Good Friday fall in 2016?

Quadratic Equations A quadratic equation is an equation involving a square. For example, x 2 - 4 = 0.

Solving by factorisation When solving quadratic equations by factorising, we use a property of zero.

For any real numbers a and b, if ab = 0 then a = 0 or b = 0

EXAMPLES Solve 1. x 2 + x - 6 = 0

Solution x2 + x - 6 = 0 (x + 3) (x - 2) = 0

Chapter 3 Equations

`

x+3=0 or x-2=0 x+3-3=0-3 x-2+2 =0 +2 x = -3 or x= 2

So the solution is x = - 3 or 2. 2. y 2 - 7y = 0

Solution y 2 - 7y = 0 y ( y - 7) = 0 ` y=0

or

y-7=0

y-7+7=0+7 y=7 So the solution is y = 0 or 7. 3. 3a 2 - 14a = - 8

Solution 3a 2 - 14a = - 8 3a 2 - 14a + 8 = - 8 + 8 3a 2 - 14a + 8 = 0 (3a - 2) (a - 4) = 0 ` 3a - 2 = 0 or 3a - 2 + 2 = 0 or 3a = 2 3a 2 = 3 3 2 a= 3 2 So the solution is a = or 4. 3

a-4 =0 a-4+4 =0+4 a=4

3.7 Exercises Solve 1.

y2 + y = 0

4.

t 2 - 5t = 0

2.

b2 - b - 2 = 0

5.

x 2 + 9x + 14 = 0

3.

p 2 + 2p - 15 = 0

6.

q2 - 9 = 0

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7.

x2 - 1 = 0

17. 5x - x 2 = 0

8.

a 2 + 3a = 0

18. y 2 = y + 2

9.

2x 2 + 8x = 0

19. 8n = n 2 + 15

10. 4x 2 - 1 = 0

20. 12 = 7x - x 2

11. 3x 2 + 7x + 4 = 0

21. m 2 = 6 - 5m

12. 2y 2 + y - 3 = 0

22. x (x + 1) (x + 2) = 0

13. 8b 2 - 10b + 3 = 0

23. (y - 1) (y + 5) (y + 2) = 0

14. x 2 - 3x = 10

24. (x + 3) (x - 1) = 32

15. 3x 2 = 2x

25. (m - 3) (m - 4) = 20

16. 2x 2 = 7x - 5

Application 1 2 at where u is the 2 initial velocity and a is the acceleration. Find the time when the displacement will be zero, given u = - 12 and a = 10. A formula for displacement s at time t is given by s = ut +

2 s = ut + 1 at 2 2 0 = -12t + 1 (10) t 2

= -12t + 5t

2

= t (-12 + 5t ) ` t = 0 or

-12 + 5t = 0

-12 + 12 + 5t = 0 + 12 5t = 12 5t 12 = 5 5 t = 2.4 So displacement will be zero when t = 0 or 2.4.

Solving by completing the square Not all trinomials will factorise, so other methods need to be used to solve quadratic equations.

Chapter 3 Equations

121

EXAMPLES Solve 1. x 2 = 7

Solution x2 = 7 x=! 7 = ! 2.6 2. ] x + 3 g2 = 11

Solution ] x + 3 g2 = 11

Take the square root of both sides.

x + 3 = ! 11 x + 3 - 3 = ! 11 - 3 x = ! 11 - 3 = 0.3, - 6.3

3. ^ y - 2 h2 = 7

Solution ^ y - 2 h2 = 7 y-2=! 7 y-2+2=! 7+2 y=! 7+2 = 4.6, - 0.6

To solve a quadratic equation like x 2 - 6x + 3 = 0, which will not factorise, we can use the method of completing the square.

You learnt how to complete the square in Chapter 2.

EXAMPLES Solve by completing the square 1. x 2 - 6x + 3 = 0 (give exact answer)

Solution x 2 - 6x + 3 = 0 x 2 - 6x = - 3

Halve 6, square it and add to both sides of the equation.

2

c 6 m = 32 = 9 2 CONTINUED

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Maths In Focus Mathematics Extension 1 Preliminary Course

x 2 - 6x + 9 = - 3 + 9 ] x - 3 g2 = 6 `

x-3=! 6 x-3+3=! 6+3 x=! 6+3

2. y 2 + 2y - 7 = 0 (correct to 3 significant figures)

Solution y 2 + 2y - 7 = 0 y 2 + 2y = 7

2

c 2 m = 12 = 1 2

y 2 + 2y + 1 = 7 + 1 ^ y + 1 h2 = 8 `

y+1=! 8 y + 1 - 1 = ! 8 -1 y = ! 8 -1 = !2 2 - 1 y = 1.83 or - 3.83

3.8 Exercises 1.

Solve by completing the square, giving exact answers in simplest surd form (a) x 2 + 4x - 1 = 0 (b) a 2 - 6a + 2 = 0 (c) y 2 - 8y - 7 = 0 (d) x 2 + 2x - 12 = 0 (e) p 2 + 14p + 5 = 0 (f) x 2 - 10x - 3 = 0 (g) y 2 + 20y + 12 = 0 (h) x 2 - 2x - 1 = 0 (i) n 2 + 24n + 7 = 0 (j) y 2 - 3y + 1 = 0

2.

Solve by completing the square and write your answers correct to 3 significant figures (a) x 2 - 2x - 5 = 0 (b) x 2 + 12x + 34 = 0 (c) q 2 + 18q - 1 = 0 (d) x 2 - 4x - 2 = 0 (e) b 2 + 16b + 50 = 0 (f) x 2 - 24x + 112 = 0 (g) r 2 - 22r - 7 = 0 (h) x 2 + 8x + 5 = 0 (i) a 2 + 6a - 1 = 0 (j) y 2 - 40y - 3 = 0

Solving by formula Completing the square is difficult with harder quadratic equations, for example 2x 2 - x - 5 = 0. Completing the square on a general quadratic equation gives the following formula.

Chapter 3 Equations

For the equation ax 2 + bx + c = 0 x=

-b !

b 2 - 4ac 2a

Proof Solve ax 2 + b + c = 0 by completing the square. ax 2 + bx + c = 0 ax 2 bx c 0 a + a +a=a bx c x2 + a + a = 0 c c bx c x2 + a + a - a = 0 - a bx c x2 + a = - a

2 2 2 b b ' 2l = c b m = b 2 a 2a 4a

bx c b2 b2 x2 + a + 2 = - a + 2 4a 4a c b2 b 2 cx + m = -a + 2 2a 4a - 4ac + b 2 = 4a 2 - 4ac + b 2 b x+ =! 2a 4a 2 2 b - 4ac =! 2a b 2 - 4ac b b b x+ =! 2a 2a 2a 2a b 2 - 4ac -b x= ! 2a 2a 2 - b ! b - 4ac = 2a

EXAMPLES 1. Solve x 2 - x - 2 = 0 by using the quadratic formula.

Solution a = 1, b = -1, c = - 2 b 2 - 4ac 2a - (-1) ! (-1) 2 - 4 (1) (-2) = 2 (1 ) 1! 1+8 = 2

x=

-b !

CONTINUED

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1! 9 2 1!3 = 2 = 2 or - 1

1! 3 gives two 2 1+ 3 separate solutions, 2 1- 3 and . 2

=

x =

2. Solve 2y 2 - 9y + 3 = 0 by formula and give your answer correct to 2 decimal places.

Solution a = 2, b = -9, c = 3 -b ! b 2 - 4ac 2a - ] -9 g ! ] -9 g2 - 4 ] 2 g ] 3 g y= 2] 2 g 9 ! 81 - 24 = 4 9 ! 57 = 4 Z 4.14 or 0.36

x=

These solutions are irrational.

3.9 1.

Exercises

Solve by formula, correct to 3 significant figures where necessary (a) y 2 + 6y + 2 = 0 (b) 2x 2 - 5x + 3 = 0 (c) b 2 - b - 9 = 0 (d) 2x 2 - x - 1 = 0 (e) - 8x 2 + x + 3 = 0 (f) n 2 + 8n - 2 = 0 (g) m 2 + 7m + 10 = 0 (h) x 2 - 7x = 0 (i) x 2 + 5x = 6 (j) y 2 = 3y - 1

2.

Solve by formula, leaving the answer in simplest surd form (a) x 2 + x - 4 = 0 (b) 3x 2 - 5x + 1 = 0 (c) q 2 - 4q - 3 = 0 (d) 4h 2 + 12h + 1 = 0 (e) 3s 2 - 8s + 2 = 0 (f) x 2 + 11x - 3 = 0 (g) 6d 2 + 5d - 2 = 0 (h) x 2 - 2x = 7 (i) t 2 = t + 1 (j) 2x 2 + 1 = 7x

Class Investigation Here is a proof that 1 = 2. Can you see the fault in the proof? x2 - x2 = x2 - x2 x(x - x) = (x + x) (x - x) x=x+x x = 2x 1=2 `

Chapter 3 Equations

125

Further Inequations Inequations involving pronumerals in the denominator can be solved in several ways. Here is one method. You will use a different method in Chapter 10.

EXAMPLES 1 1. Solve x 1 3.

Solution 1 is undefined. 0

x!0 1 Solve x = 3. 1 x #x=3#x 1 = 3x 3x 1 = 3 3 1 =x 3 1 1 is not a solution of the inequation x 1 3. 3 1 Place x = 0 and x = on a number plane and test x values on either side 3 of these values in the inequation. x=

-3

-2

-1

0 1 3

1

Test for x 1 0, say x = -1 Substitute into the inequation: 1 x 13 1 13 -1 -1 1 3 So x 1 0 is part of the solution. 1 1 Test for 0 1 x 1 , say x = 3 10 1 13 1 10 10 1 3 1 So 0 1 x 1 is not part of the solution. 3 1 Test for x 2 , say x = 1 3 Substitute into the inequation:

2

3

4

5

(true)

(false)

CONTINUED

Circle these values as they are not included in the solution.

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Maths In Focus Mathematics Extension 1 Preliminary Course

1 13 1 113

(true)

1 is part of the solution. 3 1 Solution is x 1 0, x 2 . 3

So x 2

-2

-3

2. Solve

-1

0 1 3

1

2

3

4

5

6 $ 1. x+3

Solution 6 is undefined. 0

x ! -3 Solve

6 = 1. x+3

6 # (x + 3) = 1 # (x + 3) x+3 6 =x+3 6-3 =x+ 3-3 3=x

Circle x = - 3 and fill in x = 3 since it is a part of the solution.

6 $ 1. x+3 Place x = - 3 and x = 3 on a number plane and test values on either side in the inequation. x = 3 is a solution of the inequation

-3

-2

-1

0

1

2

3

4

5

Test for x 1 - 3, say x = - 4 Substitute into the inequation: 6 $1 x+3 6 $1 -4 + 3 -6 $ 3

(false)

So x 1 - 3 is not part of the solution. Test for - 3 1 x # 3, say x = 0 6 $1 0+3 2$1

(true)

So - 3 1 x # 3 is part of the solution. Test for x $ 3, say x = 4 Substitute into the inequation: 6 $1 4+3 6 $1 7 So x $ 3 is not part of the solution.

(false)

Chapter 3 Equations

Solution is - 3 1 x # 3 -2

-3

3. Solve

-1

0

1

2

3

4

5

y2 - 6 # 1. y

Solution y!0 y2 - 6 = 1. y 2 y -6 y #y=1#y y2 - 6 = y y2 - y - 6 = y - y y2 - y - 6 = 0 ^y - 3h^y + 2h = 0 y - 3 = 0, y+2 =0 y - 3 + 3 = 0 + 3, y + 2 - 2 = 0 - 2 y = 3, y = -2 Solve

Sketch these on a number line and test values on either side. -3

-2

-1

0

1

Test for y # - 2, say y = - 3 Substitute into the inequation: y2 - 6 #1 y 2 ]-3 g - 6 #1 -3 -1 # 1

2

3

4

5

(true)

So y # - 2 is part of the solution. Test for - 2 # y 1 0, say y = -1 ] -1 g2 - 6 #1 -1 5#1

(false)

So - 2 # y 1 0 is not part of the solution. Test 0 1 y # 3, say y = 1 12 - 6 #1 1 -5 # 1 So 0 1 y # 3 is part of the solution.

(true)

Test y $ 3, say y = 4 CONTINUED

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42 - 6 #1 4 1 2 #1 2 So y $ 3 is not part of the solution. The solution is y # - 2, 0 1 y # 3 -3

3.10

-2

-1

0

(false)

2

1

3

Exercises

Solve 1.

1 y 11

16.

3x + 1 1 $ x-4 3

2.

1 x 22

17.

8p + 7 25 2p - 9

3.

3 x 12

18.

3 x-2 # 5x + 1 4

4.

2 m $7

19.

7t + 4 $ -1 3t - 8

5.

3 x 2 -5

20.

5m + 4 1 1 4 2m

6.

2 # -1 b

21.

x2 - 5 1 -4 x

7.

1 24 x -1

22.

n2 + 8 $6 n

8.

1 1 -5 z+3

23.

x 2 - 15 22 x

9.

3 $4 x-2

24.

m2 - 8 #4 m +1

10.

-1 16 2-x

25.

4 $x x-3

11.

5 # -9 x+4

26.

2x 2 # -1 3x - 2

12.

2 25 3x - 4

27.

3 #x x-2

13.

-3 12 2a + 5

28.

n+5 2n n-3

14.

x 25 2x - 1

29.

3x 2 1 -2 7x + 4

15.

y 12 y +1

30.

2 x ( x - 4) #7 x -1

4

5

Chapter 3 Equations

129

Quadratic Inequations Solving quadratic inequations is similar to solving quadratic equations, but you need to do this in two stages. The first is to solve the equation and then the second step is to look at either the number line or the number plane for the inequality.

To solve a quadratic inequation: 1. Factorise and solve the quadratic equation 2. Test values in the inequality

In Chapter 10 you will look at how to use the number plane to solve these quadratic inequations. Here are some examples of solving quadratic inequations using the number line.

EXAMPLES Solve 1. x 2 + x - 6 2 0

Solution Be careful: x 2 + x - 6 2 0 does not mean x - 2 2 0 and x + 3 2 0.

First solve x + x - 6 = 0 (x - 2 ) (x + 3 ) = 0 ` x = 2 or -3 2

Now look at the number line. -4

-3

-2

-1

0

1

2

3

4

Choose a number between - 3 and 2, say x = 0. Substitute x = 0 into the inequation. x2 + x - 6 2 0 02 + 0 - 6 2 0 -6 2 0

(false)

So the solution is not between -3 and 2. ` the solution lies either side of -3 and 2. Check by choosing a number on either side of the two numbers. Choose a number on the RHS of 2, say x = 3. CONTINUED

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Maths In Focus Mathematics Extension 1 Preliminary Course

Substitute x = 3 into the inequation. 32 + 3 - 6 2 0 620 So the solution is on the RHS of 2. Choose a number on the LHS of -3, say x = -4 Substitute x = -4 into the inequation

(true)

(- 4) 2 + ( - 4) - 6 2 0 620 So the solution is on the LHS of -3. -4

-3

-2

-1

(true)

0

1

2

3

4

1

2

3

4

This gives the solution x 1 -3, x 2 2. 2. 9 - x 2 $ 0

Solution First solve 9 - x 2 = 0 (3 - x) (3 + x) = 0 ` x = !3 -4

-3

-2

-1

0

Choose a number between -3 and 3, say x = 0. Substitute x = 0 into the inequation. Check numbers on the RHS and LHS to verify this.

9 - x2 $ 0 9 - 02 $ 0 9$0

(true)

So the solution is between -3 and 3, that is -3 # x # 3. On the number line: -4

-3

-2

-1

0

1

2

3

4

Earlier in the chapter you learned how to solve inequations with the unknown in the denominator. Some people like to solve these using quadratic inequations. Here are some examples of how to do this.

Chapter 3 Equations

131

EXAMPLES Solve 1 1. x 1 3

Solution x 2 is positive, so the inequality sign does not change.

x!0 First, multiply both sides by x 2 . 1 x 13 x 1 3x 2 0 1 3x 2 - x Now, solve

3x 2 - x = 0 x(3x - 1) = 0 x = 0 or -2

1 3 -1

0 1 3

1

2

By checking on each side of 0 and 1 , for 0 1 3x 2 - x, the solution is 3 x 1 0, x 2 1 . 3 2.

3 $2 x+5

Solution 2

(x + 5) is positive, so the inequality sign does not change.

x ! -5 First, multiply both sides by (x + 5)2 . 3 $2 x+5 3 ( x + 5 ) $ 2 ( x + 5) 2 0 $ 2 ( x + 5 ) 2 - 3 ( x + 5) 0 $ ( x + 5 ) [ 2 ( x + 5) - 3 ] 0 $ ( x + 5 ) ( 2 x + 7) Now, solve (x + 5) (2x + 7) = 0 ` x + 5 = 0 or 2x + 7 = 0 x = -5 -6

-5

Check this factorisation carefully.

x cannot be -5 as this would give 0 in the denominator.

x = -3 1 2 -4 -3 1 -3 2

-2

1 Check by choosing a number on each side of -5 and -3 for 2 1 0 $ (x + 5) (2x + 7) that the solution is -5 1 x # -3 . 2

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3.11

Exercises

Solve 1.

x 2 + 3x 1 0

21. x 2 1 2x

2.

y 2 - 4y 1 0

22. 2a 2 - 5a + 3 # 0

3.

n2 - n $ 0

23. 5y 2 + 6y $ 8

4.

x2 - 4 $ 0

24. 6m 2 2 15 - m

5.

1 - n2 1 0

25. 3x 2 # 7x - 4

6.

n 2 + 2n - 15 # 0

1 26. x 2 2

7.

c2 - c - 2 2 0

8.

x + 6x + 8 # 0

9.

x 2 - 9x + 20 1 0

2

3 27. x # 6 28.

1 15 y+1

29.

1 $2 n-3

30.

3 $ -1 x+5

31.

1 17 5x - 2

32.

4 $ -5 x-5

33.

x #5 x+1

34.

2x + 1 21 x-2

35.

2x - 3 $6 5x + 3

10. 4b 2 + 10b + 4 $ 0 11. 1 - 2a - 3a 1 0 2

12. 2y 2 - y - 6 2 0 13. 3x 2 - 5x + 2 $ 0 14. 6 - 13b - 5b 1 0 2

15. 6x 2 + 11x + 3 # 0 16. y 2 + y # 12 17. x 2 2 16 18. a 2 # 1 19. x 2 1 x + 6 20. x $ 2x + 3 2

Simultaneous Equations Two equations, each with two unknown pronumerals, can be solved together to find one solution that satisfies both equations. There are different ways of solving simultaneous equations. The elimination method adds or subtracts the equations. The substitution method substitutes one equation into the other.

Chapter 3 Equations

Linear equations These equations can be solved by either method. Many students prefer the elimination method.

EXAMPLES Solve simultaneously 1. 3a + 2b = 5 and 2a - b = -6

Solution

] 2 g # 2: ] 1 g + (3):

3a + 2b = 5 2a - b = -6

(1) (2)

4a - 2b = -12 3a + 2b = 5 7a = - 7 a = -1

(3) (1)

Substitute a = -1 in (1) 3 (-1) + 2b = 5 -3 + 2b = 5 2b = 8 b=4 ` solution is a = -1, b = 4 2. 5x - 3y = 19 and 2x - 4y = 16

Solution

(1) # 4: ( 2 ) # 3: (3) - (4):

5x - 3y = 19 2x - 4y = 16 20x - 12y = 76 6x - 12y = 48 14x = 28 x=2

Substitute x = 2 in (2) 2 ( 2) - 4 y 4 - 4y - 4y y

= 16 = 16 = 12 = -3

( 1) ( 2) (3) (4 )

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3.12

Exercises

Solve simultaneously 1.

a - b = -2 and a + b = 4

12. 3a - 4b = -16 and 2a + 3b = 12

2.

5x + 2y = 12 and 3x - 2y = 4

3.

4p - 3q = 11 and 5p + 3q = 7

13. 5p + 2q + 18 = 0 and 2p - 3q + 11 = 0

4.

y = 3x - 1 and y = 2x + 5

5.

2x + 3y = -14 and x + 3y = -4

6.

7t + v = 22 and 4t + v = 13

16. 5s - 3t - 13 = 0 and 3s - 7t - 13 = 0

7.

4x + 5y + 2 = 0 and 4x + y + 10 = 0

17. 3a - 2b = - 6 and a - 3b = - 2

8.

2x - 4y = 28 and 2x - 3y = -11

18. 3k - 2h = -14 and 2k - 5h = -13

9.

5x - y = 19 and 2x + 5y = -14

10. 5m + 4n = 22 and m - 5n = -13 11. 4w 1 + 3w 2 = 11 and 3w 1 + w 2 = 2

14. 7x 1 + 3x 2 = 4 and 3x 1 + 5x 2 = - 2 15. 9x - 2y = -1 and 7x - 4y = 9

19. 2v 1 + 5v 2 - 16 = 0 and 7v 1 + 2v 2 + 6 = 0 20. 1.5x + 3.4y = 7.8 and 2 . 1 x - 1 . 7y = 1 . 8

PROBLEM A group of 39 people went to see a play. There were both adults and children in the group. The total cost of the tickets was $939, with children paying $17 each and adults paying $29 each. How many in the group were adults and how many were children? (Hint: let x be the number of adults and y the number of children.)

Non-linear equations In questions involving non-linear equations there may be more than one set of solutions. In some of these, the elimination method cannot be used. Here are some examples using the substitution method.

Chapter 3 Equations

EXAMPLES Solve simultaneously 1. xy = 6 and x + y = 5

Solution xy = 6 x+y=5 From (2): y=5-x Substitute (3) in (1) x (5 - x) = 6

( 1) (2 ) (3 )

5x - x 2 = 6 0 = x 2 - 5x + 6 0 = (x - 2 ) (x - 3 ) ` x - 2 = 0 or x - 3 = 0 x = 2 or x = 3 Substitute x = 2 in (3) y=5-2=3 Substitute x = 3 in (3) y=5-3=2 ` solutions are x = 2, y = 3 and x = 3, y = 2 2. x 2 + y 2 = 16 and 3x - 4y - 20 = 0

Solution x 2 + y 2 = 16 3x - 4y - 20 = 0 From ] 2 g: 3x - 20 = 4y 3x - 20 =y 4 Substitute (3) into (1) 3x - 20 2 m = 16 x2 + c 4 9x 2 - 120x + 400 n = 16 x2 + d 16 16x 2 + 9x 2 - 120x + 400 = 256 25x 2 - 120x + 144 = 0 (5x - 12)2 = 0 ` 5x - 12 = 0 5x = 12 x = 2.4 Substitute x = 2.4 into ] 3 g 3 (2.4) - 20 4 = -3.2 So the solution is x = 2.4, y = -3.2. y=

(1) ( 2)

(3)

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3.13

Exercises

Solve the simultaneous equations. 1.

y = x 2 and y = x

11. y = x - 1 and y = x 2 - 3

2.

y = x 2 and 2x + y = 0

12. y = x 2 + 1 and y = 1 - x 2

3.

x 2 + y 2 = 9 and x + y = 3

13. y = x 2 - 3x + 7 and y = 2x + 3

4.

x - y = 7 and xy = -12

14. xy = 1 and 4x - y + 3 = 0

5.

y = x 2 + 4x and 2x - y - 1 = 0

15. h = t 2 and h = ] t + 1 g2

6.

y = x 2 and 6x - y - 9 = 0

16. x + y = 2 and 2x 2 + xy - y 2 = 8

7.

x = t 2 and x + t - 2 = 0

17. y = x 3 and y = x 2 + 6x

8.

m 2 + n 2 = 16 and m + n + 4 = 0

18. y = | x | and y = x 2

9.

xy = 2 and y = 2x

19. y = x 2 - 7x + 6 and 24x + 4y - 23 = 0

10. y = x 3 and y = x 2

20. x 2 + y 2 = 1 and 5x + 12y + 13 = 0

Equations with 3 unknown variables Four unknowns need 4 equations, and so on.

Three equations can be solved simultaneously to find 3 unknown pronumerals.

EXAMPLE Solve simultaneously a - b + c = 7, a + 2b - c = -4 and 3a - b - c = 3.

Solution a-b +c=7 a + 2b - c = - 4 3a - b - c = 3 (1) + (2): a-b+c=7 a + 2b - c = - 4 2a + b =3 (1) + (3): a- b+c=7 3a - b - c = 3 4a - 2b = 10 or 2a - b =5 (4) + (5): 2a + b =3 4a =8 a=2

(1 ) (2) (3)

( 4)

(5)

Chapter 3 Equations

Substitute a = 2 in (4) 2 ( 2) + b = 3 4+b=3 b = -1 Substitute a = 2 and b = -1 in (1) 2 - (-1) + c = 7 2 +1 + c = 7 3+c=7 c=4 ` solution is a = 2, b = -1, c = 4

3.14

You will solve 3 simultaneous equations in later topics (for example, in Chapter 10).

Exercises

Solve the simultaneous equations. 1.

x = - 2, 2x - y = 4 and x - y + 6z = 0

2.

a = - 2, 2a - 3b = -1 and a - b + 5c = 9

3.

2a + b + c = 1, a + b = - 2 and c = 7

4.

a + b + c = 0, a - b + c = - 4 and 2a - 3b - c = -1

5.

x + y - z = 7, x + y + 2z = 1 and 3x + y - 2z = 19

6.

x - y - z = 1, 2x + y - z = -9 and 2x - 3y - 2z = 7

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7.

2p + 5q - r = 25, 2p - 2q - r = -24 and 3p - q + 5r = 4

8.

2x - y + 3z = 9, 3x + y - 2z = -2 and 3x - y + 5z = 14

9.

3h + j - k = -3, h + 2j + k = -3 and 5h - 3j - 2k = -13

10. 2a - 7b + 3c = 7, a + 3b + 2c = -4 and 4a + 5b - c = 9

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Test Yourself 3 1.

Solve (a) 8 = 3b - 22 a a+2 (b) =9 4 3 (c) 4 (3x + 1) = 11x - 3 -4 (d) #3 x+3 (e) 3p + 1 # p + 9

2.

3.

The compound interest formula is r n m . Find correct to 2 A = P c1 + 100 decimal places. (a) A when P = 1000, r = 6 and n = 4 (b) P when A = 12 450, r = 5.5 and n = 7 Complete the square on (a) x 2 - 8x (b) k 2 + 4k

9.

Solve -2 1 3y + 1 # 10, and plot your solution on a number line.

10. Solve correct to 3 significant figures (a) x 2 + 7x + 2 = 0 (b) y 2 - 2y - 9 = 0 (c) 3n 2 + 2n - 4 = 0 11. The surface area of a sphere is given by A = 4rr 2 . Evaluate to 1 decimal place (a) A when r = 7.8 (b) r when A = 102.9 12. Solve

x-3 3 - 2 9. 7 4

13. Solve x 2 - 11x + 18 2 0. 14. Solve the simultaneous equations x 2 + y 2 = 16 and 3x + 4y - 20 = 0. 4 3 rr . 3 Evaluate to 2 significant figures (a) V when r = 8 (b) r when V = 250

15. The volume of a sphere is V =

4.

Solve these simultaneous equations. (a) x - y + 7 = 0 and 3x - 4y + 26 = 0 (b) xy = 4 and 2x - y - 7 = 0

5.

Solve (a) 3 x + 2 = 81 (b) 16 y = 2

6.

Solve (a) 3b - 1 = 5

(a) x 2 - 6x + 9 = 0

(b) 5g - 3 = 3g + 1

(c) x - 2 = 7 - x

(c) 2x - 7 $ 1

(d) x 2 - x + 4 = 0

7.

8.

The area of a trapezium is given by A = 1 h (a + b). Find 2 (a) A when h = 6, a = 5 and b = 7 (b) b when A = 40, h = 5 and a = 4. Solve 2x 2 - 3x + 1 = 0 by (a) factorisation (b) quadratic formula.

16. Which of the following equations has (i) 2 solutions (ii) 1 solution (iii) no solutions? (b) 2x - 3 = 7

(e) 2x + 1 = x - 2 17. Solve simultaneously a + b = 5, 2 a + b + c = 4, a - b - c = 5. 18. Solve 3n + 5 2 5, and plot the solution on a number line. 19. Solve

3 4 =x x+1

^ x ! 0, -1 h .

Chapter 3 Equations

20. Solve 9 2x + 1 = 27 x .

(k) 27 2x - 1 = 9 (l) 4b - 3 # 5 (m) 3x + 2 = 2x - 3 (n) 4t - 5 = t + 2 (o) x 2 1 2x + 3 (p) m 2 + m $ 6 2t - 3 (q) 15 t y+1 (r) 22 y-1 n (s) $3 2n - 4 3x - 2 (t) # -1 2x + 1

21. Solve (a) 2 ^ 3y - 5 h 2 y + 5 (b) n 2 + 3n # 0 (c) 3 2x - 1 = 27 (d) 5x 3 - 1 = 39 (e) 5x - 4 = 11 (f) 2t + 1 $ 3 (g) x 2 + 2x - 8 # 0 (h) 8 x + 1 = 4 x (i) y 2 - 4 2 0 (j) 1 - x 2 # 0

Challenge Exercise 3 1 . a2

1.

Find the value of y if a 3y - 5 =

2.

Solve x 2 a .

3.

The solutions of x 2 - 6x - 3 = 0 are in the form a + b 3 . Find the values of a and b.

4.

5. 6.

7.

8.

9.

2

2

2 1 = 1 correct to 3 x -1 x +1 significant figures. (x ! ! 1) y2 - 6 Solve # 1. y Solve

11. Solve ] x - 4 g ] x - 1 g # 28. 3

12. Solve x 2 =

1 . 8

13. The volume of a sphere is given by 4 V = rr 3 . Find the value of r when 3 V = 51.8 (correct to three significant figures). 14. Solve x - 3 + x + 4 = x - 2 . 15. Find the solutions of x 2 - 2ax - b = 0 by completing the square.

Factorise x 5 - 9x 3 - 8x 2 + 72. Hence solve x 5 - 9x 3 - 8x 2 + 72 = 0.

16. Solve

Solve simultaneous equations y = x 3 + x 2 and y = x + 1.

17. Given A = P c 1 +

Find the value of b if x 2 - 8x + b 2 is a perfect square. Hence solve x 2 - 8x - 1 = 0 by completing the square.

18. Solve 3x 2 = 8 (2x - 1) and write the solution in the simplest surd form.

Considering the definition of absolute x-3 value, solve = x, where x ! 3. 3-x

10. Solve t + 2 + 3t - 1 1 5.

6y 2 # - 3. 3y - 2

r n m , find P 100 correct to 2 decimal places when A = 3281.69, r = 1.27 and n = 30.

19. Solve

5x + 3 2 2 x. x+4

20. Solve 3y - 1 + 2y + 3 2 5.

139

4 Geometry 1 TERMINOLOGY Altitude: Height. Any line segment from a vertex to the opposite side of a polygon that is perpendicular to that side

Polygon: General term for a many sided plane figure. A closed plane (two dimensional) figure with straight sides

Congruent triangles: Identical triangles that are the same shape and size. Corresponding sides and angles are equal. The symbol is /

Quadrilateral: A four-sided closed figure such as a square, rectangle, trapezium etc.

Interval: Part of a line including the endpoints

Similar triangles: Triangles that are the same shape but different sizes. The symbol is zy

Median: A line segment that joins a vertex to the opposite side of a triangle that bisects that side

Vertex: The point where three planes meet. The corner of a figure

Perpendicular: A line that is at right angles to another line. The symbol is =

Vertically opposite angles: Angles that are formed opposite each other when two lines intersect

Chapter 4 Geometry 1

INTRODUCTION GEOMETRY IS USED IN many areas, including surveying, building and graphics.

These fields all require a knowledge of angles, parallel lines and so on, and how to measure them. In this chapter, you will study angles, parallel lines, triangles, types of quadrilaterals and general polygons. Many exercises in this chapter on geometry need you to prove something or give reasons for your answers. The solutions to geometry proofs only give one method, but other methods are also acceptable.

DID YOU KNOW? Geometry means measurement of the earth and comes from Greek. Geometry was used in ancient civilisations such as Babylonia. However, it was the Greeks who formalised the study of geometry, in the period between 500 BC and AD 300.

Notation In order to show reasons for exercises, you must know how to name figures correctly. •B The point is called B.

The interval (part of a line) is called AB or BA.

If AB and CD are parallel lines, we write AB < CD.

This angle is named +BAC or +CAB. It can sometimes be named +A. ^

Angles can also be written as BAC or BAC.

This triangle is named 3ABC.

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To name a quadrilateral, go around it: for example, BCDA is correct, but ACBD is not.

Producing a line is the same as extending it.

This quadrilateral is called ABCD.

Line AB is produced to C.

+ABD and +DBC are equal.

DB bisects +ABC.

AM is a median of D ABC.

AP is an altitude of D ABC.

Types of Angles Acute angle

0c1 xc1 90c

Chapter 4 Geometry 1

Right angle

A right angle is 90c. Complementary angles are angles whose sum is 90c.

Obtuse angle

90c1 xc1180c

Straight angle

A straight angle is 180c. Supplementary angles are angles whose sum is 180c.

Reflex angle

180c1 xc1 360c

Angle of revolution

An angle of revolution is 360c.

Vertically opposite angles

+AEC and +DEB are called vertically opposite angles. +AED and +CEB are also vertically opposite angles.

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Vertically opposite angles are equal. That is, +AEC = +DEB and +AED = +CEB.

Proof Let +AEC = xc Then +AED = 180c - xc (+CED straight angle, 180c) Now +DEB = 180c - (180c - xc) (+AEB straight angle, 180c) = xc Also +CEB = 180c - xc (+CED straight angle, 180c) ` +AEC = +DEB and +AED =+CEB

EXAMPLES Find the values of all pronumerals, giving reasons. 1.

Solution x + 154 = 180 (+ABC is a straight angle, 180c) x + 154 - 154 = 180 - 154 ` x = 26 2.

Solution 2x + 142 + 90 2x + 232 2x + 232 - 232 2x 2x 2 x

= 360 (angle of revolution, 360c ) = 360 = 360 - 232 = 128 128 = 2 = 64

Chapter 4 Geometry 1

3.

Solution y + 2y + 30 = 90 3y + 30 3y + 30 - 30 3y 3y 3 y

(right angle, 90c)

= 90 = 90 - 30 = 60 60 3 = 20 =

4.

Solution x + 50 = 165 x + 50 - 50 = 165 - 50 x = 115 y = 180 - 165 = 15 w = 15

(+WZX and +YZV vertically opposite)

(+XZY straight angle, 180c) (+WZY and +XZV vertically opposite)

5.

CONTINUED

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Solution a = 90 b + 53 + 90 = 180 b + 143 = 180 b + 143 - 143 = 180 - 143 b = 37 d = 37 c = 53

(vertically opposite angles) (straight angle, 180c)

(vertically opposite angles) (similarly)

6. Find the supplement of 57c 12l.

Solution Supplementary angles add up to 180c. So the supplement of 57c 12l is 180c - 57c 12l = 122c 48l. 7. Prove that AB and CD are straight lines. A

D

(x + 30)c C

(6x + 10)c

(2x 2 + 10)c E (5x + 30)c B

Solution 6x + 10 + x + 30 + 5x + 30 + 2x + 10 = 360 ^ angle of revolution h 14x + 80 - 80 = 360 - 80 14x = 280 14x 280 = 14 14 x = 20 +AEC = (20 + 30)c = 50c +DEB = (2 # 20 + 10)c = 50c These are equal vertically opposite angles. ` AB and CD are straight lines

Chapter 4 Geometry 1

4.1 Exercises 1.

Find values of all pronumerals, giving reasons. (a)

yc

(i)

133c

(b)

(j)

(c) 2.

Find the supplement of (a) 59c (b) 107c 31l (c) 45c 12l

3.

Find the complement of (a) 48c (b) 34c 23l (c) 16c 57l

4.

Find the (i) complement and (ii) supplement of (a) 43c (b) 81c (c) 27c (d) 55c (e) 38c (f) 74c 53l (g) 42c 24l (h) 17c 39l (i) 63c 49l (j) 51c 9l

5.

(a) Evaluate x. (b) Find the complement of x. (c) Find the supplement of x.

(d)

(e)

(f)

(g)

(h) (2x + 30)c 142c

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6.

Find the values of all pronumerals, giving reasons for each step of your working.

8.

(a) Prove that CD bisects +AFE. 9.

Prove that AC is a straight line. D C

(b) (3x + 70)c (110 - 3x)c B

(c) A

10. Show that +AED is a right angle. A

(d)

B

(50 - 8y)c

(e)

C

(5y - 20)c

E

(f)

7.

Prove that AC and DE are straight lines.

(3y + 60)c

D

Chapter 4 Geometry 1

149

Parallel Lines When a transversal cuts two lines, it forms pairs of angles. When the two lines are parallel, these pairs of angles have special properties.

Alternate angles

Alternate angles form a Z shape. Can you find another set of alternate angles?

If the lines are parallel, then alternate angles are equal.

Corresponding angles

Corresponding angles form an F shape. There are 4 pairs of corresponding angles. Can you find them?

If the lines are parallel, then corresponding angles are equal.

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Cointerior angles Cointerior angles form a U shape. Can you find another pair?

If the lines are parallel, cointerior angles are supplementary (i.e. their sum is 180c).

Tests for parallel lines

If alternate angles are equal, then the lines are parallel.

If +AEF = +EFD, then AB < CD.

If corresponding angles are equal, then the lines are parallel.

If +BEF = +DFG, then AB < CD.

If cointerior angles are supplementary, then the lines are parallel.

If +BEF + +DFE = 180c, then AB < CD.

Chapter 4 Geometry 1

If 2 lines are both parallel to a third line, then the 3 lines are parallel to each other. That is, if AB < CD and EF < CD, then AB < EF.

EXAMPLES 1. Find the value of y, giving reasons for each step of your working.

Solution +AGF = 180c - 125c = 55c

(+FGH is a straight angle)

`

(+AGF, +CFE corresponding angles, AB < CD)

y = 55c

2. Prove EF < GH.

Solution +CBF = 180c - 120c (+ABC is a straight angle) = 60c ` +CBF = +HCD = 60c But +CBF and +HCD are corresponding angles ` EF < GH

Can you prove this in a different way?

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Think about the reasons for each step of your calculations.

4.2 Exercises 1.

Find values of all pronumerals. (a)

(h)

(i)

(b) (j)

(c)

2.

Prove AB < CD. (a)

(d)

(b) (e)

(c)

A

(f)

(g)

B

104c

C 76c

D

E

Chapter 4 Geometry 1

A

(d)

(e)

B 138c

B

52c

E C

C

E 128c

D

23c

F 115c

G

H

F

Types of Triangles Names of triangles A scalene triangle has no two sides or angles equal.

A right (or right-angled) triangle contains a right angle.

The side opposite the right angle (the longest side) is called the hypotenuse. An isosceles triangle has two equal sides. The angles (called the base angles) opposite the equal sides in an isosceles triangle are equal.

An equilateral triangle has three equal sides and angles.

A D

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All the angles are acute in an acute-angled triangle.

An obtuse-angled triangle contains an obtuse angle.

Angle sum of a triangle

The sum of the interior angles in any triangle is 180c, that is, a + b + c = 180

Proof

Let +YXZ = ac, +XYZ = bc and +YZX = cc Draw line AB < YZ Then +BXZ = cc (+BXZ, +XZY alternate angles, AB < YZ) +AXY = bc (similarly) +YXZ + +AXY + +BXZ = 180c (+AXB is a straight angle) ` a + b + c = 180

Chapter 4 Geometry 1

Class Investigation 1. 2. 3. 4. 5.

Could you prove the base angles in an isosceles triangle are equal? Can there be more than one obtuse angle in a triangle? Could you prove that each angle in an equilateral triangle is 60c? Can a right-angled triangle be an obtuse-angled triangle? Can you find an isosceles triangle with a right angle in it?

Exterior angle of a triangle

The exterior angle in any triangle is equal to the sum of the two opposite interior angles. That is, x+y=z

Proof

Let +ABC = xc , +BAC = yc and +ACD = zc Draw line CE < AB zc = +ACE + +ECD +ECD = xc +ACE = yc ` z=x+y

(+ECD,+ABC corresponding angles, AB < CE) (+ACE,+BAC alternate angles, AB < CE)

EXAMPLES Find the values of all pronumerals, giving reasons for each step. 1.

CONTINUED

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Solution x + 53 + 82 = 180 (angle sum of D 180c) x + 135 = 180 x + 135 - 135 = 180 - 135 x = 45 2.

Solution +A = +C = x x + x + 48 = 180 2x + 48 = 180 2x + 48 - 48 = 180 - 48 2x = 132 132 2x = 2 2 x = 66

(base angles of isosceles D) (angle sum in a D 180c)

3.

Solution y + 35 = 141 (exterior angle of D) y + 35 - 35 = 141 - 35 ` y = 106 This example can be done using the interior sum of angles. +BCA = 180c - 141c = 39c y + 39 + 35 = 180 y + 74 = 180 y + 74 - 74 = 180 - 74 ` y = 106

(+BCD is a straight angle 180c) (angle sum of D 180c)

Chapter 4 Geometry 1

Think of the reasons for each step of your calculations.

4.3 Exercises 1.

Find the values of all pronumerals. (a)

(h)

(b) (i)

(j) (c)

(d)

(k)

(e)

(f)

(g)

157

2.

Show that each angle in an equilateral triangle is 60c.

3.

Find +ACB in terms of x.

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4.

5.

6.

Prove AB < ED.

(d)

8.

Prove D IJL is equilateral and D JKL is isosceles.

9.

In triangle BCD below, BC = BD. Prove AB ED.

Show D ABC is isosceles.

Line CE bisects +BCD. Find the value of y, giving reasons.

A B C

46c E 88c

D

7.

Evaluate all pronumerals, giving reasons for your working. (a)

10. Prove that MN QP . 32c

M

(b) 75c

O

73c

Q

(c)

P

N

Chapter 4 Geometry 1

Congruent Triangles Two triangles are congruent if they are the same shape and size. All pairs of corresponding sides and angles are equal. For example:

We write D ABC / D XYZ.

Tests To prove that two triangles are congruent, we only need to prove that certain combinations of sides or angles are equal.

Two triangles are congruent if • SSS: all three pairs of corresponding sides are equal • SAS: two pairs of corresponding sides and their included angles are equal • AAS: two pairs of angles and one pair of corresponding sides are equal • RHS: both have a right angle, their hypotenuses are equal and one other pair of corresponding sides are equal

EXAMPLES 1. Prove that DOTS / DOQP where O is the centre of the circle.

CONTINUED

The included angle is the angle between the 2 sides.

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Solution S: A: S:

OS = OQ +TOS = +QOP OT = OP

`

by SAS, DOTS / DOQP

(equal radii) (vertically opposite angles) (equal radii)

2. Which two triangles are congruent?

Solution To find corresponding sides, look at each side in relation to the angles. For example, one set of corresponding sides is AB, DF, GH and JL. D ABC / D JKL (by SAS) 3. Show that triangles ABC and DEC are congruent. Hence prove that AB = ED.

Solution A: +BAC = +CDE A: +ABC = +CED S: AC = CD

(alternate angles, AB < ED) (similarly) (given)

` by AAS, D ABC / D DEC ` AB = ED

(corresponding sides in congruent D s)

Chapter 4 Geometry 1

4.4 Exercises 1.

Are these triangles congruent? If they are, prove that they are congruent. (a)

2.

Prove that these triangles are congruent. (a)

B

(b)

Y 4.7

m

110c

2.3

4.7

m

m

Z

110 c C

A 2

.3 m

(b)

X

(c)

(c)

(d)

(d) (e)

(e)

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3.

A

Prove that (a) Δ ABD is congruent to Δ ACD (b) AB bisects BC, given D ABC is isosceles with AB = AC.

D

B

4.

Prove that triangles ABD and CDB are congruent. Hence prove that AD = BC.

C

(a) Prove that TABC and TADC are congruent. (b) Show that +ABC = +ADC. The centre of a circle is O and AC is perpendicular to OB.

7.

A

5.

In the circle below, O is the centre of the circle. A

O

D

B O

C

B

C

(a) Prove that TOAB and TOCD are congruent. (b) Show that AB = CD. 6.

(a) Show that TOAB and TOBC are congruent. (b) Prove that +ABC = 90c. ABCF is a trapezium with AF = BC and FE = CD. AE and BD are perpendicular to FC.

8.

In the kite ABCD, AB = AD and BC = DC.

F

A

B

E

D

C

(a) Show that TAFE and TBCD are congruent. (b) Prove that +AFE = +BCD.

Chapter 4 Geometry 1

9.

The circle below has centre O and OB bisects chord AC.

10. ABCD is a rectangle as shown below. A

B

D

C

C O B

A

(a) Prove that TOAB is congruent to TOBC. (b) Prove that OB is perpendicular to AC.

(a) Prove that TADC is congruent to TBCD. (b) Show that diagonals AC and BD are equal.

Investigation The triangle is used in many structures, for example trestle tables, stepladders and roofs. Find out how many different ways the triangle is used in the building industry. Visit a building site, or interview a carpenter. Write a report on what you find.

Similar Triangles Triangles, for example ABC and XYZ, are similar if they are the same shape but different sizes. As in the example, all three pairs of corresponding angles are equal. All three pairs of corresponding sides are in proportion (in the same ratio).

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Maths In Focus Mathematics Extension 1 Preliminary Course

We write: D ABC