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- 1. Algebraic Methods 2 © University of Wales Newport 2009 This work is licensed under a Creative Commons Attribution 2.0 License . Mathematics 1 Level 4
- 2. <ul><li>The following presentation is on the second part of the Algebraic Methods 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 1 st year undergraduate programme. </li></ul><ul><li>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. </li></ul><ul><li>Contents </li></ul><ul><li>Factorisation </li></ul><ul><li>Factorisation by grouping </li></ul><ul><li>Factorisation of Quadratic Expressions </li></ul><ul><li>Factorising Quadratics </li></ul><ul><li>Difference between two squares </li></ul><ul><li>Solving Quadratic Equations </li></ul><ul><li>Completing the square </li></ul><ul><li>Roots of a Quadratic Equations </li></ul><ul><li>Special cases </li></ul><ul><li>Exponential functions </li></ul><ul><li>Logarithmic Functions </li></ul><ul><li>The Logarithm Laws </li></ul><ul><li>Logarithms to Base 10 </li></ul><ul><li>Natural Logarithms (to the base e) </li></ul><ul><li>Solving Exponential Equations using Logarithms </li></ul><ul><li>Credits </li></ul><ul><li>In addition to the resource below, there are supporting documents which should be used in combination with this resource. Please see: </li></ul><ul><li>KA Stroud & DJ Booth, Engineering Mathematics, 8 th Editon, Palgrave 2008. </li></ul><ul><li>http://www.mathcentre.ac.uk/ </li></ul><ul><li>Derive 6 </li></ul>Algebraic Methods 2
- 3. 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) = x 2 – x + 3x – 3 = x 2 + 2x – 3 Factorisation is going from: x 2 + 2x – 3 to (x + 3)(x - 1) Algebraic Methods 2
- 4. 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
- 5. 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) mn 3 – mn Answer mn mn(n 2 – 1) 5x + 10x 2 - 15x 3 Answer 5x 5x(1 + 2x - 3x 2 )
- 6. Examples 1. 5a + 5b 2. 3a - 12 3. b 2 + 4b 4. ab - ay 5. 2b – 4b 2 6. 4x 2 + 16x 7. a 3 + a 2 + a 8. 2a 3 - 4a 2 + a 9. 4b 3 – 8b 2 + 12b 10. 36c 3 – 9c Algebraic Methods 2
- 7. Factorisation by grouping This 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
- 8. Examples 1. px + py + qx + qy 2. px - py - qx + qy 3. a(b + 1) – 3(b + 1) 4. x 2 (y – 1) – 6(y – 1) 5. 2ab – 4ac + bd -2dc 6. am + bn + bm + an Algebraic Methods 2
- 9. Factorisation of Quadratic Expressions Quadratic implies that there is a square of the unknown term in the expression: Quadratic expression: x 2 + 3x – 4 no equals Quadratic equation: x 2 + 3x – 4 = 0 terms both sides of the =. Consider the example x 2 + 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
- 10. Factorising Quadratics Factorising a quadratic means putting it into 2 brackets For example, ax 2 + 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, b 1 and b 2 3) split bx into b 1 x + b 2 x, so the original quadratic is re-written as ax 2 + b 1 x + b 2 x + 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
- 11. Factorisation of Quadratic Expressions What 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 x 2 + 3x + 4x + 12 = x(x +3) + 4(x + 3) Giving us (x + 4)(x + 3) Algebraic Methods 2
- 12. Factorisation of Quadratic Expressions e.g. x 2 - 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 = +2 Result +3 and -5 So x 2 + 3x – 5x -15 = x(x + 3) – 5(x + 3) Giving us (x – 5)(x + 3) e.g. 10x 2 + 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 -5 Result +25 and -6 So 10x 2 + 25x - 6x – 15 = 5x(2x + 5) -3(2x + 5) Giving us (5x – 3)(2x + 5)
- 13. Factorisation of Quadratic Expressions e.g. 12x 2 + 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 -9 So 12x 2 +20x -9x -15 4x(3x + 5) – 3(3x +5) Giving us (4x – 3)(3x + 5) Algebraic Methods 2
- 14. Examples 1. x 2 + 5x + 6 2. x 2 + 7x + 12 3. x 2 – x - 12 4. x 2 - 2x – 15 5. x 2 + 8x + 15 6. x 2 – 8x + 15 7. x 2 + 2x + 1 8. x 2 - 4x – 21 9. x 2 + 4x – 21 10. x 2 – 12x + 20 11. x 2 – 4x – 5 12. x 2 + 8x + 7 13. x 2 - 11x + 30 14. x 2 – 11x + 28 15. x 2 -4x + 4 16. 3x 2 + 7x – 4 17. 3x 2 – 5x + 2 18. 2x 2 + 7x – 4 19. 2x 2 + 7x + 6 20. 6x 2 -13x + 2 Algebraic Methods 2
- 15. Difference between two squares You 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 x 2 – y 2 (x + y)(x – y) e.g. Factorise x 2 – 9 (x + 3)(x – 3) e.g. Factorise 9x 2 – 25 (3x + 5)(3x – 5) Algebraic Methods 2
- 16. Difference between two squares It may be necessary before the factorisation is carried out to look for common terms: e.g. Factorise 18c 2 – 50 2(3c + 5)(3c – 5) e.g. Factorise 3x 2 – 108 3(x + 6)(x – 6) 1. x 2 - 4 2. x 2 - 49 3. x 2 – 64 4. x 2 - 100 5. 2x 2 - 200 6. 4x 2 – 25 7. 9x 2 -36y 2 8. a 2 – b 2 9. 25x 2 – 100y 2 10. a 2 + b 2 = 2(9c 2 – 25) = 3(x 2 – 36) Algebraic Methods 2
- 17. Solving Quadratic Equations As has been stated earlier: Quadratic expression: x 2 + 3x – 4 no equals Quadratic equation: x 2 + 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. x 2 + 3x – 4 = 0 +4 -4 +2 -1 +1 -2 +3 -3 0 1 x -4 = -4 Result +4 -1 So x 2 + 4x – x - 4 x(x + 4) - 1(x + 4) Giving us (x – 1)(x + 4) Algebraic Methods 2
- 18. Solving Quadratic Equations We 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 x 2 + 3x – 4 1 2 + 3 x 1 – 4 = 0 x = -4 x 2 + 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? x 2 + 2.5x – 1.5 = 0 we can scale it by multiplying both sides – in this case by 2 Algebraic Methods 2
- 19. Solving Quadratic Equations 2x 2 + 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 2x 2 + 6x – x – 3 2x(x + 3) – 1(x + 3) (2x – 1)(x + 3) = 0 Algebraic Methods 2
- 20. Completing the square Solve x 2 – 4 x – 8 = 0 1 x -8 = -8 +8 -8 +4 -4 -1 +1 -2 +2 +7 -7 +2 -2 no result Rearrange the equation x 2 – 4 x – 8 = 0 x 2 – 4 x = 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 x 2 – 4 x + 4 = 12 Algebraic Methods 2
- 21. Completing the square The 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
- 22. Roots of a Quadratic Equations All quadratic equations will have the form: ax 2 + 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: Let us try an equation which we can factorise to check that it works: Consider the example x 2 + 7x + 12 = 0 In this case the factors are: (x + 3)(x + 4) = 0 x = -3 or x = -4 Algebraic Methods 2
- 23. Roots of a Quadratic Equations a = 1, b = 7 and c = 12 Try the example: x 2 – 4 x – 8 = 0 Algebraic Methods 2
- 24. Special cases Not 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. If b 2 – 4ac is positive then we will have a pair of real values for the answer. If b 2 – 4ac is zero then there will be a single value as the answer equal to –b/2a. If b 2 – 4ac is negative then there will be no real answer to the equation (the answer will be a complex number). Algebraic Methods 2
- 25. Examples <ul><li>Solves the following quadratic equations using any method. </li></ul><ul><li>2x 2 – 5x – 3 = 0 2. 2x 2 – 7x – 1 = 0 </li></ul><ul><li>3x 2 – 2x – 6 = 0 4. 3x 2 + 2x – 1 = 0 </li></ul><ul><li>4x 2 + 6x – 2 = 0 6. 4x 2 + 6x – 1 = 0 </li></ul><ul><li>2x 2 – 5x + 1 = 0 8. 3x 2 + 8x + 2 = 0 </li></ul><ul><li>4x 2 – 6x = -1 10. 5x 2 + 8x = -2 </li></ul><ul><li>11. 2x 2 = 6x – 3 12. 5x 2 + 3 = 9x </li></ul>Algebraic Methods 2
- 26. Exponential functions Exponential functions have the form: f(x) = b X where b is the base and x is the exponent (or power ). If b is positive, the function continuously increases in value. A special property of exponential functions is that the slope of the function also continuously increases as x increases. Calculators have a button labelled x y Algebraic Methods 2
- 27. Exponential functions Example of an Exponential Function Consider the function f ( x ) = 2 X . In this case, we have an exponential function with base 2. Some typical values for this function would be: On the next slide is the graph of y = 2 X . Algebraic Methods 2 x -2 -1 0 1 2 3 (x) 0.25 0.5 1 2 4 8
- 28. Exponential functions <ul><li>Note: </li></ul><ul><li>That as x increases, y also increases. </li></ul><ul><li>That as x increases, the slope of the graph also increases. </li></ul><ul><li>That the curve passes through (0, 1). All exponential curves of the form f ( x ) = b X pass through (0, 1), if b > 0. </li></ul><ul><li>The curve does not pass through the x -axis. It just gets closer and closer to the x -axis as we take smaller and smaller x -values. </li></ul>Algebraic Methods 2
- 29. Logarithmic Functions A logarithm is simply an exponent that is written in a special way. For example, we know that the following exponential equation is true: 3 2 = 9 In this case, the base is 3 and the exponent is 2. We can write this equation in logarithm form (with identical meaning) as follows: log 3 9 = 2 We say this as "the logarithm of 9 to the base 3 is 2". What we have effectively done is to move the exponent down on to the main line. This was done historically to make multiplications and divisions easier, but logarithms are still very handy in mathematics. Algebraic Methods 2
- 30. Logarithmic Functions The logarithmic function is defined as: f ( x ) = log b x The base of the logarithm is b . The 2 most common bases that we use are base 10 and base e , which are referred to as Log (base 10) and Ln (base e ). The logarithmic function has many real-life applications, in acoustics, electronics, earthquake analysis and population prediction. Algebraic Methods 2
- 31. Examples <ul><li>1. Write in logarithm form: 8 = 2 3 </li></ul><ul><li>2. Write in exponential form: log 10 1000 = 3 </li></ul><ul><li>Find b if </li></ul><ul><li>4. Evaluate y = 9 x if x = 0.5 </li></ul><ul><li>5. Express 8 2 = 64 in logarithmic form. </li></ul><ul><li>6. Express log 11 121 = 2 in exponential form. </li></ul><ul><li>7. Determine the unknown: log 10 0.01 = x </li></ul><ul><li>8. Determine the unknown: log b (1/4) = -1/2 </li></ul>Algebraic Methods 2
- 32. The Logarithm Laws Since a logarithm is simply an exponent which is just being written down on the line, we expect the logarithm laws to work the same as the rules for indices, and luckily, they do. Exponents Logarithms b m × b n = b m + n log b xy = log b x + log b y b m ÷ b n = b m-n log b ( x/y ) = log b x − log b y ( b m ) n = b mn log b ( x n ) = n log b x b 1 = b log b ( b ) = 1 b 0 = 1 log b (1) = 0 Algebraic Methods 2
- 33. Examples In these examples, I am not including any base, since the laws hold for logarithms with any positive integer base. 1. Expand log 7 x 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 x 5 . 4. All of these are equal to 1: log 6 6 = log 10 10 = log x x = log a a = 1 5. All of these are equivalent to 0: log 7 1 = log 10 1 = log e 1 = log x 1 = 0 Algebraic Methods 2
- 34. 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 : log 2 x + log 2 y = 1 Algebraic Methods 2
- 35. 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. One example is acoustics. Our 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. We write log x to mean log 10 x. 1. Find the logarithm of 5623 to base 10. Write this in exponential form. 2. Find the antilogarithm of -6.9788
- 36. Natural Logarithms (to the base e) 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 log e x (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
- 37. Solving Exponential Equations using Logarithms Solve the equation 3 X = 12.7. We can estimate the answer before we start to be somewhere between 2 and 3, because 3 2 = 9 and 3 3 = 27. But how do we find the answer? First we take the logarithm of both sides of the given equation: log 3 X = log 12.7 x log 3 = log 12.7 Now divide both sides by log 3: x = (log 12.7)/(log 3) = 2.3135 Algebraic Methods 2
- 38. Solving Exponential Equations using Logarithms Two populations of bacteria are growing at rates of 5 t +2 and e 2 t respectively (at time t ). At what time are the populations the same? This problem requires us to solve the equation: 5 t +2 = e 2 t We need to use log e because of the base e on the right hand side. ln (5 t+ 2 ) = ln ( e 2 t ) ( t + 2) ln 5 = 2 t ln e Now, ln e = 1 , and we need to collect t terms together: t ln 5 + 2 ln 5 = 2 t t (ln 5 - 2) = - 2 ln 5 So is the required time.
- 39. Solving Exponential Equations using Logarithms 1. Solve 5 X = 0.3 2. Solve 3 log(2 x - 1) = 1 3. Solve for x : log 2 x + log 2 7 = log 2 21 4. 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?
- 40. <ul><li>This resource was created by the University of Wales Newport and released as an open educational resource through the Open Engineering Resources project of the HE Academy Engineering Subject Centre. The Open Engineering Resources project was funded by HEFCE and part of the JISC/HE Academy UKOER programme. </li></ul><ul><li>© 2009 University of Wales Newport </li></ul><ul><li>This work is licensed under a Creative Commons Attribution 2.0 License . </li></ul><ul><li>The JISC logo is licensed under the terms of the Creative Commons Attribution-Non-Commercial-No Derivative Works 2.0 UK: England & Wales Licence. All reproductions must comply with the terms of that licence. </li></ul><ul><li>The HEA logo is owned by the Higher Education Academy Limited may be freely distributed and copied for educational purposes only, provided that appropriate acknowledgement is given to the Higher Education Academy as the copyright holder and original publisher. </li></ul><ul><li>The name and logo of University of Wales Newport is a trade mark and all rights in it are reserved. The name and logo should not be reproduced without the express authorisation of the University. </li></ul>Algebraic Methods 2

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