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YEAR 11 2 Unit Second Edition CAMBRIDGE Mathematics BILL PENDER DAVID SADLER JULIA SHEA DEREK WARD COLOUR VERSION WITH STUDENT CD-ROM Now in colour with an electronic version of the book on CD

CAMBRIDGE UNIVERSITY PRESS Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo, Delhi Cambridge University Press 477 Williamstown Road, Port Melbourne, VIC 3207, Australia www.cambridge.edu.au Information on this title: www.cambridge.org/9780521147170 © Bill Pender, David Sadler, Julia Shea, Derek Ward 2009 First edition 1999 Reprinted 2001, 2004 Second edition 2005 Colour version 2009 Cover design by Sylvia Witte Typeset by Aptara Corp. Printed in China by Printplus National Library of Australia Cataloguing in Publication data Bill Pender Cambridge mathematics 2 unit : year 11 / Bill Pender … [et al.] . 2 nd ed. 9780521147170 (pbk.) Includes index. For secondary school age. Mathematics. Mathematics--Problems, exercises, etc. Sadler, David. Shea, Julia. Ward, Derek. 510 ISBN 978-0-521-14717-0 paperback 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 publication, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL) under the Act. For details of the CAL licence for educational institutions contact: Copyright Agency Limited Level 15, 233 Castlereagh Street Sydney NSW 2000 Telephone: (02) 9394 7600 Facsimile: (02) 9394 7601 Email: info@copyright.com.au Reproduction and Communication for other purposes Except as permitted under the Act (for example a fair dealing for the purposes of study, research, criticism or review) no part of this publication may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher at the address above. Cambridge University Press has no responsibility for the persistence or accuracy of URL S for external or third-party internet websites referred to in this publication and does not guarantee that any content on such websites is, or will remain, accurate or appropriate. Information regarding prices, travel timetables and other factual information given in this work are correct at the time of first printing but Cambridge University Press does not guarantee the accuracy of such information thereafter. Student CD-ROM licence Please see the file 'licence.txt' on the Student CD-ROM that is packed with this book.

Contents Preface .......................................... vi How to Use This Book .................................. vii About the Authors .................................... x Chapter One — Methods in Algebra .......................... 1 1A ArithmeticwithPronumerals ....................... 1 1B ExpandingBrackets ............................ 4 1C Factoring .................................. 6 1D AlgebraicFractions ............................. 9 1E FactoringtheSumandDifferenceofCubes ............... 13 1F Solving Linear Equations . . ........................ 14 1G Solving Quadratic Equations....................... 17 1H Solving Simultaneous Equations . . .................... 20 1I CompletingtheSquare ........................... 23 1J ChapterReviewExercise .......................... 25 Chapter Two — Numbers and Surds .......................... 27 2A IntegersandRationalNumbers ...................... 27 2B TerminatingandRecurringDecimals ................... 30 2C RealNumbersandApproximations .................... 33 2D SurdsandtheirArithmetic ........................ 36 2E FurtherSimplificationofSurds ...................... 38 2F RationalisingtheDenominator ...................... 40 2G ChapterReviewExercise .......................... 42 Chapter Three — Functions and their Graphs ..................... 45 3A FunctionsandRelations .......................... 45 3B ReviewofLinearGraphs .......................... 50 3C ReviewofQuadraticGraphs ........................ 52 3D Higher Powers ofxandCircles ...................... 55 3E TwoGraphsthathaveAsymptotes .................... 57 3F TransformationsofKnownGraphs .................... 59 3G ChapterReviewExercise .......................... 63 Chapter Four — Graphs and Inequations ........................ 66 4A InequationsandInequalities ........................ 66 4B Solving Quadratic Inequations....................... 69 4C InterceptsandSign ............................. 70 4D OddandEvenSymmetry ......................... 74 4E TheAbsoluteValueFunction ....................... 77

  iv Contents 4F UsingGraphstoSolveEquationsandInequations ............ 81 4G RegionsintheNumberPlane ....................... 85 4H ChapterReviewExercise .......................... 89 Chapter Five — Trigonometry .............................. 92 5A TrigonometrywithRight-AngledTriangles ................ 92 5B Problems Involving Right-Angled Triangles ............... 97 5C TrigonometricFunctionsofaGeneralAngle ...............101 5D TheQuadrant,theRelatedAngleandtheSign .............105 5E GivenOneTrigonometricFunction,FindAnother............111 5F TrigonometricIdentities ..........................113 5G TrigonometricEquations ..........................117 5H TheSineRuleandtheAreaFormula ...................123 5I TheCosineRule ..............................129 5J Problems Involving General Triangles ...................133 5K ChapterReviewExercise ..........................138 Appendix—ProvingtheSine,CosineandAreaRules .........141 Chapter Six — Coordinate Geometry ..........................143 6A LengthsandMidpointsofIntervals ....................143 6B GradientsofIntervalsandLines ......................147 6C EquationsofLines .............................153 6D FurtherEquationsofLines .........................156 6E PerpendicularDistance ...........................161 6F Lines Through the Intersection of Two Given Lines...........164 6G CoordinateMethodsinGeometry .....................167 6H ChapterReviewExercise ..........................170 Appendix—TheProofsofTwoResults .................172 Chapter Seven — Indices and Logarithms .......................174 7A Indices ....................................174 7B FractionalIndices ..............................179 7C Logarithms .................................182 7D The Laws for Logarithms.........................185 7E Equations Involving Logarithms and Indices ...............188 7F Graphs of Exponential and Logarithmic Functions ...........190 7G ChapterReviewExercise ..........................193 Chapter Eight — Sequences and Series ........................195 8A SequencesandHowtoSpecifyThem ...................195 8B ArithmeticSequences ............................199 8C GeometricSequences ............................203 8D Solving Problems about APs and GPs..................207 8E AddingUptheTermsofaSequence ...................211 8F SumminganArithmeticSeries ......................214 8G SummingaGeometricSeries .......................218 8H TheLimitingSumofaGeometricSeries .................222 8I RecurringDecimalsandGeometricSeries ................227 8J ChapterReviewExercise ..........................228

  Contents v Chapter Nine — The Derivative .............................230 9A TheDerivative—GeometricDefinition .................230 9B TheDerivativeasaLimit .........................233 9C A Rule for Differentiating Powers ofx..................237 9D Tangents and Normals — The Notationdy dx...............241 9E DifferentiatingPowerswithNegativeIndices ...............246 9F DifferentiatingPowerswithFractionalIndices ..............249 9G TheChainRule ...............................251 9H TheProductRule..............................254 9I TheQuotientRule .............................257 9J LimitsandContinuity ...........................259 9K Differentiability . . .............................263 9L ChapterReviewExercise ..........................265 Appendix—ProvingSomeRulesforDifferentiation ..........267 Chapter Ten — The Quadratic Function ........................269 10AFactoringandtheGraph ..........................269 10BCompletingtheSquareandtheGraph ..................274 10CTheQuadraticFormulaeandtheGraph .................278 10DEquationsReducibletoQuadratics ....................280 10EProblemsonMaximisationandMinimisation ..............281 10FTheTheoryoftheDiscriminant ......................285 10G Definite and Indefinite Quadratics....................289 10HSumandProductofRoots .........................292 10I QuadraticIdentities ............................295 10J ChapterReviewExercise ..........................298 Appendix—IdenticallyEqualQuadratics ................300 Chapter Eleven — Locus and the Parabola .......................301 11AALocusanditsEquation .........................301 11BTheGeometricDefinitionoftheParabola ................305 11CTranslationsoftheParabola ........................309 11DChapterReviewExercise ..........................312 Chapter Twelve — The Geometry of the Derivative ...................313 12AIncreasing,DecreasingandStationaryataPoint ............313 12BStationaryPointsandTurningPoints...................318 12CSecondandHigherDerivatives ......................322 12DConcavityandPointsofInflexion .....................324 12EAReviewofCurveSketching .......................329 12FGlobalMaximumandMinimum......................333 12GApplicationsofMaximisationandMinimisation .............335 12HPrimitiveFunctions .............................341 12I ChapterReviewExercise ..........................346 Answers to Exercises ..................................348 Index ...........................................415

  Preface This textbook has been written for students in Years 11 and 12 taking the 2 Unit calculus course ‘Mathematics’ for the NSW HSC. The book covers all the content of the course at the level required for the HSC examination. There are two volumes — the present volume is roughly intended for Year 11, and the next volume for Year 12. Schools will, however, differ in their choices of order of topics and in their rates of progress. Although the Syllabus has not been rewritten for the new HSC, there has been a gradual shift of emphasis in recent examination papers. •The interdependence of the course content has been emphasised. •Graphs have been used much more freely in argument. •Structured problem solving has been expanded. •There has been more stress on explanation and proof. This text addresses these new emphases, and the exercises contain a wide variety of different types of questions. There is an abundance of questions and problems in each exercise — too many for any one student — carefully grouped in three graded sets, so that with proper selection, the book can be used at all levels of ability in the 2 Unit course. This new second editionhas been thoroughly rewritten to make it more acces- sible to all students. The exercises now have more early drill questions to reinforce each new skill, there are more worked exercises on each new algorithm, and some chapters and sections have been split into two so that ideas can be introduced more gradually. We have also added a review exercise to each chapter. We would like to thank our colleagues at Sydney Grammar School and Newington College for their invaluable help in advising us and commenting on the successive drafts. We would also like to thank the Headmasters of our two schools for their encouragement of this pro ject, and Peter Cribb, Chris Gray and the team at Cambridge University Press, Melbourne, for their support and help in discussions. Finally, our thanks go to our families for encouraging us, despite the distractions that the pro ject has caused to family life. Dr Bill Pender Sub ject Master in Mathematics Sydney Grammar School College Street Darlinghurst NSW 2010 David Sadler Mathematics Sydney Grammar SchoolJulia Shea Director of Curriculum Newington College 200 Stanmore Road Stanmore NSW 2048 Derek Ward Mathematics Sydney Grammar School

  How to Use This Book This book has been written so that it is suitable for the full range of 2 Unit students, whatever their abilities and ambitions. The Exercises: No-one should try to do all the questions!We have written long exercises so that everyone will find enough questions of a suitable standard — each student will need to select from them, and there should be plenty left for revision. The book provides a great variety of questions, and representatives of all types should be attempted. Each chapter is divided into a number of sections. Each of these sections has its own substantial exercise, subdivided into three groups of questions: Foundation:These questions are intended to drill the new content of the sec- tion at a reasonably straightforward level. There is little point in proceeding without mastery of this group. Development:This group is usually the longest. It contains more substantial questions, questions requiring proof or explanation, problems where the new content can be applied, and problems involving content from other sections and chapters to put the new ideas in a wider context. Challenge:Many questions in recent 2 Unit HSC examinations have been very demanding, and this section is intended to match the standard of those recent examinations. Some questions are algebraically challenging, some re- quire more sophistication in logic, some establish more difficult connections between topics, and some complete proofs or give an alternative approach. The Theory and the Worked Exercises: All the theory in the course has been properly developed, but students and their teachers should feel free to choose how thor- oughly the theory is presented in any particular class. It can often be helpful to learn a method first and then return to the details of the proof and explanation when the point of it all has become clear. The main formulae, methods, definitions and results have been boxed and num- bered consecutively through each chapter. They provide a bare summary only, and students are advised to make their own short summary of each chapter using the numbered boxes as a basis. The worked examples have been chosen to illustrate the new methods introduced in the section. They should provide sufficient preparation for the questions in the following exercise, but they cannot possibly cover the variety of questions that can be asked.

  viii How to Use This Book The Chapter Review Exercises: A Chapter Review Exercise has been added to each chapter of the second edition. These exercises are intended only as a basic review of the chapter — for harder questions, students are advised to work through more of the later questions in the exercises. The Order of the Topics: We have presented the topics in the order that we have found most satisfactory in our own teaching. There are, however, many effective orderings of the topics, and apart from questions that provide links between topics, the book allows all the flexibility needed in the many different situations that apply in different schools. We have left Euclidean geometry and probability until Chapter Seven of the Year 12 volume for two reasons. First, we believe that functions and calculus should be developed as early as possible because these are the fundamental ideas in the course. Secondly, the courses in Years 9 and 10 already develop most of the work in Euclidean geometry and probability, at least in an intuitive fashion, so that revisiting them in Year 12, with a greater emphasis now on proof in geometry, seems an ideal arrangement. Many students, however, will want to study geometry in Year 11. The publishers have therefore made this chapter available free on their website at www.cambridge.edu.au/education/2unitGeometry The two geometry chapters from the 3 Unit volume are also on the website. The Structure of the Course: Recent examination papers have made the interconnec- tions amongst the various topics much clearer. Calculus is the backbone of the course, and the two processes of differentiation and integration, inverses of each other, are the basis of most of the topics. Both processes are introduced as geo- metrical ideas — differentiation is defined using tangents, and integration using areas — but the subsequent discussions, applications and exercises give many other ways of understanding them. Besides linear functions, three groups of functions dominate the course: The Quadratic Functions:(Covered in the Year 11 volume) These func- tions are known from earlier years. They are algebraic representations of the parabola, and arise naturally when areas are being considered or a constant acceleration is being applied. They can be studied without calculus, but calculus provides an alternative and sometimes quicker approach. The Exponential and Logarithmic Functions:(Covered in the Year 12 volume) Calculus is essential for the study of these functions. We have begun the topic with the exponential function. This has the great advantage of emphasising the fundamental property that the exponential function with baseeis its own derivative — this is the reason why it is essential for the study of natural growth and decay, and therefore occurs in almost every application of mathematics. The logarithmic function, and its relationship with the rectangular hyperbolay=1/x, has been covered in a separate chapter. The Trigonometric Functions:(Covered in the Year 12 volume) Calculus is also essential for the study of the trigonometric functions. Their definitions, like the associated definition ofπ, are based on the circle. The graphs of the sine and cosine functions are waves, and they are essential for the study of all periodic phenomena.

  How to Use This Book ix Thus the three basic functions in the course,x 2,e xand sinx, and the related numb erseandπ, can all be developed from the three most basic degree-2 curves — the parabola, the rectangular hyperbola and the circle. In this way, everything in the course, whether in calculus, geometry, trigonometry, coordinate geometry or algebra, can easily be related to everything else. Algebra and Graphs: One of the chief purposes of the course, stressed heavily in re- cent examinations, is to encourage arguments that relate a curve to its equation. Algebraic arguments are constantly used to investigate graphs of functions. Con- versely, graphs are constantly used to solve algebraic problems. We have drawn as many sketches in the book as space allowed, but as a matter of routine, stu- dents should draw diagrams for most of the problems they attempt. It is because sketches can so easily be drawn that this type of mathematics is so satisfactory for study at school. Theory and Applications: Although this course develops calculus in a purely mathe- matical way, using geometry and algebra, its content is fundamental to all the sciences. In particular, the applications of calculus to maximisation, motion, rates of change and finance are all parts of the syllabus. The course thus allows students to experience a double view of mathematics, as a system of pure logic on the one hand, and an essential part of modern technology on the other. Limits, Continuity and the Real Numbers: This is a first course in calculus, and rigorous arguments about limits, continuity or the real numbers would be quite inappro- priate. Any such ideas required in this course are not difficult to understand intuitively. Most arguments about limits need only the limit lim x→∞ 1/x=0and occasionally the sandwich principle. Introducing the tangent as the limit of the secant is a dramatic new idea, clearly marking the beginning of calculus, and is quite accessible. The functions in the course are too well-behaved for continuity to be a real issue. The real numbers are defined geometrically as points on the number line, and any properties that are needed can be justified by appealing to intuitive ideas about lines and curves. Everything in the course apart from these subtle issues of ‘foundations’ can be proven completely. Technology: There is much discussion about what role technology should play in the mathematics classroom and what calculators or software may be effective. This is a time for experimentation and diversity. We have therefore given only a few specific recommendations about technology, but we encourage such investigation, and to this new colour version we have added some optional technology resources which can be accessed via the student CD in the back of the book. The graphs of functions are at the centre of the course, and the more experience and intuitive understanding students have, the better able they are to interpret the mathemat- ics correctly. A warning here is appropriate — any machine drawing of a curve should be accompanied by a clear understanding of why such a curve arises from the particular equation or situation.

 

  About the Authors Dr Bill Pender is Sub ject Master in Mathematics at Sydney Grammar School, where he has taught since 1975. He has an MSc and PhD in Pure Mathemat- ics from Sydney University and a BA (Hons) in Early English from Macquarie University. In 1973–74, he studied at Bonn University in Germany, and he has lectured and tutored at Sydney University and at the University of NSW, where he was a Visiting Fellow in 1989. He has been involved in syllabus development since the early 1990s — he was a member of the NSW Syllabus Committee in Mathematics for two years and of the subsequent Review Committee for the 1996 Years 9–10 Advanced Syllabus, and was later involved in the writing of the 2004 K–10 Mathematics Syllabus. He has recently been appointed to the Education Advisory Committee of the Australian Mathematical Sciences Institute and will be involved in writing the proposed AMSI National Mathematics Textbook. He is a regular presenter of inservice courses for AIS and MANSW, and plays piano and harpsichord. David Sadler is Second Master in Mathematics at Sydney Grammar School, where he has taught since 1980. He has a BSc from the University of NSW and an MA in Pure Mathematics and a DipEd from Sydney University. In 1979, he taught at Sydney Boys’ High School, and he was a Visiting Fellow at the University of NSW in 1991. Julia Shea is now Director of Curriculum at Newington College, having been appointed Head of Mathematics there in 1999. She has a BSc and DipEd from the University of Tasmania, she taught for six years at Rosny College, a State Senior College in Hobart, and was a member of the Executive Committee of the Mathematics Association of Tasmania for five years. She then taught for five years at Sydney Grammar School before moving to Newington College. Derek Ward has taught Mathematics at Sydney Grammar School since 1991 and is Master in Charge of Statistics. He has an MSc in Applied Mathematics and a BScDipEd, both from the University of NSW, where he was subsequently Senior Tutor for three years. He has an AMusA in Flute, and is a lay clerk at St James’, King Street, where he sings counter-tenor. He also does occasional solo work at various venues.

 ff The mathematician’s patterns, like the painter’s or the poet’s, must be beautiful. The ideas, like the colours or the words, must fit together in a harmonious way. Beauty is the first test. — The English mathematician G. H. Hardy (1877–1947)

CHAPTERONE Methods in Algebra Fluency in algebra, and particularly in factoring, is absolutely vital for everything in this course. Much of this chapter will be a review of earlier work, but several topics will probably be quite new, including: •the sum and difference of cubes in Section 1E •three simultaneous equations in three variables in Section 1H. 1AArithmetic with Pronumerals Apronumeralis a symbol that stands for a number. The pronumeral may stand for a known number, or for an unknown number, or it may be ava r i a b l e, standing for any one of a whole set of possible numbers. Pronumerals, being numbers, can therefore take part in all the operations that are possible with numbers, such as addition, subtraction, multiplication and division (except by zero). Like and Unlike Terms: Analgebraic expressionconsists of pronumerals, numbers and the operations of arithmetic. Here is an example: x 2+2x+3x 2−4x−3 This particular algebraic expression can besimplifiedby combininglike terms. •The two like termsx 2and 3x 2can be combined to give 4x 2. •Another pair of like terms 2xand−4xcan be combined to give−2x. •This yields threeunlike terms,4x 2,−2xand−3, which cannot be combined. WORKED EXERCISE : Simplify each expression by combining like terms. (a)7a+15−2a−20(b)x 2+2x+3x 2−4x−3 SOLUTION : (a) 7a+15−2a−20 = 5a−5(b)x 2+2x+3x 2−4x−3=4x 2−2x−3 Multiplying and Dividing: To simplify a product like 3y×(−6y), or a quotient like 10x 2y÷5y, work systematically through the signs, then the numerals, and then each pronumeral in turn. WORKED EXERCISE : Simplify these products and quotients. (a)3y×(−6y)(b)4ab×7bc(c)10x 2y÷5y SOLUTION : (a) 3y×(−6y)=−18y 2 (b) 4ab×7bc=28ab 2c(c) 10x 2y÷5y=2x 2

  2 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 2U NIT YEAR 11 Index Laws: Here are the standard laws for dealing with indices. They will be covered in more detail in Chapter Seven. 1 THE INDEX LAWS : •To multiply powers of the same base, add the indices:a xay=a x+y •To divide powers of the same base, subtract the indices:a x ay=a x−y •To raise a power to a power, multiply the indices: (a x)n=a xn •The power of a product is the product of the powers: (ab) x=a xbx •The power of a quotient is the quotient of the powers: a b x =a x bx In expressions with several factors, work systematically through the signs, then the numerals, and then each pronumeral in turn. WORKED EXERCISE : Use the index laws above to simplify each expression. (a)3x 4×4x 3 (b)(48x 7y3)÷(16x 5y3)(c)(3a 4)3 (d)(−5x 2)3×(2xy) 4 (e) 2x 3y 4 SOLUTION : (a) 3x 4×4x 3=12x 7 (multiplying powers of the same base) (b) (48x 7y3)÷(16x 5y3)=3x 2 (dividing powers of the same base) (c) (3a 4)3=27a 12 (raising a power to a power) (d) (−5x 2)3×(2xy) 4=−125x 6×16x 4y4 (two powers of products) =−2000x 10y4 (multiplying powers of the same base) (e) 2x 3y 4 =16x 4 81y 4 (a power of a quotient) Exercise1A 1.Simplify: (a)5x+3x(b)5x−3x(c)−5x+3x(d)−5x−3x 2.Simplify: (a)−2a+3a+4a(b)−2a−3a+4a(c)−2a−3a−4a(d)−2a+3a−4a 3.Simplify: (a)−2x+x (b)3y−y(c)3a−7a (d)−8b+5b(e)4x−(−3x) (f )−2 ab−ba(g)−3pq+7pq (h)−5abc−(−2abc) 4.Simplify: (a)6x+3−5x (b)−2+2y−1 (c)3a−7−a+4 (d)3x−2y+5x+6y(e)−8t+12−2t−17 (f )2a 2+7a−5a 2−3a (g)9x 2−7x+4−14x 2−5x−7 (h)3a−4b−2c+4a+2b−c+2a−b−2c

  CHAPTER 1: Methods in Algebra 1A Arithmetic with Pronumerals 3 5.Simplify: (a)−3a×2(b)−4a×(−3a)(c)a 2×a 3 (d)(a 2)3 6.Simplify: (a)−10a÷5(b)−24a÷(−8a)(c)a 7÷a 3 (d)7a 2÷7a 7.Simplify: (a)5x x(b)−7x 3 x(c)−12a 2b −ab(d)−27x 6y7z2 9x 3y3z 8.Simplify: (a)t 2+t 2 (b)t 2−t 2 (c)t 2×t 2 (d)t 2÷t 2 9.Simplify: (a)−6x+3x(b)−6x−3x(c)−6x×3x(d)−6x÷3x DEVELOPMENT 10.Ifa=−2, find the value of: (a)3a+2(b)a 3−a 2 (c)3a 2−a+4(d)a 4+3a 3+2a 2−a 11.Ifx=2 andy=−3, find the value of: (a)3x+2y(b)y 2−5x(c)8x 2−y 3 (d)x 2−3xy+2y 2 12.Subtract: (a)xfrom 3x(b)−xfrom 3x(c)2afrom−4a(d)−bfrom−5b 13.Multiply: (a)5aby 2 (b)6xby−3(c)−3abya (d)−2a 2by−3ab(e)4x 2by−2x 3 (f )−3p 2qby 2pq 3 14.Divide: (a)−2xbyx (b)3x 3byx 2 (c)x 3y2byx 2y(d)a 6x3by−a 2x3 (e)14a 5b4by−2a 4b (f )−50a 2b5c8by−10ab 3c2 15.Find the sum of: (a)x+y+z,2x+3y−2zand 3x−4y+z (b)2a−3b+c,15a−21b−8cand 24b+7c+3a (c)5ab+bc−3ca,ab−bc+caand−ab+2ca+bc (d)x 3−3x 2y+3xy 2,−2x 2y−xy 2−y 3andx 3+4y 3 16.Fr o m : (a)7x 2−5x+6 take 5x 2−3x+2 (b)4a−8b+ctakea−3b+5c(c)3a+b−c−dtake 6a−b+c−3d (d)ab−bc−cdtake−ab+bc−3cd 17.Subtract: (a)x 3−x 2+x+1 fromx 3+x 2−x+1 (b)3xy 2−3x 2y+x 3−y 3fromx 3+3x 2y+3xy 2+y 3 (c)b 3+c 3−2abcfroma 3+b 3−3abc (d)x 4+5+x−3x 3from 5x 4−8x 3−2x 2+7

  4 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 2U NIT YEAR 11 18.Simplify: (a)2a 2b4×3a 3b2 (b)−6ab 5×4a 3b3 (c)(−3a 3)2 (d)(−2a 4b)3 CHALLENGE 19.What must be added to 4x 3−3x 2+2 to give 3x 3+7x−6? 20.Simplify: (a)3a×3a×3a 3a+3a+3a(b)3c×4c 2×5c 3 3c 2+4c 2+5c 2 (c)ab 2×2b 2c3×3c 3a4 a3b3+2a 3b3+3a 3b3 21.Simplify: (a)(−2x 2)3 −4x(b)(3xy 3)3 3x 2y4 (c)(−ab) 3×(−ab 2)2 −a 5b3 (d)(−2a 3b2)2×16a 7b (2a 2b)5 22.Divide the product of (−3x 7y5)4and (−2xy 6)3by (−6x 3y8)2. 1BExpanding Brackets Expanding brackets is routine in arithmetic. For example, to calculate 7×61, 7×(60 + 1) = 7×60 + 7×1, which quickly gives the result 7×61 = 420 + 7 = 427. The algebraic version of this procedure can be written as: 2 EXPANDING BRACKETS IN ALGEBRA : a(x+y)=ax+ay WORKED EXERCISE : Expand and simplify each expression. (a)3x(4x−7)(b)5a(3−b)−3b(6−5a) SOLUTION : (a) 3x(4x−7) = 12x 2−21x(b) 5a(3−b)−3b(6−5a)=15a−5ab−18b+15ab =15a+10ab−18b Expanding the Product of Two Bracketed Terms: Each pair of brackets should be ex- panded in turn and the reulting expression should then be simplified. WORKED EXERCISE : Expand and simplify each expression. (a)(x+3)(x−5)(b)(3 +x)(9 + 3x+x 2) SOLUTION : (a) (x+3)(x−5) =x(x−5) + 3(x−5) =x 2−5x+3x−15 =x 2−2x−15(b) (3 +x)(9 + 3x+x 2) =3(9+3x+x 2)+x(9 + 3x+x 2) =27+9x+3x 2+9x+3x 2+x 3 =27+18x+6x 2+x 3

  CHAPTER 1: Methods in Algebra 1B Expanding Brackets 5 Special Expansions: These three identities are important and must be memorised. Examples of these expansions occur very frequently, and knowing the formulae greatly simplifies the working. They are proven in the exercises. 3 SQUARE OF A SUM :(A+B) 2=A 2+2AB+B 2 SQUARE OF A DIFFERENCE :(A−B) 2=A 2−2AB+B 2 DIFFERENCE OF SQUARES :(A+B)(A−B)=A 2−B 2 WORKED EXERCISE :Use the three special expansions above to simplify: (a)(4x+1) 2 (b)(s−3t) 2 (c)(x+3y)(x−3y) SOLUTION : (a) (4x+1) 2=16x 2+8x+ 1 (the square of a sum) (b) (s−3t) 2=s 2−6st+9t 2 (the square of a difference) (c) (x+3y)(x−3y)=x 2−9y 2 (the difference of squares) Exercise1B 1.Expand: (a)3(x−2) (b)2(x−3) (c)−3(x−2)(d)−2(x−3) (e)−3(x+2) (f )−2(x+3)(g)−(x−2) (h)−(2−x) (i)−(x+3) 2.Expand: (a)3(x+y) (b)−2(p−q) (c)4(a+2b)(d)x(x−7) (e)−x(x−3) (f )−a(a+4)(g)5(a+3b−2c) (h) −3(2x−3y+5z) (i)xy(2x−3y) 3.Expand and simplify: (a)2(x+1)−x (b)3a+5+4(a−2) (c)2+2(x−3) (d)−3(a+2)+10(e)3−(x+1) (f )b+c−(b−c) (g)(2x−3y)−(3x−2y) (h)3(x−2)−2(x−5)(i)−4(a−b)−3(a+2b) (j)4(s−t) −5(s+t) (k)2x(x+6y)−x(x−5y) (l)−7(2a−3b+c)−6(−a+4b−2c) 4.Expand and simplify: (a)(x+2)(x+3) (b)(y+4)(y+7) (c)(t+6)(t−3) (d)(x−4)(x+2) (e)(t−1)(t−3)(f )(2a+3)(a+5) (g)(u−4)(3u+2) (h)(4p+ 5)(2p−3) (i)(2b −7)(b−3) (j)(5a−2)(3a+1)(k)(6−c)(c−3) (l)(2d−3)(4 +d) (m)(2x+3)(y−2) (n)(a−2)(5b+4) (o)(3−2m)(4−3n) DEVELOPMENT 5.(a)By expanding (A+B)(A+B), prove the special expansion (A+B) 2=A 2+2AB+B 2. (b)Similarly, prove the special expansions: (i) (A−B) 2=A 2−2AB+B 2 (ii) (A−B)(A+B)=A 2−B 2

  6 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 2U NIT YEAR 11 6.Use the special expansions to expand: (a)(x+y) 2 (b)(x−y) 2 (c)(x−y)(x+y) (d)(a+3) 2 (e)(b−4) 2 (f )(c+5) 2 (g)(d−6)(d+6) (h)(7 +e)(7−e) (i)(8 +f) 2 (j)(9−g) 2 (k)(h+ 10)(h−10) (l)(i+ 11) 2 (m)(2a+1) 2 (n)(2b−3) 2 (o)(3c+2) 2 (p)(2d+3e) 2 (q)(2f+3g)(2f−3g) (r)(3h−2i)(3h+2i)(s)(5j+4) 2 (t)(4k−5) 2 (u)(4 + 5m)(4−5m) (v)(5−3n) 2 (w)(7p+4q) 2 (x)(8−3r) 2 7.Expand: (a)−a(a 2−a−1) (b)−2x(x 3−2x 2−3x+1)(c)3xy(2x 2y−5x 3) (d)−2a 2b(a 2b3−2a 3b) 8.Simplify: (a)14− 10−(3x−7)−8x (b)4 a−2(b−c)− a−(b−2) 9.Expand and simplify: (a) t+1 t 2 (b) t−1 t 2 (c) t+1 t t−1 t CHALLENGE 10.Subtracta(b+c−a)fromthesumofb(c+a−b)andc(a+b−c). 11.Multiply: (a)a−2bbya+2b (b) 2−5xby 5 + 4x(c) 4x+ 7 by itself (d)x 2+3ybyx 2−4y(e)a+b−cbya−b (f ) 9x 2−3x+1 by 3x+1 12.Expand and simplify: (a) (a−b)(a+b)−a(a−2b) (b) (x+2) 2−(x+1) 2 (c) (a−3) 2−(a−3)(a+3)(d) (p+q) 2−(p−q) 2 (e) (2x+3)(x−1)−(x−2)(x+1) (f ) 3(a−4)(a−2)−2(a−3)(a−5) 13.Use the special expansions to find the value of: (a) 102 2 (b) 999 2 (c) 203×197 1CFactoring Factoringis the reverse process of expanding brackets, and will be needed on a routine basis throughout the course. There are four basic methods, but in every situation, common factors should always be taken out first. 4 THE FOUR BASIC METHODS OF FACTORING : • HIGHEST COMMON FACTOR :Always try this first. • DIFFERENCE OF SQUARES : This involves two terms. • QUADRATICS : This involves three terms. • GROUPING : This involves four or more terms. Factoring should continue until each factor isirreducible, meaning that it cannot be factored further.

  CHAPTER 1: Methods in Algebra1C Factor ing 7 Factoring by Taking Out the Highest Common Factor: Always look first for any common factors of all the terms, and then take out the highest common factor. WORKED EXERCISE : Factor each expression by taking out the highest common factor. (a)4x 3+3x 2 (b)18a 2b3−30b 3 SOLUTION : (a) The highest common factor of 4x 3and 3x 2isx 2, so 4x 3+3x 2=x 2(4x+3). (b) The highest common factor of 18a 2b3and 30b 3is 6b 3, so 18a 2b3−30b 3=6b 3(3a 2−5). Factoring by Difference of Squares: The expression must have two terms, both of which are squares. Sometimes a common factor must be taken out first. WORKED EXERCISE : Use the difference of squares to factor each expression. (a)a 2−36(b)80x 2−5y 2 SOLUTION : (a)a 2−36 = (a+6)(a−6) (b) 80x 2−5y 2=5(16x 2−y 2) (Take out the highest common factor.) =5(4x−y)(4x+y) (Use the difference of squares.) Factoring Monic Quadratics: A quadratic is calledmonicif the coefficient ofx 2is 1. Suppose that we want to factor the monic quadratic expressionx 2−13x+36. We l o o k f o r t w o n u m b e r s : •whose sum is−13 (the coefficient ofx), and •whose product is +36 (the constant term). WORKED EXERCISE : Factor these monic quadratics. (a)x 2−13x+36(b)a 2+12a−28 SOLUTION : (a) The numbers with sum−13 and product +36 are−9and−4, sox 2−13x+36 = (x−9)(x−4). (b) The numbers with sum +12 and product−28 are +14 and−2, soa 2+12a−28 = (a+ 14)(a−2). Factoring Non-monic Quadratics: In anon-monicquadratic like 2x 2+11x+ 12, where the coefficient ofx 2isnot1,welookfortwonumbers: •whose sum is 11 (the coefficient ofx), and •whose product is 12×2 = 24 (the constant times the coefficient ofx 2).

  8 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 2U NIT YEAR 11 WORKED EXERCISE : Factor these non-monic quadratics. (a)2x 2+11x+12(b)6s 2−11s−10 SOLUTION : (a) The numbers with sum 11 and product 12×2 = 24 are 8 and 3, so 2x 2+11x+12 = (2x 2+8x)+(3x+ 12) (Split 11xinto 8x+3x.) =2x(x+4)+3(x+ 4) (Take out the HCF of each group.) =(2x+3)(x+4).(x+ 4 is a common factor.) (b) The numbers with sum−11 and product−10×6=−60 are−15 and 4, so 6s 2−11s−10 = (6s 2−15s)+(4s−10) (Split−11sinto−15s+4s.) =3s(2s−5) + 2(2s−5) (Take out the HCF of each group.) =(3s+ 2)(2s−5).(2s−5 is a common factor.) Factoring by Grouping: When there are four or more terms, it is sometimes possible to split the expression into groups, factor each group in turn, and then factor the whole expression by taking out a common factor or by some other method. WORKED EXERCISE : Factor each expression by grouping. (a)12xy−9x−16y+12(b)s 2−t 2+s−t SOLUTION : (a) 12xy−9x−16y+12 = 3x(4y−3)−4(4y−3) (Take out the HCF of each pair.) =(3x−4)(4y−3) (4y−3 is a common factor.) (b)s 2−t 2+s−t=(s+t)(s−t)+(s−t) (Factors 2−t 2using difference of squares.) =(s−t)(s+t+1) (s−tis a common factor.) Exercise1C 1.Factor, by taking out any common factors: (a)2x+8 (b)6a−15 (c)ax−ay (d)20ab−15ac(e)x 2+3x (f )p 2+2pq (g)3a 2−6ab (h)12x 2+18x(i)20cd−32c (j)a 2b+b 2a (k)6a 2+2a 3 (l)7x 3y−14x 2y2 2.Factor, by grouping in pairs: (a)mp+mq+np+nq (b)ax−ay+bx−by (c)ax+3a+2x+6 (d)a 2+ab+ac+bc (e)z 3−z 2+z−1(f )ac+bc−ad−bd (g)pu−qu−pv+qv (h)x 2−3x−xy+3y (i)5p−5q−px+qx (j)2ax−bx−2ay+by(k)ab+ac−b−c (l)x 3+4x 2−3x−12 (m)a 3−3a 2−2a+6 (n)2t 3+5t 2−10t−25 (o)2x 3−6x 2−ax+3a 3.Factor, using the difference of squares: (a)a 2−1 (b)b 2−4 (c)c 2−9 (d)d 2−100(e)25−y 2 (f )1−n 2 (g)49−x 2 (h)144−p 2 (i)4c 2−9 (j)9u 2−1 (k)25x 2−16 (l)1−49k 2 (m)x 2−4y 2 (n)9a 2−b 2 (o)25m 2−36n 2 (p)81a 2b2−64

  CHAPTER 1: Methods in Algebra 1D Algebraic Fractions 9 4.Factor each quadratic expression. They are all monic quadratics. (a)a 2+3a+2 (b)k 2+5k+6 (c)m 2+7m+6 (d)x 2+8x+15 (e)y 2+9y+20 (f )t 2+12t+20(g)x 2−4x+3 (h)c 2−7c+10 (i)a 2−7a+12 (j)b 2−8b+12 (k)t 2+t−2 (l)u 2−u−2(m)w 2−2w−8 (n)a 2+2a−8 (o)p 2−2p−15 (p)y 2+3y−28 (q)c 2−12c+27 (r)u 2−13u+42(s)x 2−x−90 (t)x 2+3x−40 (u)t 2−4t−32 (v)p 2+9p−36 (w)u 2−16u−80 (x)t 2+23t−50 DEVELOPMENT 5.Factor each quadratic expression. They are all non-monic quadratics. (a)3x 2+4x+1 (b)2x 2+5x+2 (c)3x 2+16x+5 (d)3x 2+8x+4 (e)2x 2−3x+1 (f )5x 2−13x+6(g)5x 2−11x+6 (h)6x 2−11x+3 (i)2x 2−x−3 (j)2x 2+3x−5 (k)3x 2+2x−5 (l)3x 2+14x−5(m)2x 2−7x−15 (n)2x 2+x−15 (o)6x 2+17x−3 (p)6x 2−7x−3 (q)6x 2+5x−6 (r)5x 2+23x+12(s)5x 2+4x−12 (t)5x 2−19x+12 (u)5x 2−11x−12 (v)5x 2+28x−12 (w)9x 2−6x−8 (x)3x 2+13x−30 6.Use the techniques of the previous questions to factor each expression. (a)a 2−25 (b)b 2−25b (c)c 2−25c+ 100 (d)2d 2+25d+50 (e)e 3+5e 2+5e+25 (f )16−f 2 (g)16g 2−g 3 (h)h 2+16h+64(i)i 2−16i−36 (j)5j 2+16j−16 (k)4k 2−16k−9 (l)2k 3−16k 2−3k+24 (m)2a 2+ab−4a−2b (n)6m 3n4+9m 2n5 (o)49p 2−121q 2 (p)t 2−14t+40(q)3t 2+2t−40 (r)5t 2+54t+40 (s)5t 2+33t+40 (t)5t 3+10t 2+15t (u)u 2+15u−54 (v)3x 3−2x 2y−15x+10y (w)1−36a 2 (x)4a 2−12a+9 CHALLENGE 7.Write each expression as a product of three factors. (Take out any common factors first.) (a) 3a 2−12 (b)x 4−y 4 (c)x 3−x (d) 5x 2−5x−30(e) 25y−y 3 (f ) 16−a 4 (g) 4x 2+14x−30 (h)a 4+a 3+a 2+a(i)c 3+9c 2−c−9 (j)x 3−8x 2+7x (k)x 4−3x 2−4 (l)ax 2−a−2x 2+2 1DAlgebraic Fractions Analgebraic fractionis a fraction containing pronumerals. They are manipulated in the same way as arithmetic fractions, and factoring may play a ma jor role. Adding and Subtracting Algebraic Fractions: A common denominator is needed. Find- ing the lowest common denominator may involve factoring each denominator. 5 ADDING AND SUBTRACTING ALGEBRAIC FRACTIONS : •First factor every denominator. •Then work with the lowest common denominator.

  10 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 2U NIT YEAR 11 WORKED EXERCISE : Use a common denominator to simplify each algebraic fraction. (a)5x 6+11x 4(b)2 3x−3 5x(c)1 x−4−1 x(d)2+x x2−x−5 x−1 SOLUTION : (a)5x 6+11x 4=10x 12+33x 12 =43x 12 (b)2 3x−3 5x=10 15x−9 15x =1 15x(c)1 x−4−1 x=x−(x−4) x(x−4) =4 x(x−4) (d)2+x x2−x−5 x−1 =2+x x(x−1)−5 x−1 =2+x−5x x(x−1) =2−4x x(x−1) Cancelling Algebraic Fractions: The key step here is to factor the numerator and denominator completely before cancelling factors. 6 CANCELLING ALGEBRAIC FRACTIONS : •First factor the numerator and denominator. •Then cancel out all common factors. WORKED EXERCISE : Simplify each algebraic fraction. (a)6x+8 6(b)x 2−x x2−1 SOLUTION : (a)6x+8 6=2(3x+4) 6 =3x+4 3 (which could be written asx+ 43).(b)x 2−x x2−1=x(x−1) (x+1)(x−1) =x x+1 Multiplying and Dividing Algebraic Fractions: These processes are done exactly as for arithmetic fractions. 7 MULTIPLYING AND DIVIDING ALGEBRAIC FRACTIONS : •First factor all numerators and denominators completely. •Then cancel common factors. To divide by an algebraic fraction, multiply by its reciprocal.

  CHAPTER 1: Methods in Algebra 1D Algebraic Fractions 11 WORKED EXERCISE : Simplify these products and quotients of algebraic fractions. (a)2a a2−9×a−3 5a(b)12x x+1÷6x x2+2x+1 SOLUTION : (a)2a a2−9×a−3 5a=2a (a−3)(a+3)×a−3 5a(Factora 2−9.) =2 5(a+3)(Cancela−3anda.) (b)12x x+1÷6x x2+2x+1=12x x+1×x 2+2x+1 6x(Multiply by the reciprocal.) =12x x+1×(x+1) 2 6x(Factorx 2+2x+1.) =2(x+ 1) (Cancelx+1 and 6x.) Simplifying Compound Fractions: Acompound fractionis a fraction in which either the numerator or the denominator is itself a fraction. 8 SIMPLIFYING COMPOUND FRACTIONS : •Find the lowest common multiple of the denominators on the top and bottom. •Multiply top and bottom by this lowest common multiple. This will clear all the fractions from the top and bottom together. WORKED EXERCISE : Simplify each compound fraction. (a) 12− 13 14+ 16 (b) 1t+1 1t−1 SOLUTION : (a)12− 13 14+ 16 = 12− 13 14+ 16 ×12 12 =6−4 3+2 = 25 (b) 1t+1 1t−1= 1t+1 1t−1×t t =1+t 1−t Exercise1D 1.Simplify: (a)x x(b)2x x(c)x 2x(d)a a2 (e)3x 2 9xy(f )12ab 4a 2b 2.Simplify: (a)x 3×3 x (b)a 4÷a 2(c)x 2×3 x (d)1 2b×b 2 (e)3x 4×2 x2 (f )5 a÷10(g)2ab 3×6 ab 2 (h)8ab 5÷4ab 15

  12 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 2U NIT YEAR 11 3.Write as a single fraction: (a)x 2+x 5 (b)a 3−a 6 (c)x 8−y 12(d)2a 3+3a 2 (e)7b 10−19b 30 (f )xy 30−xy 12(g)1 x+1 2x (h)3 4x+4 3x (i)1 a−1 b(j)x+1 x (k)a+b a (l)1 x−1 x2 DEVELOPMENT 4.Simplify: (a)x+1 2+x+2 3 (b)2x−1 5+2x+3 4 (c)x+3 6+x−3 12(d)x+2 2−x+3 3 (e)2x+1 4−2x−3 5 (f )2x−1 3−2x+1 6(g)2x+1 3−x−5 6+x+4 4 (h)3x−7 5+4x+3 2−2x−5 10 (i)x−5 3x−x−3 5x 5.Factor where possible and then simplify: (a)2p+2q p+q (b)3t−12 2t−8 (c)x 2+3x 3x+9 (d)a ax+ay(e)3a 2−6ab 2a 2b−4ab 2 (f )x 2+2x x2−4 (g)a 2−9 a2+a−12 (h)x 2+2x+1 x2−1(i)x 2+10x+25 x2+9x+20 (j)ac+ad+bc+bd a2+ab (k)y 2−8y+15 2y 2−5y−3 (l)9ax+6bx−6ay−4by 9x 2−4y 2 6.Simplify: (a)1 x+1 x+1 (b)1 x−1 x+1 (c)1 x+1+1 x−1(d)2 x−3+3 x−2 (e)3 x+1−2 x−1 (f )2 x−2−2 x+3(g)x x+y+y x−y (h)a x+a−b x+b (i)x x−1−x x+1 7.Simplify: (a)8a 3b 5÷4ab 15 (b)2a 3b×5c 2 2a 2b×3b 2 2c(c)12x 2yz 8xy 3 ×24xy 2 36yz 2 (d)3a 2b 4b 3c×2c 2 8a 3÷6ac 16b 2 CHALLENGE 8.Simplify: (a)3x+3 2x×x 2 x2−1 (b)a 2+a−2 a+2×a 2−3a a2−4a+3 (c)c 2+5c+6 c2−16÷c+3 c−4(d)x 2−x−20 x2−25×x 2−x−2 x2+2x−8÷x+1 x2+5x (e)ax+bx−2a−2b 3x 2−5x−2×9x 2−1 a2+2ab+b 2 (f )2x 2+x−15 x2+3x−28÷x 2+6x+9 x2−4x÷6x 2−15x x2−49

  CHAPTER 1: Methods in Algebra 1E Factoring the Sum and Difference of Cubes 13 9.Simplify: (a)1 x2+x+1 x2−x (b)1 x2−4+1 x2−4x+4 (c)1 x−y+2x−y x2−y 2 (d)3 x2+2x−8−2 x2+x−6 (e)x a2−b 2−x a2+ab (f )1 x2−4x+3+1 x2−5x+6−1 x2−3x+2 10.Simplify: (a)b−a a−b (b)v 2−u 2 u−v (c)x 2−5x+6 2−x(d)1 a−b−1 b−a (e)m m−n+n n−m (f )x−y y2+xy−2x 2 11.Study the worked exercise on compound fractions and then simplify: (a)1− 12 1+ 12 (b)2+ 13 5− 23 (c) 12− 15 1+ 110 (d) 1720 − 34 45− 310 (e) 1x 1+ 2x (f )t− 1t t+ 1t (g)1 1b+ 1a (h) xy+ yx xy− yx (i)1− 1x+1 1x+ 1x+1 (j) 3x+2 − 2x+1 5x+2 − 4x+1 1EFactoring the Sum and Difference of Cubes The factoring of thedifference of cubesis similar to the factoring of the difference of squares. Thesum of cubescan also be factored, whereas the sum of squares cannot be factored. 9 DIFFERENCE OF CUBES :A 3−B 3=(A−B)(A 2+AB+B 2) SUM OF CUBES :A 3+B 3=(A+B)(A 2−AB+B 2) The proofs of these identities are left to the first question in the following exercise. WORKED EXERCISE : Factor each expression. (a)x 3−8(b)27a 3+1 SOLUTION : (a)x 3−8=(x−2)(x 2+2x+ 4) (Use the difference of cubesx 3−2 3.) (b) 27a 3+1 = (3a+ 1)(9a 2−3a+ 1) (Use the sum of cubes (3a) 3+1 3.) WORKED EXERCISE : (a)Simplifya 3+1 a+1.(b)Factora 3−b 3+a−b. SOLUTION : (a)a3+1 a+1=(a+1)(a 2−a+1) a+1 =a 2−a+1(b)a 3−b 3+a−b =(a−b)(a 2+ab+b 2)+(a−b) =(a−b)(a 2+ab+b 2+1)

  14 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 2U NIT YEAR 11 Exercise1E 1.(a)Prove the identityA 3−B 3=(A−B)(A 2+AB+B 2) by expanding the RHS. (b)Similarly, prove the identityA 3+B 3=(A+B)(A 2−AB+B 2). 2.Factor each expression. (a)x 3+y 3 (b)a 3−b 3 (c)y 3+1(d)k 3−1 (e)a 3+8 (f )b 3−8(g)27−t 3 (h)27 +u 3 (i)x 3+64(j)y 3−64 (k)125 +a 3 (l)125−b 3 DEVELOPMENT 3.Factor each expression. (a)8p 3+1 (b)8q 3−1 (c)u 3−64v 3 (d)t 3+64u 3 (e)27c 3+8 (f )27d 3−8 (g)64m 3−125n 3 (h)64p 3+ 125q 3 (i)216e 3−343f 3 (j)216g 3+ 343h 3 (k)a 3b3c3+ 1000 (l)729x 3−1331y 3 4.Factor each expression fully. (Remember to take out any common factors first.) (a)5x 3−5 (b)2x 3+16 (c)a 4−ab 3 (d)24t 3+81 (e)x 3y−125y (f )250p 3−432q 3 (g)27x 4+ 1000xy 3 (h)5x 3y3−5 (i)x 6+x 3y3 CHALLENGE 5.Simplify each expression. (a)x 3−1 x2−1 (b)a 2−3a−10 a3+8(c)a 3+1 6a 2 ×3a a2+a (d)x 2−9 x4−27x÷x+3 x2+3x+9 6.Simplify: (a)3 a−2−3a a2+2a+4 (b)1 x3−1+x+1 x2+x+1(c)1 x2−2x−8−1 x3+8 (d)a 2 a3+b 3+a−b a2−ab+b 2+1 a+b 1FSolving Linear Equations The first principle in solving any equation is to simplify the equation by doing the same things to both sides. Linear equations can be solved completely by following this principle 10 SOLVING LINEAR EQUATIONS : •Any number can be added to or subtracted from both sides. •Both sides can be multiplied or divided by any non-zero number. An equation involving algebraic fractions can often be reduced to a linear equation by following these steps.

  CHAPTER 1: Methods in Algebra 1F Solving Linear Equations 15 WORKED EXERCISE :Solve each equation. (a)6x+5 = 4x−9 (b)4−7x 4x−7=1 SOLUTION : (a) 6x+5 = 4x−9 −4x 2x+5 =−9 −5 2x=−14 ÷2 x=−7(b)4−7x 4x−7=1 ×(4x−7) 4−7x=4x−7 +7x 4=11x−7 +7 11 = 11x ÷11 x=1 Changing the Subject of a Formula: Similar sequences of operations allow the sub ject of a formula to be changed from one pronumeral to another. WORKED EXERCISE : Change the sub ject of each formula tox. (a)y=4x−3(b)y=x+1 x+2 SOLUTION : (a)y=4x−3 +3 y+3 = 4x ÷4 y+3 4=x x=y+3 4(b)y=x+1 x+2 ×(x+2) xy+2y=x+1 Rearranging,xy−x=1−2y Factoring,x(y−1) = 1−2y ÷(y−1) x=1−2y y−1 Exercise1F 1.Solve: (a)a−10 = 5 (b)t+3 = 1 (c)5c=−35(d)n 6=−3 (e)−2x=−20 (f )3x=2(g)−a=5 (h)x −4=−1 (i)−1−x=0(j)0·1y=5 (k)2t=t (l)− 12x=8 2.Solve: (a)2x+1 = 7 (b)5p−2=−2 (c)a 2−1=3(d)3−w=4 (e)3x−5=22 (f )4x+7 =−13(g)1−2x=9 (h)6x=3x−21 (i)−2=4+t 5(j)−13 = 5a−6 (k)19 = 3−7y (l)23−u 3=7 3.Solve: (a)3n−1=2n+3 (b)4b+3 = 2b+1 (c)5x−2=2x+10 (d)5−x=27+x(e)16 + 9a=10−3a (f )13y−21 = 20y−35 (g)13−12x=6−3x (h)3(x+7) =−2(x−9)(i)8+4(2−x)=3−2(5−x) (j)7x−(3x+ 11) = 6−(15−9x) (k)4(x+2) = 4x+9 (l)3(x−1) = 2(x+1)+x− 5

  16 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 2U NIT YEAR 11 DEVELOPMENT 4.Solve: (a)x 8=1 2 (b)a 12=2 3 (c)y 20=4 5 (d)1 x=3(e)2 a=5 (f )3=9 2y (g)2x+1 5=−3 (h)5a 3=a+1(i)7−4x 6=1 (j)5+a a=−3 (k)9−2t t=13 (l)6−c 3=c(m)1 a+4 = 1−2 a (n)4 x−1=−5 (o)3x 1−2x=7 (p)11t 8t+13=−2 5.Solve: (a)(x−3)(x+6) = (x−4)(x−5) (b)(1 + 2x)(4 + 3x)=(2−x)(5−6x)(c)(x+3) 2=(x−1) 2 (d)(2x−5)(2x+5) = (2x−3) 2 6.(a)Ifv=u+at, findawhent=4,v=20 andu=8. (b)Given thatv 2=u 2+2as, find the value ofswhenu=6,v=10 anda=2. (c)Suppose that1 u+1 v=1 t.Findv,giventhatu=−1andt=2. (d)IfS=−15,n=10 anda=−24, find,giventhatS= n2(a+). (e)The formulaF= 95C+ 32 relates temperatures in degrees Fahrenheit and Celsius. Find the value ofCthat corresponds toF= 95. (f )Suppose thatcanddare related by the formula3 c+1=5 d−1.Findcwhend=−2. 7.Solve each problem by forming, and then solving, a linear equation. (a)Three less than four times a certain number is equal to 21. Find the number. (b)Five more than twice a certain number is one more than the number itself. What is the number? (c)Bill and Derek collect Batman cards. Bill has three times as many cards as Derek, and altogether they have 68 cards. How many cards does Derek have? (d)If I paid $1.45 for an apple and an orange, and the apple cost 15 cents more than the orange, how much did the orange cost? 8.Solve: (a)x 3−x 5=2 (b)y+y 2=1 (c)a 10−a 6=1 (d)x 6+2 3=x 2−5 6(e)x 3−2=x 2−3 (f )1 x−3=1 2x (g)1 2x−2 3=1−1 3x (h)x−2 3=x+4 4(i)3 x−2=4 2x+5 (j)x+1 x+2=x−3 x+1 (k)(3x−2)(3x+2) (3x−1) 2 =1 (l)a+5 2−a−1 3=1 9.Rearrange each formula so that the pronumeral written in square brackets is the sub ject. (a)a=bc−d[b] (b)t=a+(n−1)d[n](c)p q+r=t[r] (d)u=1+3 v[v]

  CHAPTER 1: Methods in Algebra 1G Solving Quadratic Equations 17 CHALLENGE 10.Solve: (a)3 4−x+1 12=2 3−x−1 6 (b)x+1 2−x−1 3=x+1 3−x−1 2 (c)2x 5+2−3x 4=3 10−3−5x 2(d) 34(x−1)− 12(3x+2) = 0 (e)4x+1 6−2x−1 15=3x−5 5−6x+1 10 (f )7(1−x) 12−3+2x 9=5(2 +x) 6−4−5x 18 11.Solve each problem by forming, and then solving, a linear equation. (a) My father is 40 years older than me and he is three times my age. How old am I? (b) The fuel tank in my new car was 40% full. I added 28 litres and then found that it was 75% full. How much fuel does the tank hold? (c) A basketballer has scored 312 points in 15 games. How many points must he average per game in his next 3 games to take his overall average to 20 points per game? (d) A cyclist rides for 5 hours at a certain speed and then for 4 hours at a speed 6 km/h greater than his original speed. If he rides 294 km altogether, what was his first speed? 12.Rearrange each formula so that the pronumeral written in square brackets is the sub ject. (a)a 2−b 3=a[a] (b)1 f+2 g=5 h[g](c)x=y y+2[y] (d)a=b+5 b−4[b](e)c=7+2d 5−3d[d] (f )u=v+w−1 v−w+1[v] 1GSolving Quadratic Equations There are three approaches to solving a quadratic equation: •factoring •completing the square •using the quadratic formula. This section reviews factoring and the quadratic formula. Completing the square will be reviewed in Section 1I. Solving a Quadratic by Factoring: This method is the simplest, but does not always work. 11 SOLVING A QUADRATIC BY FACTORING : 1. Get all the terms on the left, then factor the left-hand side. 2. Use the principle that ifAB=0, thenA=0 orB=0. WORKED EXERCISE : Solve the quadratic equation 5x 2+34x−7 = 0 by factoring. SOLUTION : 5x 2+34x−7=0 5x 2+35x−x−7 = 0 (35 and−1 have sum 34 and product−7×5=−35.) 5x(x+7)−(x+7) = 0 (5x−1)(x+ 7) = 0 (The LHS is now factored.) 5x−1=0 orx+ 7 = 0 (One of the factors must be zero.) x= 15 orx=−7 (There are two solutions.)

  18 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 2U NIT YEAR 11 Solving a Quadratic by the Formula: This method works whether the solutions are rational numbers or involve surds. It will be proven in Chapter Ten. 12 THE QUADRATIC FORMULA : •The solutions ofax2+bx+c=0 are: x=−b+√ b2−4ac 2aorx=−b−√ b2−4ac 2a. •Always calculateb 2−4acfirst. (Later, this quantity will be called thediscriminantand given the symbol Δ.) WORKED EXERCISE : Solve each quadratic equation using the quadratic formula. (a)5x 2+2x−7=0(b)3x 2+4x−1=0 SOLUTION : (a) For 5x 2+2x−7=0, a=5,b=2 andc=−7. Henceb 2−4ac=2 2+ 140 = 144 =12 2, sox=−2+12 10or−2−12 10 =1 or−1 25.(b) For 3x 2+4x−1=0, a=3,b=4 andc=−1. Henceb 2−4ac=4 2+12 =28 =4×7, sox=−4+2√ 7 6or−4−2√ 7 6 =−2+√ 7 3or−2−√ 7 3. Exercise1G 1.Solve: (a)x 2=9 (b)y 2=25 (c)a 2−4=0(d)c 2−36 = 0 (e)1−t 2=0 (f )x 2= 94 (g)4x 2−1=0 (h)9a 2−64 = 0 (i)25y 2=16 2.Solve, by factoring: (a)x 2−5x=0 (b)y 2+y=0 (c)c 2+2c=0 (d)k 2−7k=0(e)t 2=t (f )3a=a 2 (g)2b 2−b=0 (h)3u 2+u=0(i)4x 2+3x=0 (j)2a 2=5a (k)3y 2=2y (l)12h+5h 2=0 3.Solve, by factoring: (a)x 2+4x+3 = 0 (b)x 2−3x+2 = 0 (c)x 2+6x+8 = 0 (d)a 2−7a+10 = 0 (e)t 2−4t−12 = 0 (f )c 2−10c+25 = 0(g)n 2−9n+8 = 0 (h)p 2+2p−15 = 0 (i)a 2−10a−24 = 0 (j)y 2+4y=5 (k)p 2=p+6 (l)a 2=a+ 132(m)c 2+18 = 9c (n)8t+20 =t 2 (o)u 2+u=56 (p)k 2=24+2k (q)50 + 27h+h 2=0 (r)α 2+20α=44

  CHAPTER 1: Methods in Algebra 1G Solving Quadratic Equations 19 DEVELOPMENT 4.Solve, by factoring: (a)2x 2+3x+1 = 0 (b)3a 2−7a+2 = 0 (c)4y 2−5y+1 = 0 (d)2x 2+11x+5 = 0 (e)2x 2+x−3=0 (f )3n 2−2n−5=0(g)3b 2−4b−4=0 (h)2a 2+7a−15 = 0 (i)2y 2−y−15 = 0 (j)3y 2+10y=8 (k)5x 2−26x+5 = 0 (l)4t 2+9 = 15t(m)13t+6 = 5t 2 (n)10u 2+3u−4=0 (o)25x 2+1 = 10x (p)6x 2+13x+6 = 0 (q)12b 2+3+20b=0 (r)6k 2+13k=8 5.Solve each equation, using the quadratic formula. Give exact answers, followed by ap- proximations to four significant figures where appropriate. (a)x 2−x−1=0 (b)y 2+y=3 (c)a 2+12 = 7a (d)u 2+2u−2=0(e)c 2−6c+2 = 0 (f )4x 2+4x+1 = 0 (g)2a 2+1 = 4a (h)5x 2+13x−6=0(i)2b 2+3b=1 (j)3c 2=4c+3 (k)4t 2=2t+1 (l)x 2+x+1 = 0 6.Solve, by factoring: (a)x=x+2 x (b)a+10 a=7(c)y+2 y=9 2 (d)(5b−3)(3b+1) = 1(e)5k+7 k−1=3k+2 (f )u+3 2u−7=2u−1 u−3 7.Find the exact solutions of: (a)x=1 x+2 (b)4x−1 x=x(c)a=a+4 a−1 (d)5m 2=2+1 m(e)y+1 y+2=3−y y−4 (f )2(k−1) =4−5k k+1 8.(a)Ify=px−ap 2, findp,giventhata=2,x=3 andy=1. (b)Given that (x−a)(x−b)=c, findxwhena=−2,b=4 andc=7. (c)Suppose thatS=n 2 2a+(n−1)d . Find the positive value ofnthat givesS=80 whena=4 andd=6. 9.Solve each problem by forming and then solving a suitable quadratic equation. (a)Find a positive integer that, when increased by 30, is 12 less than its square. (b)Two positive numbers differ by 3 and the sum of their squares is 117. Find the numbers. ( − 7)xcm ( + 2)xcm xcm (c)Find the value ofxin the diagram opposite. CHALLENGE 10.Solve each equation. (a)2 a+3+a+3 2=10 3 (b)k+10 k−5−10 k=11 6(c)3t t2−6=√ 3 (d)3m+1 3m−1−3m−1 3m+1=2

  20 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 2U NIT YEAR 11 11.Solve each problem by constructing and then solving a quadratic equation. (a) A rectangular area can be completely tiled with 200 square tiles. If the side length of each tile was increased by 1 cm, it would take only 128 tiles to tile the area. Find the side length of each tile. (b) The numerator of a certain fraction is 3 less than its denominator. If 6 is added to the numerator and 5 to the denominator, the value of the fraction is doubled. Find the fraction. (c) A photograph is 18 cm by 12 cm. It is to be surrounded by a frame of uniform width whose area is equal to that of the photograph. Find the width of the frame. (d) A certain tank can be filled by two pipes in 80 minutes. The larger pipe by itself can fill the tank in 2 hours less than the smaller pipe by itself. How long does each pipe take to fill the tank on its own? (e) Two trains each make a journey of 330 km. One of the trains travels 5 km/h faster than the other and takes 30 minutes less time. Find the speeds of the trains. 1HSolving Simultaneous Equations This section reviews the two algebraic approaches to solving simultaneous equa- tions — substitution and elimination. Both linear and non-linear simultaneous equations are reviewed. The methods are extended to systems of three equations in three unknowns, which will be new for most readers. Solution by Substitution: This method can be applied whenever one of the equations can be solved for one of the variables. 13 SOLVING SIMULTANEOUS EQUATIONS BY SUBSTITUTION : •Solve one of the equations for one of the variables. •Then substitute it into the other equation. WORKED EXERCISE : Solve each pair of simultaneous equations by substitution. (a)3x−2y=29 (1) 4x+y=24 (2)(b)y=x 2 (1) y=x+2 (2) SOLUTION : (a) Solving (2) fory,y=24−4x.(2A) Substituting (2A) into (1), 3x−2(24−4x)=29 x=7. Substitutingx= 7 into (1), 21−2y=29 y=−4. Hencex=7 andy=−4. (This should be checked in the original equations.) (b) Substituting (1) into (2),x 2=x+2 x 2−x−2=0 (x−2)(x+1) = 0 x=2 or−1. From (1), whenx=2,y=4, and whenx=−1,y=1. Hencex=2 andy=4, orx=−1andy= 1. (Check in the original equations.)

  CHAPTER 1: Methods in Algebra 1H Solving Simultaneous Equations 21 Solution by Elimination: This method, when it can be used, is more elegant, and can involve less algebraic manipulation. 14 SOLVING SIMULTANEOUS EQUATIONS BY ELIMINATION : Take suitable multiples of the equations so that one variable is eliminated when the equations are added or subtracted. WORKED EXERCISE : Solve each pair of simultaneous equations by elimination. (a)3x−2y=29 (1) 4x+5y=8 (2)(b)x 2+y 2=53 (1) x 2−y 2=45 (2) SOLUTION : (a) Taking 4×(1) and 3×(2), 12x−8y= 116 (1A) 12x+15y=24.(2A) Subtracting (1A) from (2A), 23y=−92 ÷23 y=−4. Substituting into (1), 3x+8 = 29 x=7. Hencex=7 andy=−4.(b) Adding (1) and (2), 2x 2=98 x 2=49. Subtracting (2) from (1), 2y 2=8 y 2=4. Hencex=7 andy=2, orx=7 andy=−2, orx=−7andy=2, orx=−7andy=−2. Systems of Three Equations in Three Variables: The key step here is to reduce the system to two equations in two variables. 15 SOLVING THREE SIMULTANEOUS EQUATIONS : Using either substitution or elimination, produce two simultaneous equations in two of the variables. WORKED EXERCISE : Solve simultaneously: 3x−2y−z=−8(1) 5x+y+3z=23 (2) 4x+y−5z=−18 (3) SOLUTION : Subtracting (3) from (2),x+8z=41.(4) Doubling (3), 8x+2y−10z=−36 (3A) and adding (1) and (3A), 11x−11z=−44 x−z=−4.(5) Equations (4) and (5) are now two equations in two unknowns. Subtracting (5) from (4), 9z=45 z=5. Substitutingz= 5 into (5),x=1 and substituting into (2),y=3. Hencex=1,y=3 andz= 5. (This should be checked in the original equations.)

  22 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 2U NIT YEAR 11 Exercise1H 1.Solve, by substituting the first equation into the second: (a)y=xand 2x+y=9 (b)y=2xand 3x−y=2 (c)y=x−1and2x+y=5(d)a=2b+1 anda−3b=3 (e)p=2−qandp−q=4 (f )v=1−3uand 2u+v=0 2.Solve, by either adding or subtracting the two equations: (a)x+y=5 andx−y=1 (b)3x−2y=7 andx+2y=−3 (c)2x+y=9 andx+y=5(d)a+3b=8 anda +2b=5 (e)4c−d=6 and 2c−d=2 (f )p−2q=4 and 3p−2q=0 3.Solve, by substitution: (a)y=2xand 3x+2y=14 (b)y=−3xand 2x+5y=13 (c)y=4−xandx+3y=8 (d)x=5y+4 and 3x−y=26(e)2x+y=10 and 7x+8y=53 (f )2x−y=9 and 3x−7y=19 (g)4x−5y=2 andx+10y=41 (h)2x+3 y=47 and 4x−y=45 4.Solve, by elimination: (a)2x+y=1 andx−y=−4 (b)2x+3y=16 and 2x+7y=24 (c)3x+2y=−6andx−2y=−10 (d)5x−3y=28 and 2x−3y=22 (e)3x+2y=7 and 5x+y=7 (f )3x+2y=0 and 2x−y=56(g)15x+2y=27 and 3x+7y=45 (h)7x−3y=41 and 3x−y=17 (i)2x +3y=28 and 3x+2y=27 (j)3x−2y=11 and 4x+3y=43 (k)4x+6y=11 and 17x−5y=1 (l)8x=5yand 13x=8y+1 DEVELOPMENT 5.Solve, by substitution: (a)y=2−xandy=x 2 (b)y=2x−3andy=x 2−4x+5 (c)y=3x 2andy=4x−x 2 (d)x−y=5 andy=x 2−11(e)x−y=2 andxy=15 (f )3x+y=9 andxy=6 (g)x 2−y 2=16 andx 2+y 2=34 (h)x 2+y 2= 117 and 2x 2−3y 2=54 6.Solve each problem by constructing and then solving a pair of simultaneous equations. (a)Find two numbers that differ by 16 and have a sum of 90. (b)I paid 75 cents for a pen and a pencil. If the pen cost four times as much as the pencil, find the cost of each item. (c)If 7 apples and 2 oranges cost $4, while 5 apples and 4 oranges cost $4·40, find the cost of each apple and orange. (d)Twice as many adults as children attended a certain concert. If adult tickets cost $8 each, child tickets cost $3 each and the total takings were $418, find the numbers of adults and children who attended. (e)A man is 3 times as old as his son. In 12 years time he will be twice as old as his son. How old is each of them now? (f )At a meeting of the members of a certain club, a proposal was voted on. If 357 members voted and the proposal was carried by a ma jority of 21, how many voted for and how many voted against?

  CHAPTER 1: Methods in Algebra 1I Completing the Square 23 7.Solve simultaneously: (a)y 4−x 3=1 andx 2+y 5=10(b)4x+y−2 3=12 and 3y−x−3 5=6 CHALLENGE 8.Solve simultaneously: (a)x=2y y=3z x+y+z=10 (b)x+2y−z=−3 3x−4y+z=13 2x+5y=−1(c) 2a−b+c=10 a−b+2c=9 3a−4c=1 (d)p+q+r=6 2p−q+r=1 p+q−2r=−9(e) 2x−y−z=17 x+3y+4z=−20 5x−2y +3z=19 (f ) 3u+v−4w=−4 u−2v+7w=−7 4u+3v−w=9 9.Solve simultaneously: (a)x+y=15 andx 2+y 2= 125 (b)x−y=3 andx 2+y 2= 185 (c) 2x+y=5 and 4x 2+y 2=17(d)x+y=9 andx 2+xy+y 2=61 (e)x+2y=5 and 2xy−x 2=3 (f ) 3x+2y=16 andxy=10 10.Set up a pair of simultaneous equations to solve each problem. (a) The value of a certain fraction becomes 15if one is added to its numerator. If one is taken from its denominator, its value becomes 17. Find the fraction. (b) Kathy paid $320 in cash for a CD player. If she paid in $20 notes and $10 notes and there were 23 notes altogether, how many of each type were there? (c) A certain integer is between 10 and 100. Its value is 8 times the sum of its digits, and if it is reduced by 45, its digits are reversed. Find the integer. (d) Two people are 16 km apart on a straight road. They start walking at the same time. If they walk towards each other, they will meet in 2 hours, but if they walk in the same direction (so that the distance between them is decreasing), they will meet in 8 hours. Find their walking speeds. 1ICompleting the Square Completing the square can be done in all situations, whereas factoring is not always possible. In particular, the quadratic formula that was reviewed in Sec- tion 1G will proven in Chapter Ten by completing the square. The review in this section is mostly restricted to monic quadratics, in which the coefficient ofx 2is 1. Chapter Ten will deal with non-monic quadratics. Perfect Squares: The expansion of the quadratic (x+α) 2is (x+α) 2=x 2+2αx+α 2. Notice that the coefficient ofxis twiceα, and the constant is the square ofα. Reversing the process, the constant term in a perfect square can be found by taking half the coefficient ofxand squaring the result. 16 COMPLETING THE SQUARE IN AN EXPRESSION x2+bx+···: Halve the coefficientbofxand square the result.

  24 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 2U NIT YEAR 11 WORKED EXERCISE : Complete the square in each expression. (a)x 2+16x+···(b)x 2−3x+··· SOLUTION : (a) The coefficient ofxis 16, half of 16 is 8, and 8 2= 64, sox 2+16x+64 = (x+8) 2. (b) The coefficient ofxis−3, half of−3is−1 12,and(−1 12)2=2 14, sox 2−3x+2 14=(x−1 12)2. Solving Quadratic Equations by Completing the Square: This process always works. 17 SOLVING QUADRATIC EQUATIONS BY COMPLETING THE SQUARE : Completethesquareinthequadraticbyaddingthesametobothsides. WORKED EXERCISE : Solve each quadratic equation by completing the square. (a)t 2+8t=20(b)x 2−x−1=0 SOLUTION : (a)t 2+8t=20 +16 t2+8t+16 = 36 (t+4) 2=36 t+4 = 6 ort+4 =−6 t=2 or−10(b)x 2−x−1=0 +1 x2−x=1 + 14 x2−x+ 14=1 14 (x− 12)2= 54 x− 12= 12 √5orx− 12=− 12 √5 x= 12+ 12 √5or 12− 12 √5 The Word ‘Algebra’: Al-Khwarizmi was a famous and influential mathematician who worked in Baghdad during the early ninth century when the Arabs excelled in science and mathematics. The Arabic word ‘algebra’ comes from the title of his most important work and means ‘the restoration of broken parts’ — a reference to the balancing of terms on both sides of an equation. Al-Khwarizmi’s own name came into modern European languages as ‘algorithm’. Exercise1I 1.What constant must be added to each expression in order to create a perfect square? (a)x 2+2x (b)y 2−6y(c)a 2+10a (d)m 2−18m(e)c 2+3c (f )x 2−x(g)b 2+5b (h)t 2−9t 2.Factor: (a)x 2+4x+4 (b)y 2+2y+1(c)p 2+14p+49 (d)m 2−12m+36(e)t 2−16t+64 (f )x 2+20x+ 100(g)u 2−40u+400 (h)a 2−24a+ 144 3.Copy and complete: (a)x 2+6x+···=(x+···) 2 (b)y 2+8y+···=(y+···) 2 (c)a 2−20a+···=(a− ···) 2 (d)b 2−100b+···=(b− ···) 2 (e)u 2+u+···=(u+···) 2 (f )t 2−7t+···=(t− ···) 2 (g)m 2+50m+···=(m+···) 2 (h)c 2−13c+···=(c− ···) 2

  CHAPTER 1: Methods in Algebra 1J Chapter Review Exercise 25 DEVELOPMENT 4.Solve each quadratic equation by completing the square. (a)x 2−2x=3 (b)x 2−6x=0 (c)a 2+6a+8 = 0(d)y 2+3y=10 (e)b 2−5b−14 = 0 (f )x 2+4x+1 = 0(g)x 2−10x+20 = 0 (h)y 2−y+2 = 0 (i)a 2+7a+7 = 0 CHALLENGE 5.Solve, by dividing both sides by the coefficient ofx 2and then completing the square: (a) 3x 2−15x+18 = 0 (b) 2x 2−4x−1=0 (c) 3x 2+6x+5 = 0(d) 2x 2+8x+3 = 0 (e) 4x 2+4x−3=0 (f ) 4x 2−2x−1=0(g) 3x 2−8x−3=0 (h) 2x 2+x−15 = 0 (i) 2x 2−10x+7 = 0 6.(a) Ifx 2+y 2+4x−2y+ 1 = 0, show that (x+2) 2+(y−1) 2=4. (b) Show that the equationx 2+y 2−6x−8y=0canbewrittenintheform (x−a) 2+(y−b) 2=c, wherea,bandcare constants. Hence finda,bandc. (c) Ifx 2+1 = 10x+12y, show that (x−5) 2= 12(y+2). (d) Find values forA,BandCify 2−6x+16y+94 = (y+C) 2−B(x+A). 1J Chapter Review Exercise 1.Simplify: (a)−8y+2y(b)−8y−2y(c)−8y×2y(d)−8y÷2y 2.Simplify: (a)−2a 2−a 2 (b)−2a 2−(−a 2)(c)−2a 2×(−a 2)(d)−2a 2÷(−a 2) 3.Simplify: (a)3t−1−t (b)−6p+3q+10p(c)7x−4y−6x+2y (d)2a 2+8a−13 + 3a 2−11a−5 4.Simplify: (a)−6k 6×3k 3 (b)−6k 6÷3k 3 (c)(−6k 6)2 (d)(3k 3)3 5.Expand and simplify: (a)4(x+3)+5(2x−3) (b)8(a−2b)−6(2a−3b) (c)−(a−b)−(a+b) (d)−4x 2(x+3)−2x 2(x−1)(e)(n+ 7)(2n−3) (f )(r+3) 2 (g)(y−5)(y+5) (h)(3x−5)(2x−3)(i)(t−8) 2 (j)(2c+ 7)(2c−7) (k)(4p+1) 2 (l)(3u−2) 2 6.Factor: (a)18a+36 (b)20b−36 (c)9c 2+36c (d)d 2−36 (e)e 2+13e+36 (f )f 2−12f+36(g)36−25g 2 (h)h 2−9h−36 (i)i 2+5i−36 (j)2j 2+11j+12 (k)3k 2−7k−6 (l)5 2−14+8(m)4m 2+4m−15 (n)n 3+8 (o)p 3−27 (p)p 3+9p 2+4p+36 (q)qt−rt−5q+5r (r)u 2w+vw−u 2x−vx

  26 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 2U NIT YEAR 11 7.Simplify: (a)x 2+x 4 (b)x 2−x 4 (c)x 2×x 4(d)x 2÷x 4 (e)3a 2b+2a 3b (f )3a 2b−2a 3b(g)3a 2b×2a 3b (h)3a 2b÷2a 3b (i)x y+y x(j)x y−y x (k)x y×y x (l)x y÷y x 8.Simplify: (a)x+4 5+x−5 3 (b)5 x+4+3 x−5 (c)x+1 2−x−4 5(d)2 x+1−5 x−4 (e)x 2−x+3 4 (f )2 x−4 x+3 9.Factor where possible, then simplify: (a)6a+3b 10a+5b (b)2x−2y x2−y 2 (c)x 2+2x−3 x2−5x+4 (d)2x 2+3x+1 2x 3+x 2+2x+1(e)a+b a2+2ab+b 2 (f )3x 2−19x−14 9x 2−4 (g)x 3−8 x2−4 (h)a 2−2a−3 a3+1 10.Solve each linear equation. (a)3x+5 = 17 (b)3(x+5) = 17 (c)x+5 3=17 (d)x 3+5 = 17(e)7a−4=2a+11 (f )7(a−4) = 2(a+11) (g)a−4 7=a+11 2 (h)a 7−4=a 2+11 11.Solve each quadratic equation by factoring the left-hand side. (a)a 2−49 = 0 (b)b 2+7b=0 (c)c 2+7c+6 = 0 (d)d 2+6d−7=0(e)e 2−5e+6 = 0 (f )2f 2−f−6=0 (g)2g 2−13g+6 = 0 (h)3h 2+2h−8=0 12.Solve, using the quadratic formula. Write the solutions in simplest exact form. (a)x 2−4x+1 = 0 (b)y 2+3y−3=0 (c)t 2+6t+4 = 0(d)3x 2−2x−2=0 (e)2a 2+5a−5=0 (f )4k 2−6k−1=0 13.Solve each quadratic by completing the square on the left-hand side. (a)x 2+4x=6 (b)y 2−6y+3 = 0(c)x 2−2x=12 (d)y 2+10y+7 = 0

CHAPTERTWO Numbers and Surds Arithmetic is the study of numbers and operations on them. This chapter reviews the arithmetic of integers, rational numbers and real numbers, with particular attention to surds. Most of this material will be familiar from earlier years, but Section 2B on recurring decimals may be new. 2AIntegers and Rational Numbers Our ideas about numbers arise from the two quite distinct sources: •Theintegersand therational numbersare developed from counting. •Thereal numbersare developed from geometry and the number line. Sections 2A and 2B deal with the integers and the rational numbers. The Integers: Countingis the first operation in arithmetic. Counting things like people in a room requireszero(if the room is empty) and all thepositive integers: 0,1,2,3,4,5,6, ... The number zero is the first number on this list, but there is no last number, because every number is followed by another number. The list is calledinfinite, which means that it never ‘finishes’. Any two of these numbers can beaddedormultiplied, and the result is another number on the list.Subtraction, however, requires thenegative integersas well: ...,−6,−5,−4,−3,−2,−1 so that calculations like 5−7=−2canbecompleted. 1 THE INTEGERS :Theintegers,orwhole numbers, are the numbers ...,−6,−5,−4,−3,−2,−1,0,1,2,3,4,5,6, ... •Two integers can be added, multiplied or subtracted to give another integer. •There are infinitely many positive and infinitely many negative integers. •The number zero is neither positive nor negative. Multiples and the LCM: Themultiplesof a positive integeraare all the products a,2a,3a,4a,5a,6a, ...For example, here are the multiples of 6 and 8. The multiples of 6 are: 6,12,18,24,30,36,42,48,54,60,66,72, ... The multiples of 8 are: 8,16,24,32,40,48,56,64,72,80,88,96, ... Thelowest common multipleorLCMof two positive integers is the smallest number that appears on both lists. In this example, the LCM of 6 and 8 is 24.

  28 CHAPTER 2: Numbers and Surds CAMBRIDGE MATHEMATICS 2U NIT YEAR 11 The key to adding and subtracting fractions is finding the LCM of their denom- inators, called thelowest common denominator: 1 6+5 8=1×4 24+5×3 24 =19 241 6−5 8=1×4 24−5×3 24 =−11 24 Divisors and the HCF: Divisionof positive integers yields a quotient and a remainder: 27÷10 = 2,remainder 7 42÷6=7,remainder 0 Because the remainder is zero in the second case, 6 is called adivisororfactor of 42. Here are the lists of all divisors of 42 and 63. The divisors of 42 are: 1,2,3,6,7,14,21,42 The divisors of 63 are: 1,3,7,9,21,63 Thehighest common factororHCFof two positive integers is the largest number that appears on both lists. In this example, the HCF of 42 and 63 is 21. The key to cancelling a fraction down to itslowest terms, and to multiplying fractions, is dividing the numerator and denominator by their HCF: 42 63=2×21 3×21=2 35 16×4 35=1 4×1 7=1 28 Prime Numbers: Aprime numberis an integer greater than 1 whose only divisors are itself and 1. The primes form a sequence whose distinctive pattern has confused every mathematician since Greek times: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79,... An integer that is greater than 1 and is not prime is called acomposite number. Without giving its proof, we shall assume theunique factorisation theorem: 2 THE UNIQUE FACTORISATION THEOREM : Every positive integer can be written as a product of prime numbers in one and only one way, apart from the order of the factors. For example, 24 = 2 3×3and30=2×3×5. The Rational Numbers: Problems like ‘Divide 7 cakes into 3 equal parts’ lead naturally tofractions, where the whole is ‘fractured’ or ‘broken’ into pieces. This gives the system ofrational numbers, which can be written as the ‘ratio’ of two integers. 3 DEFINITION OF RATIONAL NUMBERS : •Arational numberis a number that can be written as afractiona b,wherea andbare integers andb = 0. Here are some examples: 2 13=7 3−1 3=−1 330÷24 =5 43·72 =372 1004=4 1 •Every integeracan be written as a fractiona 1. Hence every integer is a rational number.

  CHAPTER 2: Numbers and Surds 2A Integers and Rational Numbers 29 The Four Operations on the Rational Numbers: Addition, multiplication, subtraction and division (except by 0) can all be carried out within the rational numbers. All except division have already been discussed. 4 DIVISION OF RATIONAL NUMBERS : •Thereciprocalof a fractiona bisb a. •To divide by a fraction, multiply by the reciprocal: 2 7÷3 4=2 7×4 3=8 21 The reciprocal of3 4is4 3. Exercise2A 1.Write down all the prime numbers:(a)less than 20,(b)between 20 and 50. 2.Write each number as a product of its prime factors. (a)10 (b)21(c)12 (d)18(e)28 (f )45(g)27 (h)40 3.Find the HCF (highest common factor) of: (a)6and8 (b)6and15(c)14 and 21 (d)12 and 20(e)18 and 27 (f )12 and 42(g)24 and 32 (h)36 and 60 4.Find the LCM (lowest common multiple) of: (a)2and5 (b)3and6(c)4and6 (d)4and7(e)6and8 (f )9and12(g)10 and 25 (h)6and15 DEVELOPMENT 5.Write each number as a product of its prime factors. (a)30 (b)36(c)39 (d)48(e)108 (f )128(g)154 (h)136 6.Find the HCF of the numerator and denominator, then use it to cancel each fraction down to lowest terms. (a)4 12 (b)8 10(c)10 15 (d)21 28(e)16 40 (f )21 45(g)24 42 (h)45 54(i)36 60 (j)54 72 7.Find the lowest common denominator, then simplify: (a)1 2+1 4 (b)3 10+2 5(c)1 2+1 3 (d)2 3−2 5(e)1 6+1 9 (f )5 12−3 8(g)7 10+2 15 (h)2 25−1 15 8.Find the value of: (a)1 4×20 (b)2 3×12(c)1 2×1 5 (d)1 3×3 7(e)2 5×5 8 (f )2÷1 3(g)3 4÷3 (h)1 3÷1 2(i)11 2÷3 8 (j)5 12÷12 3

  30 CHAPTER 2: Numbers and Surds CAMBRIDGE MATHEMATICS 2U NIT YEAR 11 CHALLENGE 9.Express in lowest terms without using a calculator: (a) 588630 (b) 4551001 (c) 500 1 000 000 10.Without using a calculator, find the value of: (a) 112 + 115 + 120 (b) 1118 − 916 (c) 12+ 14+ 15+ 125 + 1100 11.Write each number as a product of its prime factors. Then find its square root by halving the indices. (a) 576 (b) 1225 (c) 1 000 000 2BTerminating and Recurring Decimals Decimal notation extends ‘place value’ to negative powers of 10. For example: 123·456 = 1×10 2+2×10 1+3×10 0+4×10 −1 +5×10 −2 +6×10 −3 Such a number can be written as a fraction, and so is a rational number. Writing a Fraction as a Terminating or Recurring Decimal: If a rational number can be written as a fraction whose denominator is a power of 10, then it can easily be written as aterminating decimal: 325 = 12100 =0·12 and 578 350 = 578 + 6100 =578·06 If a rational number cannot be written with a power of 10 as its denominator, then repeated division will yield an infinite string of digits in its decimal representation. This string will cycle once the same remainder recurs, giving arecurring decimal. 23=0·666 666 666 666...=0·˙ 6 (which has cycle length 1) 6 37=6·428 571 428 571...=6·˙ 42857˙ 1 (which has cycle length 6) 24 3544 =24·795 454 545 45...=24·79˙ 5˙ 4 (which has cycle length 2) Writing a Recurring Decimal as a Fraction: Conversely, every recurring decimal can be written as a fraction. The following worked exercise shows the method. WORKED EXERCISE : Write as a fraction in lowest terms: (a)0·˙ 5˙ 1 (which has cycle length 2),(b)7·3˙ 28˙ 4 (which has cycle length 3). SOLUTION : (a) Letx=0·˙ 5˙ 1. Thenx=0·515 151... ×100 100x=51·515 151... Subtracting the last two lines, 99x=51 ÷99 x= 1733, so 0·˙ 5˙ 1= 1733.(b) Letx=7·3˙ 28˙ 4. Thenx=7·328 428 4... ×1000 1000x= 7328·428 428 4... Subtracting the last two lines, 999x= 7321·1 ÷999 x= 7321·1999 , so 7·3˙ 28˙ 4= 73 2119990 . 5 W RITING A RECURRING DECIMAL AS A FRACTION : If the cycle length isn, multiply by 10nand subtract.

  CHAPTER 2: Numbers and Surds 2B Terminating and Recurring Decimals 31 Percentages: Many practical situations involving fractions, decimals and ratios are commonly expressed in terms of percentages. 6 PERCENTAGES : •To convert a fraction to a percentage, multiply by 1001%: 3 20=3 20×100 1% = 15% •To convert a percentage to a fraction, replace % by 1100 : 15%=15×1 100=3 20 Manyproblemsarebestsolvedbytheunitary method, illustrated below. WORKED EXERCISE : (a)A table marked $1400 has been discounted by 30%. How much does it now cost? (b)A table discounted by 30% now costs $1400. What was the original cost? SOLUTION : (a) 100% is $1400 ÷10 10% is $140 ×7 70% is $980 so the discounted price is $980.(b) 70% is $1400 ÷7 10% is $200 ×10 100% is $2000 so the original price was $2000. Exercise2B 1.Write as a fraction in lowest terms: (a)30%(b)80%(c)75%(d)5% 2.Write as a decimal: (a)60%(b)27%(c)9%(d)16·5% 3.Write as a percentage: (a)1 4(b)2 5(c)6 25(d)13 20 4.Write as a percentage: (a)0·32(b)0·09(c)0·225(d)1·5 5.Express each fraction as a terminating decimal by first rewriting it as a fraction with denominator 10, 100 or 1000. (a)1 2 (b)1 5(c)3 5 (d)3 4(e)1 25 (f )7 20(g)1 8 (h)5 8 6.Express each fraction as a recurring decimal by dividing the numerator by the denominator. (a)1 3 (b)2 3(c)1 9 (d)5 9(e)3 11 (f )1 11(g)1 6 (h)5 6

  32 CHAPTER 2: Numbers and Surds CAMBRIDGE MATHEMATICS 2U NIT YEAR 11 7.Express each terminating decimal as a fraction in lowest terms. (a)0·4 (b)0·25(c)0·15 (d)0·16(e)0·78 (f )0·005(g)0·375 (h)0·264 8.Express each recurring decimal as a fraction in lowest terms. (a)0·˙ 2 (b)0·˙ 7(c)0·˙ 4 (d)0·˙ 6˙ 5(e)0·˙ 5˙ 4 (f )0·˙ 8˙ 4(g)0·˙ 0˙ 6 (h)0·˙ 13˙ 5(i)0·˙ 76˙ 2 (j)0·˙ 03˙ 3 DEVELOPMENT 9.(a)Find 12% of $5. (b)Find 7·5% of 200 kg. (c)Increase $6000 by 30%. (d)Decrease 1 12hours by 20%. 10.Express each fraction as a decimal. (a)33 250 (b)1 40(c)5 16 (d)27 80(e)7 12 (f )19 11(g)2 15 (h)13 55 11.Express each decimal as a fraction. (a)1·˙ 6 (b)3·˙ 2˙ 1(c)2·˙ 42˙ 3 (d)1·˙ 07˙ 4(e)0·2˙ 3 (f )0·1˙ 5(g)0·6˙ 3˙ 8 (h)0·3˙ 4˙ 5 12.(a)Steve’s council rates increased by 5% this year to $840. What were his council rates last year? (b)Joanne received a 10% discount on a pair of shoes. If she paid $144, what was the original price? (c)Marko spent $135 this year at the Easter Show, a 12·5% increase on last year. How much did he spend last year? 13.Work out the recurring decimals for 17,27,37,47,57and 67. Is there a pattern? CHALLENGE 14.(a) Work out the recurring decimals for 111, 211, 311, 411,..., 1011. Is there a pattern? (b) Work out the recurring decimals for 113, 213, 313, 413,..., 1213. Is there a pattern? 15.Use the method for converting a recurring decimal into a fraction to prove that: (a) 0·˙ 9=1 (b) 2·4˙ 9=2·5 16.(The numbers you obtain in this question may vary depending on the calculator used.) (a) Use your calculator to express 13as a decimal by entering 1÷3. (b) Subtract 0·333 333 33 from this, multiply the result by 10 8and then take the reciprocal. (c) Show arithmetically that the final answer in part (b) is 3. Is the answer on your calculator also equal to 3? What does this tell you about the way fractions are stored on a calculator?

  CHAPTER 2: Numbers and Surds 2C Real Numbers and Approximations 33 2CReal Numbers and Approximations There are two good reasons why decimals are used so often: •Any two decimals can easily be compared with each other. •Any quantity can be approximated ‘as closely as we like’ by a decimal. A measurement is only approximate, no matter how good the instrument, and rounding using decimals is a useful way of showing how accurate it is. Rounding to a Certain Number of Decimal Places: The rules for rounding a decimal are: 7 RULES FOR ROUNDING A DECIMAL NUMBER : To round a decimal to, say, two decimal places, look at the third digit. •If the third digit is 0, 1, 2, 3 or 4, leave the second digit alone. •If the third digit is 5, 6, 7, 8 or 9, increase the second digit by 1. Always use = . .rather than = when a quantity has been rounded or approximated. For example, 3·8472 = . .3·85,correct to two decimal places. (Look at 7, the third digit.) 3·8472 = . .3·8,correct to one decimal place. (Look at 4, the second digit.) Scientific Notation and Rounding to a Certain Number of Significant Figures: The very large and very small numbers common in astronomy and atomic physics are easier to comprehend when they are written in scientific notation: 1 234 000 = 1·234×10 6 (There are four significant figures.) 0·000 065 432 = 6·5432×10 −5 (There are five significant figures.) The digits in the first factor are called thesignificant figuresof the number. It is often more sensible to round a quantity correct to a given number of significant figures rather than to a given number of decimal places. To round to, say, three significant figures, look at the fourth digit. If it is 5, 6, 7, 8 or 9, increase the third digit by 1. Otherwise, leave the third digit alone. 3·085×10 9=. .3·09×10 9,correct to three significant figures. 2·789 654×10 −29 =. .2·790×10 −29 ,correct to four significant figures. The number can be in normal notation and still be rounded this way: 31·203 = . .31·20,correct to four significant figures. There are Numbers that are Not Rational: At first glance, it would seem reasonable to believe that all the numbers on the number line are rational, because the rational numbers are clearly spread ‘as finely as we like’ along the whole number line. Between 0 and 1 there are 9 rational numbers with denominator 10: 01 1 102 10 3 10 4 10 5 10 6 108 109 10 7 10 Between 0 and 1 there are 99 rational numbers with denominator 100: 01 50 100 Most points on the number line, however, represent numbers that cannot be writ- ten as fractions, and are calledirrational numbers. Some of the most important numbers in this course are irrational, like√ 2andπand the numberethat will be introduced much later in the course.

  34 CHAPTER 2: Numbers and Surds CAMBRIDGE MATHEMATICS 2U NIT YEAR 11 The Square Root of 2 is Irrational: The number√ 2 is particularly important, because by Pythagoras’ theorem,√ 2 is the diagonal of a unit square. Here is a proof by contradiction that√ 2 is an irrational number. Suppose that√ 2 were a rational number. 1 1 2 Then√ 2 could be written as a fraction in lowest terms. That is,√ 2=a b,whereb>1since√ 2 is not an integer. Squaring, 2 =a 2 b2,whereb 2>1 becauseb>1. Sincea bis in lowest terms,a 2 b2 is also in lowest terms, which is impossible, sincea 2 b2= 2, butb 2>1. This is a contradiction, so√ 2 cannot be a rational number. The Greek mathematicians were greatly troubled by the existence of irrational numbers. Their concerns can still be seen in modern English, where the word ‘irrational’ means both ‘not a fraction’ and ‘not reasonable’. The Real Numbers and the Number Line: The integers and the rational numbers were based oncounting. The existence of irrational numbers, however, means that this approach to arithmetic is inadequate, and a more general idea of number is needed. Wehavetoturnawayfromcountingandmakeuseofgeometry. 8 DEFINITION OF THE REAL NUMBERS : •Thereal numbersare defined to be the points on the number line. •All rational numbers are real, but real numbers like√ 2andπare irrational. At this point, geometry replaces counting as the basis of arithmetic. Exercise2C 1.Write each number correct to one decimal place. (a)0·32(b)5·68(c)12·75(d)0·05(e)3·03(f )9·96 2.Write each number correct to two significant figures. (a)0·429(b)5·429(c)5·029(d)0·0429(e)429(f )4290 3.Use a calculator to find each number correct to three decimal places. (a)√ 10 (b)√ 47(c) 916 (d) 3748 (e)π (f )π 2 4.Use a calculator to find each number correct to three significant figures. (a)√ 58 (b) 3√133(c)62 2 (d)14 5 (e) 4√0·3 (f )124 −1 5.To how many significant figures are each of these numbers accurate? (a)0·04 (b)0·40(c)0·404 (d)0·044(e)4·004 (f )400

  CHAPTER 2: Numbers and Surds 2C Real Numbers and Approximations 35 6.Classify these real numbers as rational or irrational. Express those that are rational in the form ab,whereaandbare integers. (a)−3 (b)1 12 (c)√ 3(d)√ 4 (e) 3√27 (f ) 4√8(g) 49 (h)0·45 (i)12%(j)0·333 (k)0·˙ 3 (l)3 17 (m)π (n)3·14 (o)0 DEVELOPMENT 7.Use a calculator to evaluate each expression correct to three decimal places. (a)67×29 43 (b)67 + 29 43(c)67 43×29 (d)67 43 + 29(e)67 + 29 43 + 71 (f )67 + 71 43×29 8.Use Pythagoras’ theorem to find the length of the unknown side in each triangle, and state whether it is rational or irrational. (a) (b) (c) (d) (e) (f ) 9.Calculate, correct to four significant figures: (a)10 −0·4 (b)1 240−13×17 (c)√ 6·5+8·3 2·7 (d) 3√10·57×12·83(e)3·5×10 4 2·3×10 5 (f )20 000(1·01) 25 (g)11·3 √19·5−14·7 (h)323+5 14 412+6 45 (i)(87·3×10 4)÷(0·629×10 −8 ) (j)√ 3+ 3√4 4√5+ 5√6 (k) 25 4× 34 5 67 2+ 23 3 (l) 36·41−19·57 23·62−11·39 CHALLENGE Use a calculator to answer the remaining questions. Write each answer in scientific notation. 10.The speed of light is approximately 2·997 929×10 8m/s. (a) How many metres are there in a light year? Assume that there are 365 14days in a year and write your answer in metres, correct to three significant figures. (b) The nearest large galaxy is Andromeda, which is estimated to be 2 200 000 light years away. How far is that in metres, correct to two significant figures? (c) The time since the Big Bang is estimated to be 13·6 billion years. How long is that in seconds, correct to three significant figures? (d) How far would light have travelled since the Big Bang? Give your answer in metres, correct to two significant figures.

  36 CHAPTER 2: Numbers and Surds CAMBRIDGE MATHEMATICS 2U NIT YEAR 11 11.The mass of a proton is 1·6726×10 −27 kg and the mass of an electron is 9·1095×10 −31 kg. (a) Calculate, correct to four significant figures, the ratio of the mass of a proton to the mass of an electron. (b) How many protons, correct to one significant figure, does it take to make 1 kg? 12.The diameter of a proton is approximately 10 −15 metres. (a) Assuming that a proton is a sphere, calculate the volume of a proton, correct to two significant figures. (The volume of a sphere is 43πr 3.) (b) Given that density is mass divided by volume, calculate, correct to one significant figure, the density of a proton. (This, very roughly, is the density of a neutron star.) (c) Water has a density of 1 g/cm 3. How many times denser is a proton? 13.Prove that√ 3 is irrational. (Adapt the given proof that√ 2 is irrational.) 2DSurds and their Arithmetic Numbers like√ 2and√ 3 occur constantly in this course because they occur in the solutions of quadratic equations. The last three sections of this chapter review various methods of dealing with them. Square Roots and Positive Square Roots: The square of any real number is positive, except that 0 2= 0. Hence a negative number cannot have a square root, and the only square root of 0 is 0 itself. A positive number, however, has two square roots, which are the opposites of each other. For example, the square roots of 9 are 3 and−3. Note that the well-known symbol√ xdoes not mean ‘the square root ofx’.Itis defined to mean thepositivesquare root ofx(or zero, ifx=0). 9 DEFINITION OF THE SYMBOL √x: •Fo rx>0,√ xmeans thepositivesquare root ofx. •Fo rx=0,√ 0=0. •Fo rx

  CHAPTER 2: Numbers and Surds 2D Surds and their Arithmetic 37 Simplifying Expressions Involving Surds: Here are some laws from earlier years for simplifying expressions involving square roots. The first pair restate the definition of the square root, and the second pair are easily proven by squaring. 11 LAWS CONCERNING SURDS :Letaandbbe positive real numbers. Then: (a)√ a2=a(c)√ a×√ b=√ ab (b) √ a 2=a(d)√ a√b= a b Taking Out Square Divisors: A surd like√ 500 is not regarded as being simplified, because 500 is divisible by the square 100, and so√ 500 can be written as 10√ 5: √ 500 =√ 100×5=√ 100×√ 5=10√ 5. 12 SIMPLIFYING A SURD : Check the number inside the square root for divisibility by one of the squares 4,9,16,25,36,49,64,81,100,121,144, ... It is quickest to divide by the largest possible square divisor. WORKED EXERCISE : Simplify these two expressions involving surds. (a)√ 108(b)5√ 27 SOLUTION : (a)√ 108 =√ 36×3 =√ 36×√ 3 =6√ 3(b) 5√ 27 = 5√ 9×3 =5×√ 9×√ 3 =15√ 3 WORKED EXERCISE : Simplify the surds in these expressions, then collect like terms. (a)√ 44 +√ 99(b)√ 72−√ 50 +√ 12 SOLUTION : (a)√ 44 +√ 99 = 2√ 11 + 3√ 11 =5√ 11(b)√ 72−√ 50 +√ 12 =6√ 2−5√ 2+2√ 3 =√ 2+2√ 3 Exercise2D 1.Write down the value of: (a)√ 16 (b)√ 36(c)√ 81 (d)√ 121(e)√ 144 (f )√ 400(g)√ 2500 (h)√ 10 000 2.Simplify: (a)√ 12 (b)√ 18 (c)√ 20 (d)√ 27(e)√ 28 (f )√ 40 (g)√ 32 (h)√ 99(i)√ 54 (j)√ 200 (k)√ 60 (l)√ 75(m)√ 80 (n)√ 98 (o)√ 800 (p)√ 1000

  38 CHAPTER 2: Numbers and Surds CAMBRIDGE MATHEMATICS 2U NIT YEAR 11 3.Simplify: (a)√ 3+√ 3 (b)5√ 7−3√ 7 (c)2√ 5−√ 5(d)−3√ 2+√ 2 (e)4√ 3+3√ 2−2√ 3 (f )−5√ 5−2√ 7+6√ 5(g)7√ 6+5√ 3−4√ 6−7√ 3 (h)−6√ 2−4√ 5+3√ 2−2√ 5 (i)3√ 10−8√ 5−7√ 10 + 10√ 5 DEVELOPMENT 4.Simplify: (a)3√ 8 (b)5√ 12(c)2√ 24 (d)4√ 44(e)3√ 45 (f )6√ 52(g)2√ 300 (h)2√ 96 5.Write each expression as a single square root. (For example, 3√ 2=√ 9×√ 2=√ 18.) (a)2√ 5 (b)5√ 2(c)8√ 2 (d)6√ 3(e)5√ 5 (f )4√ 7(g)2√ 17 (h)7√ 10 6.Simplify fully: (a)√ 8+√ 2 (b)√ 12−√ 3 (c)√ 50−√ 18(d)√ 54 +√ 24 (e)√ 45−√ 20 (f )√ 90−√ 40 +√ 10(g)√ 27 +√ 75−√ 48 (h)√ 45 +√ 80−√ 125 (i)√ 2+√ 32 +√ 72 CHALLENGE 7.Simplify fully: (a)√ 600 +√ 300−√ 216 (b) 4√ 18 + 3√ 12−2√ 50 (c) 2√ 175−5√ 140−3√ 28 8.Find the value ofxif: (a)√ 63−√ 28 =√ x(b)√ 80−√ 20 =√ x(c) 2√ 150−3√ 24 =√ x 2EFurther Simplification of Surds This section deals with the simplification of more complicated surdic expressions. The usual rules of algebra, together with the methods of simplifying surds given in the last section, are all that is needed. Simplifying Products of Surds: The product of two surds is found using the identity √ a×√ b=√ ab . It is important to check whether the answer needs further simplification. WORKED EXERCISE :Simplify each product. (a)√ 15×√ 5(b)5√ 6×7√ 10 SOLUTION : (a)√ 15×√ 5=√ 75 =√ 25×3 =5√ 3(b) 5√ 6×7√ 10 = 35√ 60 =35√ 4×15 =35×2√ 15 =70√ 15

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