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Transcript of EAS 2.6 Sample
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Year 12Mathematics
Contents
uLake Ltdu a e tduLake Ltd
Innovative Publisher of Mathematics Texts
EAS 2.6Robert Lakeland & Carl Nugent
Algebraic Methods
x2 + 5x + 6 = 3
5 log 3
h = V r
2
? AchievementStandard .................................................. 2
Expanding............................................................. 3
Factorising............................................................. 6
Exponents .............................................................. 12
ChangingtheSubject ................................................... 16
Logarithms............................................................. 21
RationalExpressions.................................................... 29
LinearEquations....................................................... 36
LinearInequations...................................................... 42
QuadraticEquations..................................................... 45
QuadraticFormula...................................................... 50
TheNatureoftheRootsofaQuadratic..................................... 54
ExcellenceQuestionsforAlgebra.......................................... 57
PracticeExternalAssessment............................................. 60
Answers............................................................... 72
OrderForm............................................................ 79
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Year 12 Mathematics EAS 2.6 Published by NuLake Ltd New Zealand Robert Lakeland & Carl Nugent
2 EAS 2.6 Algebraic Methods
Thisachievementstandardinvolvesapplyingalgebraicmethodsinsolvingproblems.
ThisachievementstandardisderivedfromLevel7ofTheNewZealandCurriculumandisrelatedto theachievementobjectives
manipulaterational,exponential,andlogarithmicalgebraicexpressions formanduselinearandquadraticequations
Applyalgebraicmethodsinsolvingproblemsinvolves: selectingandusingmethods demonstratingknowledgeofalgebraicconceptsandterms communicatingusingappropriaterepresentations.
Relationalthinkinginvolvesoneormoreof: selectingandcarryingoutalogicalsequenceofsteps
connectingdifferentconceptsorrepresentations demonstratingunderstandingofconcepts formingandusingamodel
andrelatingfindingstoacontext,orcommunicatingthinkingusingappropriatemathematical statements.
Extendedabstractthinkinginvolvesoneormoreof: devisingastrategytoinvestigateorsolveaproblem
identifyingrelevantconceptsincontext developingachainoflogicalreasoning,orproof
formingageneralisation
andusingcorrectmathematicalstatements,orcommunicatingmathematicalinsight.
Problemsaresituationsthatprovideopportunitiestoapplyknowledgeorunderstandingof mathematicalconceptsandmethods.Situationswillbesetinreal-lifeormathematicalcontexts.
Methodsincludeaselectionfromthoserelatedto: manipulatingalgebraicexpressionsincludingrationalexpressions manipulatingexpressionswithexponents,includingfractionalandnegativeexponents determiningthenatureoftherootsofaquadraticequation solvingexponentialequations(whichmayincludemanipulatinglogarithms) formingandsolvinglinearandquadraticequations.
Achievement Achievement with Merit Achievement with Excellence Applyalgebraicmethodsin
solvingproblems. Applyalgebraicmethods,
usingrelationalthinking,insolvingproblems.
Applyalgebraicmethods,usingextendedabstractthinking,insolvingproblems.
Algebraic Methods 2.6
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Year 12 Mathematics EAS 2.6 Published by NuLake Ltd New Zealand Robert Lakeland & Carl Nugent
6 EAS 2.6 Algebraic Methods
Factorising
There are four methods of factorising an expression. The flow chart below may be helpful in identifying the correct approach for a particular expression.
No Yes
ax2 + bx + c
No Yes
Does each termhave a common
factor?
Follow thecommon factor
approach.Are there fourterms (2 pairs)which could be
grouped?
No YesAre there amaximum of
three terms whichform a quadratic?
Either yourexpression cannotbe factorised oryou have made awrong decision.
Is the coefficientin front of the x
term 1?2
No YesFollow thecomplex
quadratic (a 1)approach.
Follow the simplequadraticapproach.
Follow thegrouping of pairs
to termsapproach.
FactorisingRewritinganexpressionasaproductofitsfactorsiscalledfactorising.
Theexpression4x+36couldbewrittenas4(x+9)becausewhenwemultiplyitoutweget4x+36.
Factorisingisthereverseprocedureofexpanding.
Weareonlyrequiredtofactoriseanexpressionifitconsistsofasum(ordifference)oftwoormoreterms.Wemustthenfactorisealloftheexpressionnotjustpartofit.
Weadoptdifferentfactorisingtechniquesfordifferenttypesofexpressions.
Common FactorsWhenwestudytheexpressionbelowweseethatallthetermsthatmakeituphaveafactorincommon,sotheappropriateapproachistoextractthehighestcommonfactor.Forexample,eachtermintheexpression
4x3y2+6xy520x2y7zhas2,x,andy2ascommonfactors.Weidentifythesebyfirststudyingthenumbersatthefront(coefficients)andseeingthat2isthehighestcommonfactor,thenwestudythexcomponent(x3,xandx2)andnotexisthehighestcommonfactorandfinallytheycomponent(y2,y5andy7)andnotey2isthehighestcommonfactor.
Wewritetheseontheoutsideofthebracketsanddividethemintothecomponentofeachterm.
4x3y2+6xy520x2y7z =2xy2(2x2+3y310xy5z)
2dividesinto4,6and20togive2,3and10.
xdividesintox3,xandx2togivex2,1andx.
y2dividesintoy2,y5andy7togive1,y3andy5.
Ourexpressionbecomes
4x3y2+6xy520x2y7z=2xy2(2x2.1+3.1.y310xy5z)
=2xy2(2x2+3y310xy5z)
(x+1)(x4)=x23x4
Factorising
Expanding uLake LtduLake Ltd Innovative Publisher of Mathematics Texts
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13EAS 2.6 Algebraic Methods
Year 12 Mathematics EAS 2.6 Published by NuLake Ltd New Zealand Robert Lakeland & Carl Nugent
ExampleEvaluate641/2
Itispossibletodothisonacalculator,butbyhanditbecomes
641/2 =1
641 2/
= 164
=18
ExampleSimplify(4x2)3x(5x4)2
Weplace(5x4)2onthebottomandsimplify
(4x2)3x(5x4)2 = ( )( )45
2 3
4 2xx
=453 6
2 8xx
= 6425
6
8xx
=6425 2x
Example
Simplify4 2 5
75xx
/
Simplifyingeachlineindependently
=4 2 5
7 5xx
/
/
=4x1
=4x
4 2 575
xx
/
Example
Simplifyandwritewithapositiveexponent ( )( )
23
1
2mmn
Writingeachlinewithapositiveexponent
=
12
9 2 2m
m n
=12m
x1
9m2n2Simplifying
=1
18 3 2m n
Investigate your calculator now. Make sure you are capable of calculating negative powers and fractional indices.
On the TI-84 Plus to find 641/2 you enter:
To convert your answer to a fraction enter:
() 1
(4 ^6
ENTER2 )
On the Casio 9750 to find 641/2 you enter:
To convert your answer to a fraction enter:
6 ^ () 1
a b/c 2 ) EXE
(4
F D
MATH 1 ENTERFrac
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23EAS 2.6 Algebraic Methods
Year 12 Mathematics EAS 2.6 Published by NuLake Ltd New Zealand Robert Lakeland & Carl Nugent
AchievementUsetheequivalentlogstatement(y=bxisequivalenttologby=x)tosolvethe following.
173. log525=A 174.log8512=B
175. log264=C 176.log20.5=D
177. log96561=E 178.log2128=F
179. log3G=3 180.log10H=3
181. log2I=0 182. log2J=7
183. log8K=4 184. log10L=1
185. logM32=5 186. logN10=1
187. logP1=0 188.logQ243=5
189. logR216=3 190.logS0.25=1
191. logT0.01=2 192.logU729=3
193. log93=V 194.log1632=W
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29EAS 2.6 Algebraic Methods
Year 12 Mathematics EAS 2.6 Published by NuLake Ltd New Zealand Robert Lakeland & Carl Nugent
2430
5
2 4x yx y =
45
5
2 4x yx y Dividethroughby6
=45
34xyy Simplifythexterms
=45
133x y Simplifytheyterms
=45
3
3xy
Rational Expressions
Rational ExpressionsArationalexpressionisanalgebraicexpressionexpressedasafraction.Ourrulesforsimplifying,adding(orsubtracting)andmultiplyingarethesameastherulesweusefornumericalfractions.
Simplifying Rational ExpressionsIfthereisacommonfactoronthetop(numerator)andbottom(denominator)wecandividethroughbyit.
Ifweexaminetherationalexpression 42
2 3
5 3a cab c
we
canseethattherearefactorsincommonbetweenthetopandbottom.Ifwewritetheexpressionoutdifferentlythesebecomemoreobvious.
42
2 3
5 3a cab c =
42. aa2. 15b
. cc3
3
=2.a.15b .1
= 25ab
Iftheexpressionsonthetopandbottomaresums,thentheyhavetobefactorisedpriortosimplification.
Example
Simplify2430
5
2 4x yx y
Look to group the same factors on top of each other.
3 a b/c 4 EXE20
a b/cSHIFT
Investigate your calculator now. Make sure you are capable of simplifying both proper and improper fractions.
On the Casio 9750 the fraction part of a rational expression (e.g. 24
30) can be quickly
simplified by using the fraction key on your calculator. In this case you enter:
If the fraction is top heavy (e.g. 3024
) then the same procedure is followed, but when you get a mixed
numeral as an answer (e.g. 1 14) then you convert it
to a top heavy fraction by:
On the TI-84 Plus the fraction part of a rational expression (e.g. 24
30) can be
quickly simplified by using the FRAC function on the calculator. In this case you enter:
If the fraction is top heavy (e.g. 3024
) then the same procedure is followed.
2 a b/c 0 EXE34
2 0 MATH34
1 ENTERFrac
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Year 12 Mathematics EAS 2.6 Published by NuLake Ltd New Zealand Robert Lakeland & Carl Nugent
36 EAS 2.6 Algebraic Methods
288. ( ) ( )x
xx
x+
+2
54
6 289.( )( )
( )( )
xx
xx
++
++
23
15
290. ( ) ( )x xxx
+ 2 42 291.
3 14
2 532
( ) ( )xx
xx++
292. ( )( )
( )( )
xx
xx
+ +12 2
34 3 293.
32 1
25 4
2xx
xx( ) ( )+ + +
1.7 Linear Equations
Linear Equations ExampleSolveforx 2x+5=3(6+4x)Alinearequationisanequationwithasingle
powerofxthatcanbesimplifiedtotheform
x =constant
Tosolvealinearequationwesuccessivelysimplifytheequationusingthefollowingsteps.
1.Expandanybrackets.
4(x+3) =2x14
4x+12 =2x14
2.Collectallxtermsontheleftandconstantson theright.
4x+12 =2x14
4x2x =1412
3.Ifinmovinganytermwecrosstheequalsign weusetheoppositeoperation.
2x =26
x = 262
=13
2x+5 =3(6+4x)
Expandthebracket 2x+5 =18+12x
Subtract5 2x =12x+185
Subtract12x 2x12x =13
10x =13
Divideby10 x =1310
=1.3
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Year 12 Mathematics EAS 2.6 Published by NuLake Ltd New Zealand Robert Lakeland & Carl Nugent
44 EAS 2.6 Algebraic Methods
354. 4 25 353
( )m m m+ > 355. 3
2 334
k k k <
d) GivethemaximumamountofmoneyKizzimustinvestontermdeposit(andstillgetatotalof$1200ininterest)withthebankandalsogivetheinterestshewillreceivefromeachinvestmentperyear.
Amountontermdeposit
InterestfromUnitTrust
Interestfromtermdeposit
Merit Answer the following questions involving inequations.
356. Kizziisplanningtoinvestatotalof$15000intwodifferentareas.Sheintendstoputsomeofthemoneyontermdepositfor12monthswithabankandwillreceive6.5%interestperannum.SheintendstoinvestthebalanceinaUnitTrustwhichoffers8.5%interestperannum.
a) WriteanexpressionwhichwillgivethetotalamountofreturnfromboththetermdepositandtheUnitTrustinvestmentintermsofn,theamountofmoneyinvestedintheUnitTrust.
b) IfKizziwishestoearnatleast$1200ininterestfortheyearfromherinvestments,formaninequalityusingyourexpressionfroma)abovetorepresentthis.DoNOTsolvetheinequalityyet.
c) Solvetheinequalityfromb)toidentifytheminimumamountofmoneyKizzineedstoinvestintheUnitTrustinordertoreceiveatleast$1200interestfortheyear.
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Year 12 Mathematics EAS 2.6 Published by NuLake Ltd New Zealand Robert Lakeland & Carl Nugent
54 EAS 2.6 Algebraic Methods
The Nature of the Roots of a QuadraticThe Nature of the Roots of a Quadratic
Inthissectionweexaminewhatdeterminesthenatureoftherootsofaquadraticequation.Arootisanothernameforthesolution(orxintercept).
Sometimeswhenwesolveaquadraticequationweonlygetone(ortwoequal)solutions.Othertimeswegetnosolutionoradecimalanswerthatneedstobeapproximated.Thesolutionofaquadraticequationisgivenby
x= b b ac
a 2 4
2
Thepartofthequadraticformulathatdeterminesthenatureoftherootsisb24ac.
Forexample,ifb24acisaperfectsquareandanexactsquarerootexiststhentherootsarealwaysabletobeexpressedasafraction.b24aciscalledthediscriminantandisrepresentedbythesymbolD.
Insummaryifthediscriminantis
positiveandaperfectsquare(e.g.1,4,9etc.),thentherootsarerational.Thismeansitispossibletowriteboththerootsasafractionandthequadraticequationfactorises.
positiveandnotaperfectsquare(e.g.1.5,7,11etc.),thentwoirrationalrootsexist.Thismeansitisnotpossibletowritetherootsasfractions.
zero,thenoneroot(ortwoequalrationalroots)exists.
negative,thentherearenorealroots.
ExampleDescribetherootsofthequadraticequationx210x5=0.
D =b24ac
=(10)24.1.(5)
=120Thisispositive,butnotaperfectsquaresothequadratichastwoirrationalroots.
On the Casio 9750 from the MENU select first EQUA then the Solver (F3).
Press to delete any existing
equation.Type in the discriminant.
For the discriminant, D = B2 4AC you enter:
Enter alongside each variable (ignoring D) the A, B and C values of the quadratic you are trying to find the discriminant for. For the example x2 10x 5 = 0 enter A = 1, B = 10 and C = 5.Place the cursor alongside the variable D and press to get the answer D = 120 (the discriminant).
F2 F1
ALPHA sin SHIFT . ALPHA log ^
EXE
On the TI-84 Plus select the MATH menu. Press 0 or scroll down until the cursor is on Solver and press ENTER. To delete any existing equation press the up arrow and then the CLEAR key.
Type in the discriminant.
For the discriminant, D = B2 4AC enter D B2 + 4AC, as it needs to be in the form, function = 0.
Enter alongside each variable (ignoring D) the A, B and C values of the quadratic you are trying to find the discriminant for. For the example x2 10x 5 = 0 enter A = 1, B = 10 and C = 5.Place the cursor alongside the variable D and press to get the answer D = 120 (the discriminant)
ALPHA x1 ALPHA APPS ^ 2
ALPHA ENTER
A graphics calculator can be used to calculate the discriminant by using the Solver.
+ 4 ALPHA MATH ALPHA PRGM ENTER
2 4 ALPHA x, , T ALPHA ln
F6
D B
A C
D = B
A C
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57EAS 2.6 Algebraic Methods
Year 12 Mathematics EAS 2.6 Published by NuLake Ltd New Zealand Robert Lakeland & Carl Nugent
Excellence Questions for Algebra
441. Aninventorhasdevelopedanewtoyandhas twomarketingoptions,self-promotion oftheproductorlicenseittoacompany.
Ifhechoosesselfpromotionthereturnin thousandsofdollarsis
R=9tt211 Ifhechoosestolicenseittoacompanythen thereturnis
R=kt
Forwhatvalueofkwillthereturnfromthe toycompanyalwaysbeinexcessofthereturn fromself-promotion?
Hint:Findthevaluesforkforwhichthe
equation9tt211=kthasnosolutions.
ExcellenceAnswerthefollowingequations.
442. Findtherangeofvaluesofpforwhichthe quadraticequation
px2+(p+3)x+p=0
hasrealroots.
443. Ifp(x)isapolynomialandisdividedby (xk)andaremainderisobtained,thenthat remainderisp(k). Ifthequadraticp(x)=x23x+5givesthe sameremainderwhendividedbyx+kasit doeswhendividedbyx3kfindthevalueof k,k0.
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Year 12 Mathematics EAS 2.6 Published by NuLake Ltd New Zealand Robert Lakeland & Carl Nugent
60 EAS 2.6 Algebraic Methods
Practice External Assessment Task Algebraic Methods 2.6
Make sure you show ALL relevant working for each question.
You are advised to spend 60 minutes answering this assessment.
QUESTION ONE(a) Expandandsimplify(2x+5)(x3)(x2)
(b) Writeasasinglelogstatement0.5log36+log2
(c) i) Factorise3x213x+4
ii) Solvetheequation3x24x+1=0
iii) Forwhatvalueofkdoestheequation3x2+kx+4=0haveequalroots?
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73EAS 2.6 Algebraic Methods
Year 12 Mathematics EAS 2.6 Published by NuLake Ltd New Zealand Robert Lakeland & Carl Nugent
Page 14 cont...
108. 14109. 8110. 8111. 27x9
112. 23x113. x2114. y115. m
12
116. qk
32
117. 9a4b6
118. k2564
2
119. pq
2
3 2
120. 944
2 2p
m n121. 12 2g
122. yx9
6
123. 1613
11pq
124. x8
125.
1x4b
126. 3ba127. x
ya
a
4
128. 12q x
129. 2n1/4m1/2
130. 2x4
131. p5/4q1/2
Page 15132. 4x3133. x3
134. 14x
135. x2
9136. 110p137. 14x
Page 15 cont...
138. 6 2nm
139. 4x9/2
140. x2/3
141. a)A= p910 5
b)V=p
2780 4
c) p=1.5cm
d)Width=4.13cm
Depth=2.04cm
Heightofbottle5.33cm
Page 17142. u=vat
143. I= VR144. c=ymx
145. x= y cm
146. pirA2=
147. pirV433=
148. a ts ut2 2
2=-
149. pipiR A r
2=
+
150. h a bA2
=+
151. b hA a2= -
152. M P t34 4
=-
153. R = 100IPT
Page 18
154. b = a2 c2
155. T= 100IPR
156. h= Emg157. F= 95 32
C+
158. x= k ab c( )+
159. y= 20 103 x
Page 19
160. a = 23 1b bvv+
161. a = 5 12d bss+
162. x = cy ay+ 5
163. b =ara r
Page 20
164. r =
A
noasris+ve
165. r =
3V
3
166. x= c ab
2
167. s av u2
2 2=
-
168. r =
3S4 noasris+ve
169. x = ( )y a4252
170. y = xk2
171. b = 1215 2 2a x172. Halfsphere= pir3
2 3
TotalVol. = pi pir r h32 3 2+
Makehthesubject
r2h=V pir32 3
h = pipi
rV r3
22
3-
h = pirV r3
22 -
Page 23
173. 25 =5A A =2
174. 512 =8B B =3
175. 64 =2C C =6
176. 0.5 =2D D =1
177. 9E =6561 E =4
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EAS 2.6 1EAS 2.6 2EAS 2.6 6EAS 2.6 13EAS 2.6 23EAS 2.6 29EAS 2.6 36EAS 2.6 44EAS 2.6 54EAS 2.6 57EAS 2.6 60EAS 2.6 73