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Algebraic Methods2

© University of Wales Newport 2009 This work is licensed under a Creative Commons Attribution 2.0 License.

Mathematics 1Level 4

2Algebraic Methods 2

The following presentation is an introduction to the Algebraic Methods – part one for level 4 Mathematics. This resources is a part of the 2009/2010 Engineering (foundation degree, BEng and HN) courses from University of Wales Newport (course codes H101, H691, H620, HH37 and 001H). This resource is a part of the core modules for the full time 1st year undergraduate programme.

The BEng & Foundation Degrees and HNC/D in Engineering are designed to meet the needs of employers by placing the emphasis on the theoretical, practical and vocational aspects of engineering within the workplace and beyond. Engineering is becoming more high profile, and therefore more in demand as a skill set, in today’s high-tech world. This course has been designed to provide you with knowledge, skills and practical experience encountered in everyday engineering environments.

Contents Factorisation Factorisation by grouping Factorisation of Quadratic Expressions Factorising Quadratics Difference between two squares Solving Quadratic Equations Completing the square Roots of a Quadratic Equations Special cases Exponential functions Logarithmic Functions The Logarithm Laws Logarithms to Base 10 Natural Logarithms (to the base e) Solving Exponential Equations using Logarithms Credits

In addition to the resource below, there are supporting documents which should be used in combination with this resource. Please see:

KA Stroud & DJ Booth, Engineering Mathematics, 8th Editon, Palgrave 2008.•http://www.mathcentre.ac.uk/•Derive 6

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Factorisation

With factorisation we start with an expression and we look for simpler expressions when multiplied together give us the original.

e.g. (x + 3)(x - 1) can be multiplied to give:

(x + 3)(x - 1) = x2 – x + 3x – 3 = x2 + 2x – 3

Factorisation is going from:

x2 + 2x – 3 to (x + 3)(x - 1)

Algebraic Methods 2

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Examples

1. (a + 2)(a + 4) 2. (a + 2)(a - 4)

3. (a - 2)(a + 4) 4. (a - 2)(a - 4)

5. (a + 3)(2a - 6) 6. (a + b)2

7. (a - b)2 8. (2x + 3y)2

9. (x - y)(x + y) 10. (2x - 3y)(x + 3y)

Algebraic Methods 2

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Looking for common factors

We will look for common terms within expressions:

e.g. 2x + 2y

Question: What can I divide each term by?

Answer 2 so 2(x + y)

3a + 6 same question

Answer 3 3(a+2)

mn3 – mn

Answer mn mn(n2 – 1)

5x + 10x2 - 15x3

Answer 5x 5x(1 + 2x - 3x2)

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Examples

1. 5a + 5b 2. 3a - 12

3. b2 + 4b 4. ab - ay

5. 2b – 4b2 6. 4x2 + 16x

7. a3 + a2 + a 8. 2a3 - 4a2 + a

9. 4b3 – 8b2 + 12b 10. 36c3 – 9c

Algebraic Methods 2

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Factorisation by groupingThis is possible only if there are four terms:

e.g. ax + ay + bx + by there are two pairs a… and b…

ax + ay + bx + by look for factors

a(x + y) +b(x + y) look for factors

(x + y)(a + b)

e.g. a + b – ay – by Group by common elements

a – ay + b – by

a(1 - y) + b(1 – y)

(a + b)(1 – y)

Algebraic Methods 2

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Examples

1. px + py + qx + qy 2. px - py - qx + qy

3. a(b + 1) – 3(b + 1) 4. x2(y – 1) – 6(y – 1)

5. 2ab – 4ac + bd -2dc 6. am + bn + bm + an

Algebraic Methods 2

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Factorisation of Quadratic ExpressionsQuadratic implies that there is a square of the

unknown term in the expression:

Quadratic expression: x2 + 3x – 4 no equals

Quadratic equation:x2 + 3x – 4 = 0 terms both sides of the =.

Consider the example x2 + 7x + 12

In this case the factors are: (x + 3)(x + 4)

To arrive at the factors perform the following operations:

What is the multiplier of the square term? 1

What is the constant term? 12

Multiply them 1 x 12 = 12 Algebraic Methods 2

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Factorising QuadraticsFactorising a quadratic means putting it into 2

brackets

For example, ax2 + bx + c

Steps: 1) multiply a and c, say, a x c = d

2) look at all pairs of numbers that multiply to give the value of d, but which also add/subtract to give the value of b, say, b1 and b2

3) split bx into b1x + b2x, so the original quadratic is re-written as ax2 + b1x + b2x + c

4) factorise the above expression by grouping

5) finally, look for the form in a product of two factors, i.e. ( )( )

Algebraic Methods 2

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Factorisation of Quadratic ExpressionsWhat are the factors of 12? (what numbers can be

multiplied to give 12)

+1 x +12 +2 x +6 +3 x +4

-1 x -12 -2 x -6 -3 x -4

Now add the pairs of factors: - look for the middle number

+1 + +12 = +13 +2 + +6 = +8 +3 + +4 = +7

-1 + -12 = -13 -2 + -6 = -8 -3 + -4 = -7

The only one is +3 and +4

So we have x2 + 3x + 4x + 12 = x(x +3) + 4(x + 3)

Giving us (x + 4)(x + 3)Algebraic Methods 2

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Factorisation of Quadratic Expressionse.g. x2 - 2x – 15 1 x -15 = -15+1 x -15 +3 x -5 -1 x 15 -3 x +5+1 + -15 = -14 +3 + -5 = -2 -1 + +15 = +14 -3 +

+5 = +2Result +3 and -5So x2 + 3x – 5x -15 = x(x + 3) – 5(x + 3)Giving us (x – 5)(x + 3)

e.g. 10x2 + 19x – 15 10 x -15 = -150+150 -150 +75 -75 +50 -50 +25 -25 +15 -

15 -1 +1 -2 +2 -3 +3 -6 +6 -10

+10+149 -149 +73 -73 +47 -47 +19 -19 +5 -

5Result +25 and -6So 10x2 + 25x - 6x – 15 = 5x(2x + 5) -3(2x +

5)Giving us (5x – 3)(2x + 5)

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Factorisation of Quadratic Expressions

e.g. 12x2 + 11x – 15 12 x -15 = -180+180 -180 +90 -90 +60 -60 +45 -45 +36 -36 +30 -30 +20 -20 +18 -18

+15 -15 -1 +1 -2 +2 -3 +3 -4 +4 -5 +5 -6 +6 -9 +9 -10 +10 -

12 +12+179 -179 +88 -88 +57 -57 +41 -41 +31 -31 +24 -24 +11 -11 +8 -8

+3 -3

Result +20 -9So 12x2 +20x -9x -15 4x(3x + 5) – 3(3x +5)Giving us (4x – 3)(3x + 5)

Algebraic Methods 2

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Examples1. x2 + 5x + 6 2. x2 + 7x + 12

3. x2 – x - 12 4. x2 - 2x – 15

5. x2 + 8x + 15 6. x2 – 8x + 15

7. x2 + 2x + 1 8. x2 - 4x – 21

9. x2 + 4x – 21 10. x2 – 12x + 20

11. x2 – 4x – 5 12. x2 + 8x + 7

13. x2 - 11x + 30 14. x2 – 11x + 28

15. x2-4x + 4 16. 3x2 + 7x – 4

17. 3x2 – 5x + 2 18. 2x2 + 7x – 4

19. 2x2 + 7x + 6 20. 6x2 -13x + 2Algebraic Methods 2

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Difference between two squaresYou will recognise these as there will only be two

terms and one of the terms will be positive and the other negative.

If we have: Term1 – Term2 the factors will be:

(√Term1 + √Term2)(√Term1 - √Term2)

This is most useful when the terms are squares of numbers

e.g. Factorise x2 – y2

(x + y)(x – y)e.g. Factorise x2 – 9

(x + 3)(x – 3)e.g. Factorise 9x2 – 25

(3x + 5)(3x – 5)Algebraic Methods 2

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Difference between two squaresIt may be necessary before the factorisation is

carried out to look for common terms:e.g. Factorise 18c2 – 50

2(3c + 5)(3c – 5)

e.g. Factorise 3x2 – 108 3(x + 6)(x – 6)

1. x2 - 4 2. x2 - 49

3. x2 – 64 4. x2 - 100

5. 2x2 - 200 6. 4x2 – 25

7. 9x2 -36y2 8. a2 – b2

9. 25x2 – 100y2 10. a2 + b2

= 2(9c2 – 25)

= 3(x2 – 36)

Algebraic Methods 2

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Solving Quadratic EquationsAs has been stated earlier:

Quadratic expression: x2 + 3x – 4 no equals

Quadratic equation:x2 + 3x – 4 = 0 terms both sides of the =.

The equation will have a value or values of x which satisfy the expression. These can be found by factorising the quadratic expression and equating to zero.

x2 + 3x – 4 = 0

+4 -4 +2

-1 +1 -2

+3 -3 0

1 x -4 = -4

Result +4 -1

So x2 + 4x – x - 4

x(x + 4) - 1(x + 4)

Giving us (x – 1)(x + 4)Algebraic Methods 2

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Solving Quadratic EquationsWe therefore have (x – 1)(x + 4) = 0

For this to be true either (x – 1) = 0 or (x + 4) = 0

This gives us two results x = +1 or x = -4

Check:

x = 1 x2 + 3x – 4 12 + 3 x 1 – 4 = 0 x = -4 x2 + 3x – 4 (-4)2 + 3 x (-4) – 4 = 0 Therefore any quadratic equation that can be

factorised can be solved.

What happens if it does not factorise easily due to fractional values?

x2 + 2.5x – 1.5 = 0 we can scale it by multiplying both sides – in this case by 2 Algebraic Methods 2

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Solving Quadratic Equations2x2 + 5x - 3 = 0 2 x -3 = -6

+6 -6 +3 -3

-1 +1 -2 +2

+5 -5 +1 -1

Therefore 2x – 1 = 0 or x + 3 = 0 x = +½ or x = -3

If the equation cannot be factorised in this way then it is possible to complete the square or use an equation to solve the quadratic equation.

result +6 -1

2x2 + 6x – x – 3

2x(x + 3) – 1(x + 3)

(2x – 1)(x + 3) = 0

Algebraic Methods 2

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Completing the squareSolve x2 – 4x – 8 = 0 1 x -8 = -8

+8 -8 +4 -4

-1 +1 -2 +2

+7 -7 +2 -2 no result

Rearrange the equation

x2 – 4x – 8 = 0 x2 – 4x = 8

Now add to both sides the number that will make the left hand side a perfect square.

This is (½ x the multiplier of x)2 in our case (-4/2)2 = 4

x2 – 4x + 4 = 12Algebraic Methods 2

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Completing the squareThe left hand side will factorise to (x – 2)2

Giving us (x – 2)2 = 12

Root both sides x – 2 = ± √12

(± as any root of a positive number can have two values e.g. √4 = +2 or -2)

So x = 2 ± √12

So x = 2 ± 3.464 = +5.464 or -1.464

Algebraic Methods 2

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Roots of a Quadratic EquationsAll quadratic equations will have the form:

ax2 + bx + c = 0 where a, b and c may be positive, negative or zero.

The equation below allows us to determine the values of x which satisfy the quadratic equation:

aacbb

x2

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Let us try an equation which we can factorise to check that it works:Consider the example x2 + 7x + 12 = 0In this case the factors are: (x + 3)(x + 4) = 0

x = -3 or x = -4Algebraic Methods 2

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Roots of a Quadratic Equations

428

326

217

217

217

12121477

24 22

xorx

xorx

xxx

aacbb

x

a = 1, b = 7 and c = 12

Try the example:

x2 – 4x – 8 = 0

Algebraic Methods 2

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Special casesNot all quadratic equations produce a pair of real values. It is possible that only a single value exists or that no real number will satisfy the equation.

The way that this can be checked is by examining the square root in the equation.

acb 42

If b2 – 4ac is positive then we will have a pair of real values for the answer.

If b2 – 4ac is zero then there will be a single value as the answer equal to –b/2a.

If b2 – 4ac is negative then there will be no real answer to the equation (the answer will be a complex number). Algebraic Methods 2

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ExamplesSolves the following quadratic equations using any

method.

1. 2x2 – 5x – 3 = 0 2. 2x2 – 7x – 1 = 0

3. 3x2 – 2x – 6 = 0 4. 3x2 + 2x – 1 = 0

5. 4x2 + 6x – 2 = 0 6. 4x2 + 6x – 1 = 0

7. 2x2 – 5x + 1 = 0 8. 3x2 + 8x + 2 = 0

9. 4x2 – 6x = -1 10. 5x2 + 8x = -2

11. 2x2 = 6x – 3 12. 5x2 + 3 = 9xAlgebraic Methods 2

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Exponential functions An exponential function may be given

f(x) = bX

where b is the base, and x is the exponent (or power).

If b is positive, the function continuously increases in value as x increases. Also, the slope of the function increases continuously.

Calculators have a button labelled x y

Algebraic Methods 2

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Exponential functions

Example of an Exponential Function

Consider the function f(x) = 2X.

In this case, the exponential function has a base 2. Some typical values for this function are:

Plotting y = 2X gives a curve as follows.

x -2 -1 0 1 2 3

(x) 0.25 0.5 1 2 4 8

Algebraic Methods 2

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Exponential functions

Note:

That Y increases dramatically as x increases

That the slope of the curve also increases as x increases. That the curve passes through (0, 1). In fact, all

exponential curves, f(x) = bX, pass through (0, 1) if b > 0.

The curve does not pass through the x-axis, but asymptotes the x-axis as the x-value gets smaller and smaller. Algebraic Methods 2

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Logarithmic Functions A logarithm is simply an exponent that is written in a special manner.

For example, for the following exponential equation: 32 = 9

We know that the base is 3 and the exponent is 2. We can re-write this equation in a form:

log39 = 2

It is called as "the logarithm of 9 to the base 3 is 2". What has been effectively done is to move the exponent down on to the main line, and move the base to the bottom. This was used historically to make multiplications and divisions easier, but logarithms are still very popular in mathematics and engineering sciences. Algebraic Methods 2

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Logarithmic Functions The logarithmic function is defined as:

f(x) = logbx

The base of the logarithm is b.

The two most common bases are base 10 and base e, which are represented as

Log (base 10) – common logarithm, and

Ln (base e) – natural logarithm.

The logarithmic function has many practical applications, such as acoustics, electronics, earthquake analysis and population prediction.

Algebraic Methods 2

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Examples 1. Write in logarithm form: 8 = 23

2. Write in exponential form: log101000 = 3

3. Find b if

4. Evaluate y = 9x if x = 0.5

5. Express 82 = 64 in logarithmic form.

6. Express log11121 = 2 in exponential form.

7. Determine the unknown: log10 0.01 = x

8. Determine the unknown: logb (1/4) = -1/2Algebraic Methods 2

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The Logarithm Laws Since a logarithm is still an exponent in another form,

logarithm laws work in the same way as the rules for indices.

Exponents Logarithms

bm × bn = bm+n logb xy = logb x + logb y

bm ÷ bn = bm-n logb (x/y) = logb x − logb y

(bm)n = bmn logb (xn) = n logb x

b1 = b logb (b) = 1

b0 = 1 logb (1) = 0Algebraic Methods 2

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Examples In the following examples, the base is not specified,

since the laws hold for logarithms with any positive integer base.

1. Expand log 7x as the sum of 2 logarithms.

2. Using your calculator, show that log (20/5) = log 20 − log 5.

3. Express as a multiple of logarithms: log x5.

4. All of these are equal to 1:

log6 6 = log10 10 = logx x = loga a = 1

5. All of these are equivalent to 0:

log7 1 = log10 1 = loge1 = logx 1 = 0 Algebraic Methods 2

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Examples 1. Express as a sum, difference, or multiple of

logarithms:

2. Express as the logarithm of a single quantity. 3. Determine the exact value of:

4. Solve for y in terms of x: log2x + log2y = 1Algebraic Methods 2

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Logarithms to Base 10 Logarithms to Base 10 were used extensively for

calculation up until the calculator was adopted. The concept of logarithms is still very important in many fields of science and engineering, such as acoustics.

Calculators allow us to use logarithms to base 10. These are called common logarithms ("log" on a calculator). We normally do not include the 10 when we write logarithms to base 10.

Therefore, log x means log10 x.

1. Find the logarithm of 5623 to base 10. Write this in exponential form.

2. Find the antilogarithm of -6.9788

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The number e frequently occurs in mathematics (especially calculus) and is an irrational constant (like π). Its value is e = 2.718 281 828 ...

Apart from logarithms to base 10, we can also have logarithms to base e. These are called natural logarithms.

We usually write natural logarithms as:ln x to mean logex (that is, "log x to the base e")Natural logarithms are commonly used throughout

science and engineering. Where does this value "e" come from? The series expansion for e is

x=1

Natural Logarithms (to the base e)

...!5!4!3!2

15432

xxxx

xex

...!5

1!4

1!3

1!2

111 xe

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Solve the equation 3X = 12.7.

We can estimate the answer before we start to be somewhere between 2 and 3, because 32 = 9 and 33 = 27. But how do we find the answer?

First we take the logarithm of both sides of the given equation: log 3X = log 12.7

x log 3 = log 12.7

Now divide both sides by log 3:

x = (log 12.7)/(log 3) = 2.3135

Solving Exponential Equations using Logarithms

Algebraic Methods 2

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Two populations of bacteria are growing at rates of 5t+2 and e2t respectively (at time t). At what time are the populations the same?

This problem requires us to solve the equation:

5t+2 = e2t

We need to use loge because of the base e on the right hand side.

ln (5t+2) = ln (e2t) (t + 2) ln 5 = 2t ln e

Now, ln e = 1, and we need to collect t terms together:

t ln 5 + 2 ln 5 = 2t t (ln 5 - 2) = - 2 ln 5

So is the required time.

Solving Exponential Equations using Logarithms

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1. Solve 5X = 0.32. Solve 3 log(2x - 1) = 13. Solve for x: log2x + log2 7 = log2214. Solve for x:

5. I have the following formula:S(n) = 5500 log n + 15000

If I know S(n) = 40 million, What is the value of n

Application - World population growth The population of the earth is growing at

approximately 1.3% per year. The population at the beginning of 2000 was just over 6 billion. After how many more years will the population double to 12 billion?

Solving Exponential Equations using Logarithms

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