Algebra External Achievement Standard (4 Credits)

AlgebraAlgebra

External Achievement StandardExternal Achievement Standard

(4 Credits)(4 Credits)

ContentsContents

Linear Equations

Indices and Surds

Rational Expressions

Log & Exponential Equations

Quadratic Equations

Simultaneous EquationsLog Laws

Expanding

Discriminant

Complete the SquareRearranging Equations

Factorising

Inequations

Rational Equations

Quadratic Formula

Expanding BracketsExpanding Brackets

Hasn’t changed from Year 11 but for an Hasn’t changed from Year 11 but for an extra challenge there can be up to 3 extra challenge there can be up to 3 brackets (a cubic).brackets (a cubic).

Eg) (x + 4)(x – 6)(3x – 1) Just pick 2 brackets to start with

(x2 – 2x - 24)(3x – 1) Add in the 3rd bracket

Expand and simplify that pair

Simplify like terms

3x3 – 7x2 – 70x +24

3x3 – 6x2 – 72x – x2 + 2x + 24

Expanding BracketsExpanding Brackets

Key points:Key points: Do a common factor lastDo a common factor last

Ie)Ie) 3(x + 4)(x – 5)3(x + 4)(x – 5) = =

Remember to write out square bracketsRemember to write out square brackets

Ie)Ie) (x + 2)(x + 2)2 2 = (x + 2)(x + 2)= (x + 2)(x + 2)

Same applies in a cubicSame applies in a cubic

Ie)Ie) (2x + 5)(x – 2)(2x + 5)(x – 2)22 = (2x + 5)(x – 2)(x – = (2x + 5)(x – 2)(x – 2)2)

3(x3(x22 + 4x – 5x + 4x – 5x – 20)– 20)3(x3(x22 – 1x – 1x – 20)– 20)3x3x22 – 3x – 60 – 3x – 60

Expanding BracketsExpanding Brackets Practice ThetaPractice Theta

– Square Bracket Expansions Square Bracket Expansions Difference of 2 SquaresDifference of 2 Squares Page 3 Ex 1.3Page 3 Ex 1.3 Page 3 Ex 1.4Page 3 Ex 1.4

– Expand and simplifyExpand and simplify Page 4 Ex 1.5Page 4 Ex 1.5

– 3 Bracket expansions3 Bracket expansions Page 5 Ex 1.6Page 5 Ex 1.6

– HomeworkHomework Page 1 and 2 ex 1.01 to 1.06Page 1 and 2 ex 1.01 to 1.06

FactorisingFactorising

This is the process of putting an This is the process of putting an expression into brackets.expression into brackets.

Three Key ChecksThree Key Checks1)1) Common factorCommon factor

2)2) Co-efficient = 1Co-efficient = 1

3)3) Co-efficient Co-efficient ≠ 1≠ 1

Common FactorCommon Factor

Simplest of the checks but often Simplest of the checks but often overlooked, ask yourself;overlooked, ask yourself;

Is there a common factor?Is there a common factor?

1)1) 5x5x22 + 10x + 10x 2)2) 4x4x22 – 15 – 15

3)3) 3x3x22 – 27x + 60 – 27x + 60 4)4) xx33 – 2x – 2x22 + x + x

Common FactorCommon Factor

PracticePractice

– Page 20 ex 3.1 Page 20 ex 3.1 particularly 23 onwardsparticularly 23 onwards

– Page 21 ex 3.2Page 21 ex 3.2 2 Step factorising2 Step factorising

Coefficient = 1Coefficient = 1

This is what you are most used to from This is what you are most used to from year 11. year 11.

Factorising into 2 brackets and the number Factorising into 2 brackets and the number in front of the xin front of the x22 = 1 = 1

1)1) xx22 + 13x + 42 + 13x + 42 2)2) xx22 – 25 – 25

3)3) 3x3x22 – 27x + 60 – 27x + 60

Coefficient = 1Coefficient = 1

PracticePractice

– Page 22 ex 3.3Page 22 ex 3.3

– Page 23 ex 3.4Page 23 ex 3.4 Must take out a common factor firstMust take out a common factor first

– Homework P13, 3.01 to 3.04Homework P13, 3.01 to 3.04

Coefficient Coefficient ≠≠ 1 1

This is when the number in front of xThis is when the number in front of x22 is is not a common factor and is bigger than 1not a common factor and is bigger than 1

Two different approaches, that give the Two different approaches, that give the same result:same result:

Guess and Check Guess and Check Cambridge MethodCambridge Method


Eg)Eg) 5x5x22 + 2x – 7 + 2x – 7Guess and Check Method

Factors of 5x2 factors of -7

5x & x -7 & 1

Cambridge Method

5 x -7 = -35

Factors of -35 that add to +2

+7, -5 works so

5x2 + 7x – 5x – 7

Take a common factor out of the first pair, then the second pair

x(5x + 7) – 1(5x + 7)

(x – 1)(5x + 7)

5x

-7

x

1

-35x + 1x = -34x -7x + 5x = -2x

So correct combination is

(5x + 7)(x – 1)

Right number wrong sign so use 7 and -1

Pick and Master one method, there is no better one.


Eg)Eg) 3x3x22 – 7x + 2 – 7x + 2Guess and Check Method

Factors of 3x2 factors of +2

3x & x 2 & 1

Cambridge Method

3 x 2 = 6

Factors of 6 that add to -7

-6, -1 works so

3x2 – 6x -1x + 2


3x(x – 2) – 1(x – 2)

(3x – 1)(x – 2)

3x

2

x

1

6x + 1x = 7x


(3x – 1)(x – 2)

Right number wrong sign so use -2 and -1


Eg)Eg) 2x2x22 + 7x + 3 + 7x + 3Guess and Check Method

Factors of 2x2 factors of 3

2x & x 3 & 1

Cambridge Method

2 x 3 = 6

Factors of 6 that add to +7

6, 1 works so

2x2 + 6x + 1x + 3


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

(2x + 1)(x + 3)

2x

3

x

1

6x + 1x = 7x


(2x + 1)(x + 3)


PracticePractice– Theta Page 23 and 24 ex 3.5Theta Page 23 and 24 ex 3.5– Homework Page 14 ex 3.05, 3.06 & 3.07Homework Page 14 ex 3.05, 3.06 & 3.07

IndicesIndices(Positive and negative exponents)(Positive and negative exponents)

Indices LawsIndices Laws

axaxmm.bx.bxnn = abx = abxm+nm+n

xxmm = x = xm-nm-n

xxnn

(ax(axmm))n n = a= annxxmnmn

Where a, b, m and n are all numbersWhere a, b, m and n are all numbers


A negative indices flips the fraction overA negative indices flips the fraction over = = xx -m-m

11 = = mm

11 = = 11

xxmm

Or negative indices above the vinculum (fraction Or negative indices above the vinculum (fraction line) are positive below and vice versa.line) are positive below and vice versa.

x as a fraction is x/1

Flip the fraction, change the sign

Expand the brackets and simplify

xx --

mm

eg)eg) xx-m-m


Some PracticeSome Practice

1)1) (¾)(¾)-2-2

2)2) xx22.x.x-4-4

3)3) xx33

xx-2-2

Page 72, ex 9.3Page 72, ex 9.3

SurdsSurds(Fractional exponents)(Fractional exponents)

SurdsSurdsaam/n m/n = = nn√a√amm

eg)eg) 4 43/43/4= 4= 433

= = 44√64= 2√64= 2

Surds behave in exactly the same way as exponents, we must convert them to fractional exponents first though.

//44√√

SurdsSurds(Fractional exponents)(Fractional exponents)

Simplify and convert to surd formSimplify and convert to surd form1)1) 5 5 ¼¼ 2)2) ((44√x√x33))33

3)3) √√xx33

√√xx55

Page 73 ex 9.4Page 73 ex 9.4Page 74 ex 9.5Page 74 ex 9.5

Surds behave in exactly the same way as Surds behave in exactly the same way as exponents, we must convert them to exponents, we must convert them to fractional exponents first though.fractional exponents first though.

Rational ExpressionsRational Expressions These are expressions involving algebraic These are expressions involving algebraic

fractions that we generally want to simplify.fractions that we generally want to simplify.

The main ones are The main ones are – MultiplicationMultiplication– DivisionDivision– Addition, SubtractionAddition, Subtraction– Simplifying (Cancelling Common Factors)Simplifying (Cancelling Common Factors)

MultiplicationMultiplication Multiply the numerators, multiply the Multiply the numerators, multiply the

denominatorsdenominators

1)1) 5x5x ×× 2x2x 44 7 7

2)2) 5x + 25x + 2 ×× 2x2x xx 3 3

3)3) 22 ×× 44 x - 3x - 3 (x – 3) (x – 3)22

DivisionDivision Take the reciprocal of the second fraction Take the reciprocal of the second fraction

then multiplythen multiply

1)1) 5x5x ÷÷ 2x2x 44 7 7

2)2) 5x + 25x + 2 ÷÷ 2x2x xx 3 3

3)3) 22 ÷÷ 44 x - 3x - 3 (x – 3) (x – 3)22

Addition and SubtractionAddition and Subtraction Key thing to remember is to get a common Key thing to remember is to get a common

denominatordenominator

1)1) 5x5x + + 2x2x 44 7 7

2)2) 5x + 25x + 2 – – 2x2x xx 3 3

3)3) 22 – – 44 x - 3x - 3 (x – 3) (x – 3)22

NoteNote

It helps to look for the lowest It helps to look for the lowest common denominator rather than common denominator rather than cross multiplying straight awaycross multiplying straight away

Simplifying Algebraic FractionsSimplifying Algebraic Fractions

Eg:Eg: 1)1) (x + 2) (x + 2) 3(x + 2)3(x + 2)

2)2) xx22 + 5x – 14 + 5x – 14 (x - 2)(x - 2)

Factorise then find a common factor and cancel it Factorise then find a common factor and cancel it out.out.

You can only cancel out things that are multiplied.You can only cancel out things that are multiplied.


Eg: 1) (x + 2)3(x + 2)

Already Factorised, Cancel out common

factor

= 1 3


Eg:Eg: 2) 2) xx22 + 5x – 14 + 5x – 14 (x - 2)(x - 2)

Factorise Top line Factorise Top line

(x + 7)(x – 2)(x + 7)(x – 2) (x – 2)(x – 2)

Cancel out common Cancel out common factorfactor

= (x + 7)

Simplifying Algebraic FractionsSimplifying Algebraic FractionsTo sum upTo sum up

Step 1Step 1 Factorise top and bottomFactorise top and bottomStep 2Step 2 Cancel out common factorsCancel out common factors

Eg:Eg: 1)1) xx22 + 5x – 14 + 5x – 14 (x - 2)(x - 2)

(x + 7)(x – 2)(x + 7)(x – 2) = x + 7 = x + 7 (x – 2)(x – 2)

2)2) xx22 – 7x – 44 – 7x – 44xx2 2 + 6x + 8 + 6x + 8

(x – 11)(x + 4)(x – 11)(x + 4) (x + 2)(x + 4)(x + 2)(x + 4)

= (x – 11) (x + 2)

Rational expressionsRational expressions To make the task of simplifying the To make the task of simplifying the

expression later on easier it helps not to expression later on easier it helps not to expand the bottom line.expand the bottom line.

PracticePractice– Theta Page 16 ex 2.2,Theta Page 16 ex 2.2, 2.3 2.3 Multiplication, DivisionMultiplication, Division

– Theta Page 16 ex 2.4Theta Page 16 ex 2.4 Mixed problemsMixed problems– Theta Page 18 ex 2.5Theta Page 18 ex 2.5 Addition, SubtractionAddition, Subtraction

– Theta Page 19 ex 2.6Theta Page 19 ex 2.6– Homework Pages 9 to 12Homework Pages 9 to 12

LogsLogs

Logs were designed by Scott, John Napier Logs were designed by Scott, John Napier as a way of dealing with very difficult as a way of dealing with very difficult multiplication problems using addition.multiplication problems using addition.

It is also useful for solving equations with an It is also useful for solving equations with an unknown exponent. unknown exponent. – Eg 3Eg 3xx = 729 = 729

Logs can have different bases but the Logs can have different bases but the calculator only uses 10 and e (a special calculator only uses 10 and e (a special number like number like ππ))

Logs and ExponentsLogs and Exponents

Logs have a base, and work as belowLogs have a base, and work as below x = bx = byy

loglogbbx = yx = y Ie Ie loglog1010(3) = 0.4771 (3) = 0.4771 becausebecause 10 100.47710.4771 = 3 = 3 Quickly work out the log statements belowQuickly work out the log statements below

loglog1010(10000) = a(10000) = a loglog22(16) = b(16) = b

loglog99(81) = c(81) = c loglog33(81) = d(81) = d

LogLogee(32) = 5(32) = 5 loglog55(f) = 4(f) = 4

What Logs meanWhat Logs mean

PracticePractice– Theta Page 89 ex 11.1 Particularly Q3 onwardsTheta Page 89 ex 11.1 Particularly Q3 onwards– Homework book Page 35Homework book Page 35

Log LawsLog Laws

Because logs are very similar to exponents Because logs are very similar to exponents there are some laws that go with them that there are some laws that go with them that make them easier to simplify.make them easier to simplify.

log(ab) log(ab) ==

log(a/b) log(a/b) ==

log(alog(ann)) ==

These laws are used when simplifying and solving log These laws are used when simplifying and solving log equations so although the are given, it helps to know them.equations so although the are given, it helps to know them.

log(a) + log(b)log(a) + log(b)

log(a) – log(b)log(a) – log(b)

nlog(a)nlog(a)

Log LawsLog Laws

Some ExamplesSome Exampleslog(3) + log(5)log(3) + log(5)log(12) – log(2)log(12) – log(2)3log(2)3log(2)2log(6) – log(9)2log(6) – log(9)

PracticePracticeTheta Page 91 ex 11.2Theta Page 91 ex 11.2Homework Page 36Homework Page 36

Linear EquationsLinear Equations

The simplest type of equation it involves no The simplest type of equation it involves no exponents and if graphed would be a exponents and if graphed would be a straight line.straight line.

eg)eg) 4x – 8 = 3x + 124x – 8 = 3x + 12

Homework book Page 3Homework book Page 3

Rational EquationsRational Equations

These are equations containing fractionsThese are equations containing fractions

eg)eg) 2x – 32x – 3 = = -3-3 55 4 4

Easiest way to deal with these is to Easiest way to deal with these is to multiply through by the lowest common multiply through by the lowest common multiple of the denominators to get rid of multiple of the denominators to get rid of the fraction.the fraction.


eg)eg) 2x – 32x – 3 = = -3-3 55 4 4

20 20 ×× 2x – 32x – 3 = = -3-3 × × 2020

55 4 44 4 ×× (2x – 3) = (-3) (2x – 3) = (-3) × 5× 5

8x – 12 = 8x – 12 = -15-15 8x = 8x = -3-3 x = x = -3/8-3/8

Lowest common multiple Lowest common multiple of 5 and 4 is 20of 5 and 4 is 20

Multiply through by 20Multiply through by 20

Cancel down Cancel down 20/5 = 420/5 = 4

20/4 = 520/4 = 5

Solve the linear equationSolve the linear equation

Some people use cross multiplying, multiply each side by the other Some people use cross multiplying, multiply each side by the other sides denominator, takes you straight to the third line.sides denominator, takes you straight to the third line.


1)1) xx = 4 = 422

2)2) 3x3x = 8 = 8 55

3)3) x + 1x + 1 = = xx 55 3 3

Practice Theta page 8 ex 1.10 even no’sPractice Theta page 8 ex 1.10 even no’s

Homework book Page 4Homework book Page 4

InequationsInequations

Very similar too linear equations but have an Very similar too linear equations but have an inequality instead of an equals sign.inequality instead of an equals sign.

eg)eg) 4x – 8 4x – 8 << 3x + 12 3x + 12

The only trick is if you divide by a negative The only trick is if you divide by a negative you have to reverse the sign.you have to reverse the sign.

InequationsInequations

1)1) 4x – 8 4x – 8 << 3x + 12 3x + 12

2)2) 11 – 3x > 3211 – 3x > 32

3)3) 3x 3x ≤ 5(2x + 4)≤ 5(2x + 4)

Practice Practice – Theta Page 9 ex 1.11Theta Page 9 ex 1.11– Theta Page 9 ex 1.12 (Applications)Theta Page 9 ex 1.12 (Applications)– Homework Page 5Homework Page 5

Solving Quadratic EquationsSolving Quadratic Equations

Eg:Eg: 1)1) (x + 6)(2x – 7)(x + 6)(2x – 7) = 0 = 0

2)2) x x22 + 12x + 20 + 12x + 20 = 0= 0

3)3) xx22 + 7x + 7x = 18= 18

4)4) 3x 3x22 + 5x – 8 + 5x – 8 = 2x= 2x22 – 3x + 40 – 3x + 40

Step 1:Step 1: Make it equal to zero by adding and subtractingMake it equal to zero by adding and subtracting

Step 2:Step 2: Factorise into 1 or 2 bracketsFactorise into 1 or 2 brackets

Step 3:Step 3: SolveSolve

Solving Quadratic EquationsSolving Quadratic EquationsEg:Eg: 1)1) (x + 6)(2x – 7)(x + 6)(2x – 7) = 0 = 0

EitherEither x + 6 = 0 x + 6 = 0

x = -6 x = -6




or 2x – 7or 2x – 7 = 0= 0

2x2x = 7= 7

x x = 3.5 = 3.5


Eg:Eg: 2) 2) xx22 + 12x + 20 + 12x + 20 = 0= 0

(x + 2)(x + 10)(x + 2)(x + 10) = 0= 0


x = -2x = -2




or x + 10= 0or x + 10= 0

xx = -10= -10

Solving Quadratic EquationsSolving Quadratic EquationsEg:Eg: 3) 3) xx22 + 7x + 7x = 18= 18

xx22 + 7x – 18 + 7x – 18 = 0= 0

(x + 9)(x – 2)(x + 9)(x – 2) = 0= 0


x = -9x = -9




or x – 2 = 0or x – 2 = 0

xx = = 22

Solving Quadratic EquationsSolving Quadratic EquationsEg:Eg: 4) 4) 3x 3x22 + 5x – 8 + 5x – 8 = 2x= 2x22 – 3x + 40 – 3x + 40

xx22 + 8x – 48 + 8x – 48 = 0= 0

(x + 12)(x – 4)(x + 12)(x – 4) = 0= 0


x = -12x = -12




or x – 4 = 0or x – 4 = 0

xx = = 44


Eg:Eg: 1)1) (x + 6)(2x – 7)(x + 6)(2x – 7) = 0 = 0


x = -6 x = -6

or 2x – 7or 2x – 7 = 0= 0

2x2x = 7= 7

x x = 3.5 = 3.5




Eg:Eg: 4) 4) 3x 3x22 + 5x – 8 + 5x – 8 = 2x= 2x22 – 3x + 40 – 3x + 40

xx22 + 8x – 48 + 8x – 48 = 0= 0

(x + 12)(x – 4)(x + 12)(x – 4) = 0= 0


x = -12x = -12

or x – 4or x – 4 = 0= 0

xx = 4= 4


Practice Practice

Theta Page 56 ex 8.1 & 8.2Theta Page 56 ex 8.1 & 8.2

Theta Page 57 ex 8.3Theta Page 57 ex 8.3

Theta Page 58 ex 8.4 (Applications)Theta Page 58 ex 8.4 (Applications)

Homework book Page 23 and 24Homework book Page 23 and 24

Completing the SquareCompleting the SquareThis uses the square brackets to help us solve This uses the square brackets to help us solve complicated quadratic equations reasonably complicated quadratic equations reasonably simply.simply.

The equation The equation

(x – 2)(x – 2)22 – 8 = 0 – 8 = 0

is relatively easy to solve, the is relatively easy to solve, the answers however are difficult decimals meaning answers however are difficult decimals meaning the equation won’t factorise.the equation won’t factorise.

The trick is fitting a quadratic equation into square The trick is fitting a quadratic equation into square brackets to start with.brackets to start with.

Completing the SquareCompleting the Square

Eg:Eg: xx22 + 8x + 12 + 8x + 12 = 0= 0

xx22 + 8x + 16 + 12 + 8x + 16 + 12 = 16= 16

(x + 4)(x + 4)22 + 12 + 12 = 16= 16

(x + 4)(x + 4)22 = 4= 4

x + 4x + 4 = = √4√4

soso x x = -2= -2

Why is there only 1 solution? Why is there only 1 solution?

There used to be 2 what went wrong?There used to be 2 what went wrong?

AnswerAnswer √4√4 = 2 or -2= 2 or -2

xx = 2 – 4 or -2 – 4= 2 – 4 or -2 – 4

= -2 = -2 -6 -6

Completing the SquareCompleting the SquareA step by step guide:A step by step guide:

Eg:Eg: 2)2) xx22 + 12x + 14 + 12x + 14 = 0= 0

xx22 + 12x + 36 + 14 + 12x + 36 + 14 = 36= 36

(x + 6)(x + 6)22 + 14 + 14 = 36= 36

(x + 6)(x + 6)22 = 22= 22

x + 6x + 6 = = ±±√22√22

soso x x = -6 = -6 ± ± √22√22


Step 2:Step 2: Make the number in front of xMake the number in front of x22 a 1 by dividing a 1 by dividing

Step 3:Step 3: Pick your square bracketPick your square bracket


Completing the SquareCompleting the SquareA step by step guide:A step by step guide:

Eg:Eg: 3)3) 5x5x22 + 30x + 24 + 30x + 24 = 0= 0

xx22 + 6x + 4.8 + 6x + 4.8 = 0= 0

xx22 + 6x + 9 + 4.8 + 6x + 9 + 4.8 = 9= 9

(x + 3)(x + 3)22 + 4.8 + 4.8 = 9= 9

(x + 3)(x + 3)22 = 4.2= 4.2

x + 3x + 3 = = ±±√4.2√4.2

soso x x = -3 = -3 ± ± √4.2√4.2


Step 2:Step 2: Make the number in front of xMake the number in front of x22 a 1 by dividing a 1 by dividing




Eg:Eg: 3)3) 55xx22 + 30x + 24 + 30x + 24 = 0= 0 divide it all by divide it all by 55

xx22 + 6x + 4.8 + 6x + 4.8 = 0= 0 half 6 = 3, 3half 6 = 3, 322 = = 99,, add 9 both sidesadd 9 both sides

xx22 + 6x + 6x + 9+ 9 + 4.8 + 4.8 = = 99 factorise the square bracketfactorise the square bracket

(x + 3)(x + 3)22 + 4.8 + 4.8 = 9= 9 - 4.8 both sides- 4.8 both sides

(x + 3)(x + 3)22 = 4.2= 4.2 Square root both sidesSquare root both sides

x + 3x + 3 = = ±±√4.2√4.2 -3 both sides-3 both sides

soso x x = -3 = -3 ± ± √4.2√4.2 CalculatorCalculator

OverviewOverview


Step 2:Step 2: Make the number in front of xMake the number in front of x22 a 1, by dividing a 1, by dividing




Practice Practice

Theta Page 61 ex 8.5Theta Page 61 ex 8.5

Quadratic FormulaQuadratic FormulaIfIf ax ax22 + bx + c + bx + c = 0= 0

ThenThen xx = = -b -b ± √b± √b22 - 4ac - 4ac2a2a

Eg:Eg: if 3xif 3x22 + 5x – 9 + 5x – 9 = 0= 0

a = 3, b = 5, c = -9a = 3, b = 5, c = -9

thenthen x x = = -(5) -(5) ± √(5)± √(5)22 – 4(3)(-9) – 4(3)(-9) 2(3)2(3)

x x = = -5 -5 ± √25 – -108± √25 – -108 66

x x = (= (-5 -5 + √132)+ √132) 66

or xor x = (= (-5 -5 – √132)– √132) 66

TAKE CARE and TAKE YOUR TIME

Quadratic FormulaQuadratic FormulaOn a graphics calculatorOn a graphics calculator

Eg:Eg: if 4xif 4x22 + 8x – 7 + 8x – 7 = x= x22 – 3x – 3x

Make one side = 0 by adding and subtractingMake one side = 0 by adding and subtracting

3x3x22 + 11x – 7 + 11x – 7 = 0= 0

Menu : Menu : EquationEquation

F2 : PolynomialF2 : Polynomial

F1 : Degree 2 (xF1 : Degree 2 (x22) Degree 3 (x) Degree 3 (x33))

a = 3, b = 11, c = -7a = 3, b = 11, c = -7

F1 : SolveF1 : Solve

xx11 = 0.5529 = 0.5529

xx22 = -4.219 = -4.219

EQUA

Equation

Select TypeF1:SimultaneousF2:PolynomialF3:Solver

Polynomial

Degree?

aX2+bX+c=0 a b c . [ 0 0 0 ]

SIML POLY SOLV2 3 CLRSOLV DEL

aX2+bX+c=0 a b c . [ 3 0 0 ]

aX2+bX+c=0 a b c . [ 3 11 0 ]

aX2+bX+c=0 a b c . [ 3 11 -7 ]

3_1_11_-_-7

aX2+bX+c=0 X . 1 0.5529 2 -4.219

REPT

Quadratic FormulaQuadratic Formula A good challenge is to try and prove the A good challenge is to try and prove the quadratic formula by completing the square onquadratic formula by completing the square on

axax22 + bx + c = 0 + bx + c = 0PracticePractice

Theta Theta Page 61 ex 8.6Page 61 ex 8.6Page 62 ex 8.8 ApplicationsPage 62 ex 8.8 Applications

HwkHwk bkbk Page 24 ex 6.06Page 24 ex 6.06Page 25 and 26Page 25 and 26

DiscriminantDiscriminant

Not every quadratic equation can be solved to Not every quadratic equation can be solved to give ‘real’ solutions. give ‘real’ solutions.

At this stage in your Math careers, you cannot At this stage in your Math careers, you cannot square root a negative number so any timesquare root a negative number so any time

bb22 – 4ac – 4acis less than zero (negative) we know there are is less than zero (negative) we know there are no real solutions.no real solutions.

bb22 – 4ac is called the discriminant because it – 4ac is called the discriminant because it tells us how many solutions there will be;tells us how many solutions there will be;


bb22 – 4ac > 0 (a positive number) – 4ac > 0 (a positive number)means there will be 2 ‘real’ solutionsmeans there will be 2 ‘real’ solutions

bb22 – 4ac = 0 – 4ac = 0means there will be 1 ‘real’ solutionsmeans there will be 1 ‘real’ solutions

bb22 – 4ac < 0 (a negative number) – 4ac < 0 (a negative number)means there will be no ‘real’ solutionsmeans there will be no ‘real’ solutions


Practice examples;Practice examples;1)1) Show that the equation Show that the equation

7x7x22 + 13x + 73 = 0 + 13x + 73 = 0 has no real solutionshas no real solutions

2) For what value of k does the equation 2) For what value of k does the equation 3x3x22 + 24x + k = 0 + 24x + k = 0

have only one solution?have only one solution?


1)1) 7x7x22 + 13x + 73 = 0 + 13x + 73 = 0 a = 7,a = 7, b = 13,b = 13, c = 73c = 73

bb22 – 4ac is the discriminant – 4ac is the discriminant

131322 – 4 – 4××77××73 73 = 169 – 2044= 169 – 2044= -1875= -1875

this is less than zero so there are no solutionsthis is less than zero so there are no solutions


2)2) 3x3x22 + 24x + k = 0 + 24x + k = 0One solution when the discriminant = 0One solution when the discriminant = 0ie) bie) b22 – 4ac = 0 – 4ac = 0

242422 – 4 – 4×3×k ×3×k = 0= 0 576 – 12k576 – 12k = 0= 0

576576 = 12k= 12kkk = 48= 48

So 3xSo 3x22 + 24x + k = 0 has only one solution when + 24x + k = 0 has only one solution when k = 48k = 48


PracticePracticeTextbook Page 66 ex 8.9Textbook Page 66 ex 8.9

page 67 ex 8.10page 67 ex 8.10

Exponential EquationsExponential Equations

These are equations containing an These are equations containing an unknown exponentunknown exponent

Eg)Eg) 33xx = 243 = 243

Some can be solved by inspection (or trial Some can be solved by inspection (or trial and error) but for more difficult ones we and error) but for more difficult ones we need a more reliable method.need a more reliable method.

We can use Logs to help us out here as We can use Logs to help us out here as they can change a power into a they can change a power into a multiplicationmultiplication


Using the third log lawUsing the third log law n loga = logan loga = logann

Now apply this in reverse to our problemNow apply this in reverse to our problem 33xx = 243= 243

log(3log(3xx)= log(243))= log(243)

x log(3)x log(3) = log(243)= log(243)

xx = log(243) / log(3)= log(243) / log(3)

Calculate the answer on a calculatorCalculate the answer on a calculator

xx = 5 = 5

Log Both SidesLog Both Sides

Move the exponent out the frontMove the exponent out the front

Divide by log(3)Divide by log(3)


PracticePractice

55xx = 3125 = 3125 1212(x+4)(x+4) = 20736 = 20736log(5log(5xx) = log(3125)) = log(3125) log(12log(12(x+4)(x+4)) = log(20736)) = log(20736)

x = log(3125) / log(5)x = log(3125) / log(5) x + 4 = log(20736)/log12x + 4 = log(20736)/log12

x log(5) = log(3125)x log(5) = log(3125) (x + 4)log(12) = log(20736)(x + 4)log(12) = log(20736)

x + 4 = 4x + 4 = 4x = 0x = 0

x = 5x = 5

Rearranging EquationsRearranging Equations

I approach these in exactly the same way as I approach these in exactly the same way as I would if I had to solve an equation. I would if I had to solve an equation.

The first thing I do is decide which letter I have to get The first thing I do is decide which letter I have to get alone.alone.

Then I ask myself what I would do if everything else Then I ask myself what I would do if everything else were numbers,were numbers,

– Get Rid of FractionsGet Rid of Fractions– Group like terms (get my letter on one side and everything Group like terms (get my letter on one side and everything

else on the other)else on the other)– Simplify, (factorise)Simplify, (factorise)– Divide through by whatever I have in front of my letter (the Divide through by whatever I have in front of my letter (the

coefficient)coefficient)


The examples we are working through are The examples we are working through are 1.1. Make ‘x’ the subject ofMake ‘x’ the subject of

Y = m x + cY = m x + c

2.2. Make ‘a’ the subject ofMake ‘a’ the subject of S S = v t + 0.5 a t= v t + 0.5 a t22

3.3. Make ‘x’ the subject of Make ‘x’ the subject of 12 – y x = 3y + 4 x12 – y x = 3y + 4 x


PracticePractice– Make x the subject of Make x the subject of

Y = m x + cY = m x + c Y – c = m xY – c = m x

(Y – c)/m= x(Y – c)/m= x

– Make a the subject ofMake a the subject of S S = v t + 0.5 a t= v t + 0.5 a t22

S – v t S – v t = 0.5 a t= 0.5 a t22

2(S – v t) 2(S – v t) = a t= a t22

2(S – v t) / t2(S – v t) / t22 = a= a

What letter are we trying to get alone?What letter are we trying to get alone?

Get everything away from x.Get everything away from x.

Divide through by the coefficient (m).Divide through by the coefficient (m).


Get everything away from a.Get everything away from a.

Divide through by the coefficients.Divide through by the coefficients.


PracticePractice– Make x the subject of Make x the subject of

12 – y x = 3 y + 4 x12 – y x = 3 y + 4 x

12 = 3 y + 4 x + y x12 = 3 y + 4 x + y x

12 – 3 y= 4 x + y x 12 – 3 y= 4 x + y x

12 – 3y = x(4 + y)12 – 3y = x(4 + y)x = x = (12 – 3y)(12 – 3y) (4 + y)(4 + y)


Get x’s to one side.Get x’s to one side.

Get everything away from x’sGet everything away from x’s

Make it only one x, (by factorising)Make it only one x, (by factorising)

Divide through by the coefficients.Divide through by the coefficients.


Practice Practice Theta textbook Page 29 exercise 4.2 (Applications)Theta textbook Page 29 exercise 4.2 (Applications) Homework Page 15 and 16Homework Page 15 and 16

Theta Textbook Page 31 ex (Harder) Theta Textbook Page 31 ex (Harder) Theta Textbook Page 35 ex (Harder applications)Theta Textbook Page 35 ex (Harder applications) Homework Page 17 and 18Homework Page 17 and 18

Simultaneous EquationsSimultaneous Equations

These are still similar to year 11, still involve These are still similar to year 11, still involve only 2 equations and 2 unknowns.only 2 equations and 2 unknowns.

We now have to be able to solve both linear We now have to be able to solve both linear and non-linear simultaneous equations.and non-linear simultaneous equations.


Linear Simultaneous EquationsLinear Simultaneous Equations– The challenge this year is that any linear The challenge this year is that any linear

simultaneous equations you are asked to solve simultaneous equations you are asked to solve will not have equations given. They will just be will not have equations given. They will just be word problems. So you need to be able to word problems. So you need to be able to interpret the information given, write 2 equations interpret the information given, write 2 equations and then solve the simultaneous equations.and then solve the simultaneous equations.

– The methods to solve them haven’t changedThe methods to solve them haven’t changed


Eg Eg Tara buys 8 cakes for her family. Five of the cakes are Tara buys 8 cakes for her family. Five of the cakes are

cream and three are plain. She spends $16.25 altogether. cream and three are plain. She spends $16.25 altogether. A cream cake costs 45 cents more than a plain cake. A cream cake costs 45 cents more than a plain cake. Calculate the price of ONE cream cake.Calculate the price of ONE cream cake.

–Step one: List the variablesStep one: List the variablesc = Cost of cream cake, p = Cost of plain c = Cost of cream cake, p = Cost of plain

cakecake

–Step two: Write the EquationsStep two: Write the Equations5c + 3p 5c + 3p = 16.25= 16.25

cc = p + 0.45= p + 0.45

Simultaneous EquationsSimultaneous Equations–Step one: List the variablesStep one: List the variables

c = Cost of cream cake, p = Cost of plain cakec = Cost of cream cake, p = Cost of plain cake

–Step two: Write the EquationsStep two: Write the Equations5c + 3p 5c + 3p = 16.25= 16.25

cc = p + 0.45= p + 0.45

–Step three: Solve the EquationsStep three: Solve the EquationsSubstitution MethodSubstitution Method

Make ‘c’ the subject of one sideMake ‘c’ the subject of one side

c = p + 0.45c = p + 0.45

Substitute ‘c’ into other equationSubstitute ‘c’ into other equation

5(p + 0.45) + 3p 5(p + 0.45) + 3p = 16.25= 16.25

Solve the equationSolve the equation

Elimination MethodElimination Method

5c + 3p 5c + 3p = 16.25= 16.25

(c – p (c – p = 0.45) = 0.45) × 5× 5

5c + 3p 5c + 3p = 16.25= 16.25

-(5c – 5p-(5c – 5p= 2.25) = 2.25)

0c + 8p 0c + 8p = 14.00= 14.00


PracticePractice– Theta TextbookTheta Textbook Page 39 ex 5.2Page 39 ex 5.2

Page 40 ex 5.4Page 40 ex 5.4

Page 41 ex 5.5Page 41 ex 5.5


Non-linear Simultaneous EquationsNon-linear Simultaneous Equations– This is really no more difficult than linear This is really no more difficult than linear

simultaneous equations. The only difference is simultaneous equations. The only difference is that on or both of the equations would draw a that on or both of the equations would draw a curve if graphed. (Normally a circle or Parabola)curve if graphed. (Normally a circle or Parabola)

– Some people find it easier as the equations are Some people find it easier as the equations are given.given.

– The best method to use is Substitution, as it will The best method to use is Substitution, as it will always work, Elimination is much trickier.always work, Elimination is much trickier.


Eg Solve xEg Solve x22 + y + y22 = 25 and y = x + 1 = 25 and y = x + 1 simultaneously.simultaneously.

Step 1: Rearrange one equation to make ‘y’ or ‘x’ the Step 1: Rearrange one equation to make ‘y’ or ‘x’ the subject.subject.

Y in the second equation is easiest so use that.Y in the second equation is easiest so use that. Step 2: Substitute ‘y’ into the equationStep 2: Substitute ‘y’ into the equation

xx22 + (x + 1) + (x + 1)22 = 25 = 25 Step 3: Solve the equation to find ‘x’.Step 3: Solve the equation to find ‘x’.

x = 3 or -4x = 3 or -4 Step 4: Substitute ‘x’ into one of the original Step 4: Substitute ‘x’ into one of the original

equations to find ‘y’.equations to find ‘y’.y = 4 or -3y = 4 or -3


Eg Find the x-coordinates of the point of Eg Find the x-coordinates of the point of intersection of y = 2x – 1 and xintersection of y = 2x – 1 and x22 + y + y22 – 4x – 5= 0. – 4x – 5= 0.

Step 1: Rearrange one equation to make ‘y’ or ‘x’ the subject.Step 1: Rearrange one equation to make ‘y’ or ‘x’ the subject.Y in the second equation is easiest so use that.Y in the second equation is easiest so use that.

Step 2: Substitute ‘y’ into the equationStep 2: Substitute ‘y’ into the equationxx22 + (2x – 1) + (2x – 1)22 – 4x – 5 = 0 – 4x – 5 = 0

Step 3: Solve the equation to find ‘x’.Step 3: Solve the equation to find ‘x’.x = 2 or -0.4x = 2 or -0.4

Step 4: Substitute ‘x’ into one of the original equations to find ‘y’.Step 4: Substitute ‘x’ into one of the original equations to find ‘y’.y = 3 or -1.8y = 3 or -1.8

Documents

Algebra External Achievement Standard (4 Credits)