CHAPTER 3 – More on operations 3.1 EXPONENTIATION We have seen multiplication is repeated addition. For example 2 + 2 + 2 = 3(2). EXPONENTIATION is repeated multiplication. We write: 2 x 2 x 2 = 23. The 3 called the exponent is placed as a superscript to the 2. We read this as ‘2 the power of 3’. 23 = 8 because (2)(2)(2) = 8. We also say 2 ‘raised to the power’ 3 is 8. Note 3 to the power of 1 is 31 = 3. (just one copy of 3) The index of 1 is silent and redundant. We do not explicitly write it. E3.1 Evaluate without the use of a calculator: (i) 24 = (ii) 32 = (iii) -53 = (iv) 61 = (v) 7 to the power of 2 = (vi) -1101 A3.1 (i) 24 = 16 (ii) 32 = 9 (iii) -53 = -125 (iv) 61 = 6 (v) 7 to the power of 2 = 72 = 49 (vi) -1101 = -1 3.2 Here are some examples of exponentiations. 1. 2. 3. We know the area of a square of 5m by 5m is given in square meters by the equation, Area = 5x5 = 52 = 25. The volume of a cube of 2m by 2m by 2m is given in cubic metres by the equation, Volume = 2x2x2 = 23 = 8. An investment of £1 doubles in value every year. Then its value will be £21 = £2 after 1 year, £22 = £4 after 2 years, £23 = £8 after 3 years etc. E3.2 An investment of £1 doubles in value every year. Give in exponent form its value after 10 years. Evaluate this expression. A3.2 3.3 V10 = £210 = £1024 When we write 23 = 8 the 2 is called the base and the 3 is called the Power or Index or Exponent. E3.3 Complete the Table (1st row has been done for you) Base Index Expression Value 2 3 23 8 3 2 43 5 125 10 1 1 What 2 other terms could have been used for the heading of the second column? A3.3 Base 2 3 4 5 1 3.4 Index 3 2 3 3 10 Expression 23 32 43 53 110 Value 8 9 64 125 1 Now let’s look at the case where the base is a negative number. For example (-3)(-3) = 9 We can write this as (-3)2 = 9. Note the result is a positive number. Note writing (-3)(-3) as -32 can be ambiguous. It could mean the square of (-3)(-3) = 9 or it could mean the ‘negative of the square of 32 = -9’ so avoid it. Check what you calculator says when you key in -32. E3.4 A3.4 3.5 Evaluate the following: Verify using your calculator. (i) (-2)3 = (ii) (-5)2 (-3)4 = (i) (-2)3 = (-2)(-2)(-2) = -8 (ii) (-5)2 = 25 (-3)4 = -27 Now note (-1)1 = -1, a negative result (-1)2 = (-1)(-1) = 1, a positive result (-1)3 = (-1)(-1)(-1) = -1, a negative result (-1)4 = (-1)(-1)(-1)(-1) = 1, appositive result With a negative base, when the Index or Exponent is odd the result is negative. With a negative base, when the Index or Exponent is even the result is positive. E3.5 Evaluate (i) (-1)99 = (ii) (-1)256 = A3.5 3.6 (i) (-1)99 = -1 (ii) (-1)256 = 1 Now if we have an expression like 2x2x2x3x3x3x3 we can use exponentiation to write it as 23x34 or (23)(34) It value is 8x81 = 648. E3.6 Write in exponent form and then evaluate: (i) 2x2x4x4x4 = (ii) 3x5x5 = (ii) 2x3x3x2x3 = A3.6 (i) 2x2x4x4x4 = (22)(43) = 4 x 64 = 256 (ii) 3x5x5 = 3(52) = 3(25) = 75 (iii) 2x3x3x2x3 =(22)(33) = 4 x 27 = 128 3.7 Now we can have fractions with repeated multiplication in the numerator and denominator. For example: 2 x 2 x 2 23 8 2 3x3 3 9 E3.7 Complete the Table (1st row has been done for you) Fraction 3 x3 5 x5 3 x 3 x3 2 x2 3 3 55 Index form Value 9 32 25 52 32 (52 ) 22 3? 7? 27 49 9 25 A3.7 Fraction 3 x3 5 x5 3 x 3 x3 2 x2 Index form Value 9 32 2 25 5 27 33 4 22 3.8 3 x 3 x 5 x5 225 32 (52 ) 2 4 2 x2 2 3 27 3 x3 x3 3 2 7 x7 49 7 3 3 (3)2 9 25 55 52 Now let’s look at the following. (4)(4)(4)(4)(4) 45 (i) 3 (4)(4)(4) 4 (4)(4)(4)(4)(4) (4)(4)(4 )((4 )((4 ) (4)(4) 42 but using cancellati on 42 (4)(4)(4) ((4 )((4 )((4 ) 1 1 Note denominato r in not a zero but a silent 1. so 45 42 43 so we get the rule : Another example: E3.8 A3.8 3.9 E3.9 A3.9 3.10 Simplify 45 45 - 3 4 2 3 4 310 3107 33 7 3 7 25 7 ............. 23 7 7 25 7.25 23 7 2 23 7 43 So how does 5 simplify? 4 (4)(4)(4) (4 )(4 )(4 ) 1 1 2 (4)(4)(4)(4)(4) (4 )(4 )(4 )(4)(4) (4)(4) 4 (4)(4)(4) 43 Using the rule from above we get also get 5 4 3-5 4 -2 (4)(4)(4)(4)(4) 4 1 1 Hence 4 -2 2 16 4 Simplify expressing the final result with a positive and negative exponent. 7 49 7 39 (i) 59 = (ii) 49 = 7 7 49 7 1 739 49 59 10 7 10 (i) 59 7 (ii) 49 739 49 7 10 7110 7 7 7 Now we have a definition for a negative index. We call this relationship a Reciprocal relationship The reciprocal involves moving the quantity (base with index) from the numerator to the denominator. Consider 2-1. We get 2-1 = reciprocal of 2 i.e. 1 1 , which we write as , because the index of 1 1 2 2 is silent. 1 23 8 8. Note we don’t have to write the denominator 1 1 2 3 of 1 explicitly. It is silent. Similarly we get E3.10 Evaluate without a calculator: (i) 2 4 1 3 2 A3.10 1 1 (i) 2 4 4 2 16 (ii) 1 32 9 3 2 1 Let’s use the reciprocal key on our calculator. (ii) 3.11 Find the reciprocal key on your calculator. It will appear either as 1 x or as x-1 Now key in the sequence: 2 followed by the reciprocal key. What is the result? The result should be a half. Press the reciprocal key several times. What do you see? You should the value in the display toggle between 0.5 and 2. So we get an important rule that if a quantity migrates from the numerator to the denominator the base remains the same but the exponent (power or index) changes sign. E3.11 Use your reciprocal key to answer these: (i) what is the reciprocal of 5? (ii) What is the reciprocal of 0.2? A3.11 (i) reciprocal of 5 = 1/5 = 0.2 (ii) reciprocal of 0.2 = 1/0.2 = 5 3.12 Note that only the base and the index migrate from numerator to denominator viz: 3(25 ) 3 . 25 Also 4 4(25 ) 3(2-5 ) 3 E3.12 Write the following with positive exponents: (i) 2 5 = 2 5 (ii) 4 = 3 (iii)4(3-2) = 4(5 3 ) (iv) 5(2 2 ) 3(25 )(53 ) 11(7 2 ) A3.12 1 (i) 2 5 = 5 2 5 2 34 (ii) 4 5 3 2 4 (iii)4(3-2) = 2 3 3 4(5 ) 4(22 ) 4(22 ) (iv) = 4 5 5(2 2 ) 5(53 ) (v) 3(25 )(53 ) 3(25 )(7 2 ) 11(7 2 ) 11(53 ) More on Rooting We have already looked at square roots. Now let’s consider cube roots. (v) 3.13 We know 2x2x2 = 23 = 8. We say the cube of 2 is 8. The inverse is: The cube root of 8 is 2. We write these as : 23 = 8 and the inverse as 3 8 2. Also note the cube of a negative quantity is negative because (-2)(-2)(-2) = -8 (-2)3 = -8 and the inverse as E3.13 8 2 . Complete the table: The first has been done for you. 3 2 =8 Inverse 3 82 3 5 3 64 3 (-1) = A3.13 3 2 =8 Inverse 3 82 53 = 125 3 3 3 3 1 1 64 -4 Roots as Exponents Now we know that 32 = 3 x 3 = 9. (-1) = -1 (-4)3 = -64 3.14 125 5 3 If we show the ‘silent’ 1 exponents we can write 31 x 31 = 31+1 = 32 = 9. If we insert an exponent of a half we get. 3 3 3 1 2 1 2 1 1 2 2 31 3 (√3)(√3) = 3. This tells us that But 1 3 32 Now note that if we put in the silent value of 2 in front of the root sign and the silent exponent of 1 we get: 2 1 31 3 2 The change a root to an exponent the rule is write the number in front of the root sign in the denominator and the index value inside the root sign in the numerator. So other examples are: E3.14 1 4 16 16 4 5 23 2 5 3 Complete the Table: The first row has been done for you. Root Form Exponent Form 4 53 3 54 2 57 3 72 5 2 7 A3.14 Root Form Exponent Form 4 53 3 54 7 52 3 72 2 57 57 (5) 2 Now let us use our calculator to evaluate roots. 2 7 3.15 Now we know without using a calculator that the cube root of 8 is 2. i.e. 3 82 To verify this with a calculator we use convert the root to an exponent and then use the exponent key on the calculator. Find the exponent key on your calculator. The exponent key will appear as a xy key as x□ key. So to verify 8 2 key in the sequence: 3 8 then x□ then ( 1 / 3 ) i.e. one third in brackets then finally the = or Ans key. The calculator should show the result as 2. Another example is 2 216 2 216 3 36 Express in index form: 3 E3.15 27 2 4 256 (ii) (i) 4 256 256 4 (ii) A3.15 3.16 3 (i) (iii) 6 (2)12 2 1 3 272 27 3 12 (iii) 6 (2)12 (2) 6 (2)2 Mixed mode expressions Mixed mode expressions are expressions in which we have a mix of add, subtract, exponentiation etc. In evaluating these expressions we use the following Prioritisation hierarchy: Brackets-----------------------------Highest Priority Calculated first EXPONENTIATION MULTIPLY or DIVIDE ADD or SUBTRACTION -----------------------Lowest Priority, done last. Our prioritisation order is easily remembered using the acronymn: BEDMAS B- Brackets , E- Exponentiation, D-Division, M-Multiplication, A-Addition, S-Subtraction Note M & D and A & S can be swapped. Consider the step wise evaluation of the following expression: = = = = E3.16 2 + 4(6) + 3(2+1)2 2 + 4(6) + 3(32) 2 + 4(6) + 3(9) 2 + 24 + 27 53 i.e. evaluate bracket first then exponentiation then the 2 multiplications lastly the summations. Complete the following stepwise evaluation: Expression 3(2+3) + 42 = = = = Comment Evaluate bracket Do exponentiation Do multiplication Do summation Stepwise evaluation may seem tedious but it trains you in good discipline, vital when manipulating complex algebraic expressions. A3.16 3.17 Expression Comment 2 3(2+3) + 4 = 3(5) + 42 Evaluate bracket = 3(5) + 16 Do exponentiation = 15 + 16 Do multiplication = 31 Do summation It is important to use and interpret brackets correctly. Where brackets are not necessary there should be removed. Consider the expression: 9 + (6 x 2). The answer is 21. As the brackets contains a multiplication which would have been done before the addition anyway, the brackets play no part and are redundant. In other words 9 + (6x2) is the exactly the same as 9 + 6x2. Now compare 6x22 with (6x2)2 6x22 = 6x4 = 24 but (6x2)2 = 122 = 144. So the brackets are not redundant and must not be removed. E3.17 Rewrite the following by removing redundant brackets: (i) (6)2 + 4(2+3) (ii) (5+4) + (3x2) (iii) (2+3) - (2+4) (iv) (-2)(3)(4) (v) 5+2((2x3)+4) (vi) (5 - (2x3)) - 4x(5)2 2 A3.17 (i) 6 + 4(2+3) (ii) (5+4) + 3x2 (iii) 2+3 - (2+4) (iv) -2(3)(4) (v) 5+2(2x3+4) (vi) 5 - 2x3 - 4x52 3.18 It is equally important to insert brackets correctly where necessary. For example ‘the sum of the squares of 2 and 3’ is given by the expression, 22 + 32 (no brackets needed) whereas ‘the square of the sum of 2 and 3’ is given by (2+3)2, where the brackets are needed. Expression in words Expression mathematically The square of the product of 2 and 3 (2x3)2 The product of the squares of 2 and three 22x32 or (22)(32) The reciprocal of the sum of 2 and 3 1 23 The sum of the reciprocals of 2 and 3 1 1 2 3 The sum of the roots of 2 and 3 2 3 The root of the sum of 2 and 3 23 E3.18 Write expressions for (i) the product of the roots of 3 and 5 (ii) the root of the sum of 2,3,6 and 9 (iii) the reciprocal of the product of 5 and 10 (iv) the reciprocal of the square of 0.1 (v) 2 raised to the power 4 added to square of 4 (vi) the cube root of the product of 3 and 8 (vii) the sum of 5 and 2 taken away from -15 (viii) the absolute value of –6 added to the absolute value of -10 (ix) the mean (average) of 0,4 and 8 (x) the sum of the roots of 4 and 9 (xi) the square of the product of 3 and 5. A3.18 (i) (ii) ( 3 )( 5 ) 23 69 1 (iii) 5(10) (iv) (v)24 + 42 1 0.12 (vi) (3 (6)(8) (vii) -15 – (5 + 2) (viii) |–6| + |-10| 048 (ix) 3 (x) 4 9 (xi) (3(5))2 3.19 Fractions:Reducing a fraction to its lowest terms We know that 84 is the same as or equivalent to 12 So we can reduce 4 8 to its lowest terms to get a half. Reducing a fraction to its lowest terms is achieved by dividing the numerator (top) and denominator (bottom) by a common factor. Example:Consider 2310 3696 Numerator and denominato r are even so cancel by thefactor 2 to get 2310 1155 . 3696 1848 Note the digits 1155 add up to 12 a factor of 3. and the digits 1848 add up to 21 a factor of 3. So cancel by the factor 3 to get 1155 385 . 1848 616 Now note both divide by 7. So we get 385 55 . 616 88 Now both divide by 11 to give 55 5 . 88 8 E3.19 Reduce the fraction to its lowest terms by dividing by 10, 3, 7 and 13. 30030 35490 A3.19 3.20 30030 3003 1001 143 11 35490 3549 1183 169 13 Calculator work: Reducing fractions to their lowest terms There is a key on your calculator that may look like this: s D It toggles between fraction and decimal. 30030 to its lowest terms 35490 Key in the following sequence: 30030, ,35490,=,<s/D>. To reduce E3.20 Use the calculator to reduce the fraction 10010 to its lowest terms. 18018 A3.20 10010 5 18018 9 Fractions:Addition, Subtraction, Multiplication and division 3.21 We want a fast way of adding fractions. 2 4 Consider the sum: 3 5 2(5) 4(3) 10 12 22 The answer is 3(5) 15 15 This is top heavy (numerator bigger than denominator) so we write it as a 7 proper fraction 1 . 15 Generalise it to: TopLeft TopRight BottomLeft BottomRight TopLeft(BottomRight) TopRight(BottomLeft) BottomLeft(BottomRight) With subtraction: replace + with ─ in the above to get TopLeft TopRight BottomLeft BottomRight TopLeft(BottomRight) TopRight(BottomLeft) BottomLeft(BottomRight) Example: 3 2 3(3) 2(4) 9 8 1 4 3 4(3) 12 12 E3.21 1 2 2 3 Evaluate: 1 2 (ii) 2 3 A3.21 1 2 1(3) 2(2) 5 2 3 2(3) 6 1 2 1(3) 2(2) 1 (ii) 2 3 2(3) 6 For multiplication 3.22 (i) (i) TopLeft TopRight x BottomLeft BottomRight TopLeft(TopRight) BottomLeft(BottomRight) Example : 2(3) 2 3 2 3 x7 3(7) 7 For Division TopLeft TopRight BottomLeft BottomRight TopLeft or BottomLeft TopRight BottomRight We leave the left hand fraction unchanged but change the divide to a multiply and take the reciprocal (invert numerator and denominator) of the right hand fraction Example E3.22 2 3 3 7 2 7 2(3) 2 x 3 3 3(7) 7 2 3 3 7 Evaluate 1 2 15 = 1 2 15 = 1 2 15 = 1 2 1 5 = A3.22 1 2 1 5 1 2 1 5 1 2 7 10 3 10 1 10 = = 1 5 = 1 2 1 5 12 15 12 x 15 52 2 12 3.23 Calculator work : fractional arithmetic Locate a key on your calculator that looks like this To evaluate 1 2 key in the sequence: 2 3 Use this key to do the above sum. This answer is displayed as the fraction, seven sixths. Now select the SHIFT version of this key. The display will now show the proper fraction, one and one sixth. E3.23 Use your calculator to do the following evaluations leaving your answer in fraction form. 12 5 (i ) 23 6 4 6 2 (ii ) x 5 11 3 A3.23 The answers are 2 12 5 3 (i ) 2 3 6 4 2 4 6 2 4 (ii ) x 5 11 3 495 Exercise 3. 1. Fill in ? below: (i) (iv) (vii) (x) 2 - 3(?) = 11 (ii) 3(2 + ?)= 0 ?3 = –27 (v) 2 x 33 =? – (4 + ?) = –5 (viii) ? 11 =-3 √?=9 (iii) 16 – 2? = 0 3(6 - ?) = -15 ? – 81 = -70 (vi) (ix) 1.Without using a calculator evaluate: (i) 12 – 8 – (12 – 8) (ii) 2x8 – 5x2 (iii) 23 - 32 (iv) 2x3 + 3x42 (v) √(32+42) 2.Put a pair of brackets in the left hand side of each of the following to give correct statements: (i) 2 – 7 – 9 + 3 = -17 (ii) 8 – 2 + 3 – 4 = -1 (iii) 7 – 2 – 6 + 10 = 1 4. Without a calculator work out the following: 3 2 12 2 (ii) 4 5 3 5 7 (i) 5. Use your calculator to do the following evaluations leaving your answer in fraction form. 12 5 (i ) 23 6 4 6 2 (ii ) x 5 11 3 2 6 Write the following with positive exponents. Do not evaluate. 1 10-3 = 2 10 3(10-1) = 4 102 0.25(10-3) = 1 4 (103 ) 3 4 (104 ) 4 -2 -3 4(3 )(5 )(4 ) = 7. Evaluate using a calculator: (i) 4 146415 (ii) 8. Write down at sight the value of (i) (iii) 3 10 3 2 100 15 100 3 13312 (ii) 100 15100 4 4.1 Starting Algebra - From Numbers to letters Consider the summation: a + a + a = 3(a) or just 3a. Note the three terms of the LHS (a + a + a) have been ‘merged’ into one term, 3a. We call the 3a ‘the term in a’ This means the lettered factor is a. The 3 is called the coefficient of this term. Another example below: coefficient E4.1 A4.1 4.2 5n Term in n Write b + b + b + b as 1 term. What is the coefficient? 4b the coefficient is 4. The expression can be a combination of repeated additions as shown below: a + a + a + b + b + b + b can be rewritten as 3(a) + 4(b) or 3a + 4b. Although 4b + 3a is also true. We usually arrange them in alphabetic order. E4.2 A4.2 4.3 Reduce to 2 terms: b + a + b + b + a + a + b = The first term of the RHS is a term in … with coefficient … The second term is a term in … with coefficient … b + a + b + b + a + a + b = 3a + 4b The first term of the RHS is a term in a with coefficient 3 The second term is a term in b with coefficient 4. We can also have a mix of numbers with letters: 1 + 1 + 1 + c + c +c + c = 3(1) + 4(c) which we write as 3 + 4c E4.3 A4.3 4.4 2+2+2+a+a+a+a= 2+b+b+2+2+a+b+2+a= 2 + 2 + 2 + a + a + a + a = 3(2) + 4a = 6 + 4a 2 + b + b + 2 + 2 + a + b + 2 + a = 4(2) + 2a + 3b = 8 + 2a + 3b Now lets consider multiplications involving letters (or algebraic quantities) When we multiply two algebraic quantities (for example a multiplied by b) we write: a x b = ab or a(b) or (a)(b) or (a)b Although ba is also correct we usually arrange the letters in alphabetic order. Note that ab is one term made up of the two factors a and b. Here are some more examples: 3 x b = 3b i.e one term with two factors 5 x a x b = 5ab i.e. one term with 3 factors 4 x a x3 x b = 12ab (note we can do some arithmetic with the number factors) E4.4 Complete the following: 2xaxbxc= How many terms? How many factors? Simplify using multiplication on the numbers: A4.4 4.5 2xaxbx3xc= 2abc ; 1 term with 4 factors. 2 x a x b x 3 x c = 6abc What about the multiplication a x a ? This repeated multiplication so we use exponents. We write the result as: a x a = a2. When we have repeated multiplication we write the number of repeats as a number in the index (Top right) position: p x p x p x p = (p)(p)(p)(p) = p4 The base is p the index is 4. So y5 in expanded form is (y)(y)(y)(y)(y) x2y3 = x(x)(y)(y)(y) 3a3b2 = 3(a)(a)(a)(b)(b) 3a3b2 + x2y3 in expanded form is 3(a)(a)(b)(b) + x(x)(y)(y)(y) E4.5 A4.5 Write 2pq2 + 3p2q in expanded form: Write in index form: 3(a)(a)(b)(b)(b) + 2(a)(b)(b) 2pq2 + 3p2q = 2pqq + 3ppq 3(a)(a)(b)(b)(b) + 2(a)(b)(b) = 2a2b3 + 2ab2 4.6 Now consider the expression we used above: 3a3b2 + x2y3 Here are some things we can say: ●The expression is made up two terms, the first term is 3a3b2, the second term is x2y3 ●The first term is a term in a3b2 and the second is a term in x2y3 ● The coefficient of the first term is 3 ●The coefficient of the second term is 1 (silent) E4.6 Consider the expression 3p2 + 4pq +2q2 and fill in the blanks below: ●The expression is made up … terms, the first term is …., the second term is … and the 3rd term is …. ●The first term is a term in …. and the second is a term in … and the 3rd term is a term in ….. ● The coefficient of the first term is … ●The coefficient of the second term is …. ●The coefficient of the third term is …. A4.6 Consider the expression 3p2 + 4pq +2q2. ●The expression is made up 3 terms, the first term is 3p2, the second term is 4pq and the 3rd term is 2q2. ●The first term is a term in p2 and the second is a term in pq and the 3rd term is a term in q2. ● The coefficient of the first term is 3 ●The coefficient of the second term is 4. ●The coefficient of the third term is 2. 4.7 Writing and Reading algebraic expressions A great skill in understanding mathematics is to be able to interchange between algebraic expressions and words in ordinary language. For example the sum of a, b and c is expressed algebraically as: a + b + c. Here are more expressions: In words The product of p and q The Square of p Take a from b The reciprocal of f E4.7 Algebraically pq p2 b–a 1 or f-1 f 3 The cube root of r r Complete the following:The first row has been done for you. In words Algebraically The difference of p and q p–q P3 Take c from b abc The fourth root of r A4.7 4.8 The difference of p and q p–q The cube of P P3 Take c from b b–c The product of a, b and c abc 4 The fourth root of r r Then we can construct more complex expressions. See the examples below. In words The root of the product of p and q Algebraically pq The product of the roots of p and q p q The sum of the squares of u and v u + v2 The square of the sum of u and v (u+v)2 The sum of the reciprocals of u and v 1 1 u v The reciprocal of the sum of u and v 1 uv Complete the Table: The first row has been done for you. In words Algebraically 2 E4.8 The cube root of the product of p,q and r p q r (u+v+w)2 A4.8 The sum of the reciprocals of r1, r2 and r3 In words The cube root of the product of p,q and r 4.9 Algebraically p q r The square of the sum of u,v and w. (u+v+w)2 The sum of the reciprocals of r1, r2 and r3 1 1 1 r1 r2 r3 Substitution As the first step the most common activity you will be doing in algebra is to replace letters by numbers and then remove redundant brackets. For example: Evaluate 3(y + 2) when y = 3. First we replace the y with (3) and then remove redundant brackets viz: 3(y + 2) = 3((3) + 2)……………..replace y by (3) = 3(3 + 2)……………………………..remove redundant brackets = 3(5) = 15……………………………simplify and evaluate. Another example : Evaluate 3(x + 4) – 3xy when x = 2 and y = -2 E4.9 A4.9 4.10 3(x + 4) – 3xy =3((2) + 4) – 3(2)(-2)……….replace x with (2) and y with (-2) =3(2 + 4) – 3(2)(-2)………….remove redundant brackets =3(6) + 12 = 30. Evaluate 3(x - y) – 3xy when x = -2 and y = -3 3(x - y) – 3xy when x = -2 and y = -3 =3((-2) – (-3)) – 3(-2)(-3) =3(-2 – (-3)) – 3(-2)(-3) =3(1) – 18 = -15 Often the substitutions are into more complex formulae. For example, the Present Value, PV of an annuity of £S per annum for n years when the interest rate is r is given by the formula: 1 n 1 S 1 r PV 1 r 1 1 1 r Find the Present value of an annuity of £10,000 per annum for 10 years when the interest rate is 7%. Do not be intimidated by the appearance of the equation. Let your calculator do the hard work. Step 1: Replace S with (10000), r with 0.07 (not 7%), and n with (10) to get (10) 1 1 (10000) 1 (0.07) PV = 1 (.07) 1 1 1 (0.07) Remove the redundant brackets. 1 10 1 10000 1 0.07 PV = 1 0.07 1 1 1 0.07 Simplify a little by doing the additions: 1 10 1 10000 1.07 PV = 1.07 1 1 1.07 Then carefully evaluate using your calculator to get PV = £70 235.82 E4.10 The Present Value, PV of an annuity of £S per annum for n years when the interest rate is r is given by the formula: PV S (1 r)n 1 . 1 r n r Find the Present value of an annuity of £10,000 per annum for 10 years when the interest rate is 7%. A4.10 4.11 PV = £70 235.82 Verification One use of substitution is to check whether a numerical result is correct. For example if we wish to check if x=3 is a solution of the equation, 3x2 - 4x -15 = 0 we substitute the result x = 3 and check if the left hand side (LHS) equals the right hand side (RHS): LHS = 3x2 - 4x -15 Substitute x = 3 to get 3(3)2 – 4(3) – 15 =27 – 12 – 15 = 0 = RHS. So x=3 is the solution of the equation. E4.11 Use Verification to test if x= - 0.5 is a solution of A4.11 0.5x2 + x + 0.5 = 0 LHS = 0.5(-0.52) + (-0.5) + 0.5 = 0.125 RHS = 0 So x = -0.5 is not a solution. Now let’s try another equation. 4.12 Is 2 a solution of the equation: x3 + 2x = x2 + 4. Substitute x = 2 in the LHS: = = = (2)3 + 2(2)…………….replace x with (2) 23 + 2(2).....………….remove redundant brackets 8 + 4 ………………..stepwise evauation 12 Now Substitute x = 2 in the RHS: x2 + 4 (2)2 + 4……………….….replace x with 2 in brackets 22 + 4.……………….…. remove redundant brackets = = =8 LHS ≠ RHS so x=2 is not a solution of this equation. E4.12 Is -2 a solution of the equation: x3 + 2x = -x4 + 4. A4.12 4.13 LHS = (-2)3 + 2(-2) = -8 + -4 = -12 RHS = -(-2)4 + 4 = -16 + 4 = -12 So x = -2 is a solution. Summation of algebraic expressions Now consider the two expressions E1 and E2: Suppose E1 = 2x2y + 3xy2 and E2 = x2y + 4xy2 Suppose we want to add the two expressions, E1 + E2 We write E1 + E2 = (2x2y + 3xy2) + (x2y + 4xy2) with brackets for each expression. Then we remove the redundant brackets: E1 + E2 = 2x2y + 3xy2 + x2y + 4xy2 Note the RHS side is made up of 4 terms and the coefficient of the third term x2y is the silent 1. The first and third are both terms in x2y ; we say there are like terms ; Using the rule of repeated addition these two terms can be merged to the single term 3x2y by adding the coefficients; The second and fourth are both terms in xy2; they are like terms so using repeated addition these two terms can be merged to the term 7xy2. So we get : E1 + E2 = 2x2y + 3xy2 + x2y + 4xy2 = 3x2y + 7xy2. E4.13 Now attempt the following: Give E1 = 10 + 2x3y2 + 4x2y3 + xy4 ; E2 = 5x + x3y2 + 2x2y3 + 3xy4; E3 = 6 + 6x + 4x2y3 + 2y4 Find: E1 + E2 = E2 + E3 = E1 + E3 = A4.13 Now attempt the following: Give E1 = 10 + 2x3y2 + 4x2y3 + xy4 ; E2 = 5x + x3y2 + 2x2y3 + 3xy4; E3 = 6 + 6x + 4x2y3 + 2y4 Find: E1 + E2 = 10 + 2x3y2 + 4x2y3 + xy4 + 5x + x3y2 + 2x2y3 + 3xy4 = 10 + 5x + 3x3y2 + 6x2y3 + 4 xy4 E2 + E3 = 5x + x3y2 + 2x2y3 + 3xy4 + 6 + 6x + 4x2y3 + 2y4 = 6 + 11x + 6x2y3 + 3xy4 + x3y2 + 2y4 E1 + E3 = 10 + 2x3y2 + 4x2y3 + xy4 + 6 + 6x + 4x2y3 + 2y4 =16 + 6x + 2x3y2 + 8x2y3 + xy4 + 2y4 E1 + E2 + E3 = 4.14 Reminder about the danger when removing brackets when there is a negative sign in front of the brackets. Consider – (a + b + c) Are the brackets redundant? Can they be removed? - the brackets are NOT redundant. They show that all the 3 terms are being negated by the – in front of the brackets. So if we remove the brackets we must change the expression as follows: – (a + b + c) = – a – b – c. Also: – (a + b – c) = – a – b + c Another Example: 4 – (a + 2b) = 4 – a – 2b Here are more examples: – (2 – a) = -2 + a – (2 + -c) = -2 – -c = -2 + c – (2 + c) = -2 – c – (2 + a – b) = -2 – a + b – (a + a + b + b + b) = - (2a + 3b) = - 2a – 3b E4.14 A4.14 4.15 Are these Correct? Correct any errors – (a + 4b – 5c) = - a – 4b + 5c – (a + a + a + b x b x b) = - (3a + b3) = - 3a + b3 – (-a + -a + -a - b x b x b) = - (-3a - b3) = -3a + b3 2b + 2a – (a + b – c) = 2b + 2a – a – b + c = a + b + c – (a + 4b – 5c) = - a – 4b + 5c – (a + a + a + b x b x b) = - (3a + b3) = - 3a + b3 – (-a + -a + -a - b x b x b) = - (-3a - b3) = -3a + b3 2b + 2a – (a + b – c) = 2b + 2a – a – b + c = a + b + c They are all correct. Now suppose we have a bracket after a coefficient e.g. 3(x + y + z). The 3 is the coefficient of the bracket. Are the brackets redundant? No as explained below. Compare 3x + y + z with 3(x + y + z). In 3x + y + z the coefficient of 3 belongs only to the x in the first term. In 3(x + y + z) the coefficient of 3 belongs to all three terms in bracket i.e. it belongs to the x term the y term and the z term. So if we wish to remove the brackets we get: 3(x + y + z) = 3x + 3y + 3z. Also 3(x – y – z) = 3x – 3y – 3z. Also 3(2x + 4y – 3z) = 6x + 12y – 9z E4.15 4.16 E4.16 The coefficient belongs to each of the terms in the bracket. Remove the brackets in the expressions: (i) 3(p + q + r) (ii) 4(a – 2b – c) (i) 3(p + q + r) = 3p + 3q + 3r (ii) 4(a – 2b – c) = 4a – 8b – 4c Track the following manipulations: (i) 3(2a + 3b – 4c) + 2(2a + 3b – c) = 6a + 9b – 12c + 4a + 6b – 2c (remove brackets) = 10a + 15b – 14c (collect like terms) (ii) 3a(2a – 4b) + 2b(2a + 3b) = 6a2 – 12ab + 4ab + 6b2 (remove brackets) = 6a2 – 8ab + 6b2 (collect like terms) Simplify the following expressions by removing the brackets and collecting like terms: (i) 4a(a – 2b) + 3b(2a – 2b) (ii) A4.16 4a2(1 – 2a + 2b) + 3b2(2a – 2b) (i) 4a(a – 2b) + 3b(2a – 2b) 2 = 4a – 8ab + 6ab – 6b2 = 4a2 – 2ab – 6b2 (ii) 4a2(1 – 2a + 2b) + 3b2(2a – 2b) 2 =4a – 8a3 + 8a2b + 6ab2 – 6b3. 4.17 Now let’s consider the effect of a negative coefficient in front of a bracket. We know that –(a + b – c) simplifies to -a – b + c. E4.17 A4.17 So –2(a + b – c) simplifies to -2a – 2b + 2c. Simplify by removing the brackets: (i) – (a – 2b + c) = (ii) – 3(a – b + c) = (iii) 4 – 3(a – b + c) = (i) – (a – 2b + c) = -a + 2b – c 4.18 (ii) – 3(a – b + c) = -3a + 3b – 3c (iii) 4 – 3(a – b + c) = 4 – 3a + 3b – 3c Track the following manipulations: (i) 3(2a + 3b – 4c) – 2(2a + 3b – c) = 6a + 9b – 12c – 4a – 6b + 2c = 2a + 3b – 10c (ii) 3a(2a – 4b) – 2b(2a + 3b) = 6a2 – 12ab – 4ab – 6b2 = 6a2 – 16ab – 6b2 E4.18 (remove brackets) (collect like terms) (remove brackets) (collect like terms) Simplify the following expressions by removing the brackets and collecting like terms: (i) 4a(a – 2b) - 3b(2a – 2b) (ii) -4a2(1 – 2a + 2b) + 3b2(2a – 2b) (iii) -2a(a – b) – (2a + 3b) 4.19 (i) 4a(a – 2b) - 3b(2a – 2b) = 4a2 – 8ab – 6ab + 6b2 (ii) -4a2(1 – 2a + 2b) + 3b2(2a – 2b) = -4a2 + 8a3 – 8a2b + 6ab2 – 6b3 (iii) -2a(a – b) – (2a + 3b) = -2a2 + 2ab – 2a – 6ab = -2a2 – 4ab – 2a Rules of Indices We know from our work on exponents that: 32 x 35 = 32+5 = 37. A common mistake is to write: 32 x 35 = 310 Note the multiply sign is not between the exponents. So remember to add the exponents. So we get the algebraic equivalent: (am)(an) = am+n So (a3)(b2)(a2)(b4) = a3+2b2+4 = a5b6 Another example: (3xy2)(5x2y) = 15x3y3. E4.19 A4.19 4.20 2x 2 5x 2 (ii) 3 y 7y (i) (2x2y3)(5x3y4) = (2x2y3)(5x3y4) = 10 x5y7 2x 2 5x 10x 3 2 (ii) 3 3 y 7y 21 y We also know from our work on exponents that: (i) 1 3-2 53 2-1. Also -3 2 2 5 3 i.e. when the base and the index migrate from the numerator(denominator) to the denominator(numerator) the index changes sign. So we get the algebraic equivalent: 1 x m y-n -n x . Also xn yn x -m Example: 1 2x -3 2y 2 3 x . Also x -3 5y- 2 5x 3 Example: Simplify 2x 3 y 2 with positive exponents only. 5xy 5 Migrate the x from the denominator to the numerator and the y2 from the numerator to the denominator to get: 2x 3 y 2 2x 3-1 2x 2 = 5xy 5 5y5- 2 5y3 E4.20 A4.20 4.21 Simplify with positive exponents only: 3x 3 y3 (i) 4x 5 y 2 3x 3 y3 8xy -2 (ii) 5 2 2 4x y 9x y 3x 3 y3 3y3- 2 3y 2 (i) 5-3 5 2 4x 4x 4x y 3x 3 y3 8xy -2 2x 31y3- 2 2 (ii) 5 2 2 5 2 2 1 3 2 3x y 4x y 9x y 3x y Now consider the numerical expression 2 . 3 4 This means 4 copies of 23 multiplied together. 4 So 23 23 23 23 23 23 3 3 3 212 . 4 E4.21 So we can abstract the result: 23 23x 4 212. Simplify (ii) 2 3 (iii) 2 3 2 (i) 75 3 3 4 2 3 4 A4.21 2 3 (ii) 2 3 =(212)(38) (iii) 2 3 2 =(212)(36)(215) = (227)(36) 3 (i) 75 =715 3 4 2 3 4 4.22 5 3 2 3 5 3 Now the algebraic equivalent is: x m n x mn . E4.22 Example : x a Simplify y 3 4 2 3 4 2 3 5 3 = 5xy (ii) x y =x12y8 (iii) a b b =a12b6b15 = a12b21 (i) 5x y 3 3 4 4.23 x axb x ab (ii) x y (iii) a b b (i) 5x A4.22 b 4 2 2 3 5 3 Roots and Exponents We have seen earlier that expressions in root form can be written in 1 exponent form or index form. For example x x 2 . More generally we can write the result in algebraic form as m n xn x m Note carefully which number is in the numerator and which number is in the denominator of the index. 3 E4.23 2 Example: x x 3 Complete the Table: The first row has been done for you. 2 Root Form Exponent Form 3 x3 x2 2 3 y4 (negative exponent) 1 3 x 2 y3 2 2 3x 3 y 14 A4.23 Root Form Exponent Form 3 x2 x3 4 y3 y4 1 (negative exponent) 3 x2 2 y3 x 3 3 y2 33 x 2 4 y 4.24 2 3 2 2 3x 3 y 14 Algebraic fractions Remember our rule for adding two fractions: TopLeft TopRight BottomLeft BottomRight TopLeft(BottomRight) TopRight(BottomLeft) BottomLeft(BottomRight) So we get for example: a c ad bc b d bd Another example: 3 5 3y 5x x y xy E4.24 A4.24 4.25 2 x 2 (i) x 2 4y 3x 5 2 4y 10 - 12xy (ii) 3x 5 15x 2 5 Now consider the sum: x 2x 2 5 4x 5x We get as the first step. x 2x 2x 2 (i) y 3 3x y 6 xy 3 3x (ii) But before we proceed notice that the two terms in the numerator and the one term in the denominator all have the common factor x. So we can simplify the result by cancelling x from each term to get: 4x 5x 4 5 9 . 2x 2 2x 2x Here’s another example: 2 5 2 3x 6x 12x 2 15x 18x 2 Now notice we can cancel each term by 3x to get: 12 x 2 15x 18x 3 4x 5 6x 2 E4.25 A4.25 3x x 5y 10y 2 3x x (i) 5y 10y 2 (i) 30xy 2 5xy 50 y3 6xy x 10 y 2 30xy 2 5x 2 y 50x 3 y3 6y x 10x 2 y 2 2 5 Now consider the sum: x 1 x - 2 4.26 3 1 2 5x y 10xy 2 3 1 (ii) 2 5x y 10xy 2 (ii) Track the following sequence: 2 5 x 1 x - 2 2( x - 2) 5(x 1) (x 1)(x - 2) 2 x - 4 5x 5 (x 1)(x - 2) 7x 1 (x 1)(x - 2) E4.26 A4.26 (i) 1 3 2x 3 3 - 2x (ii) 3 1 2x 3 3 - 2x (i) 1 3 (3 - 2x) 3(2x 3) 3 - 2x 6x 9 12 4x 2x 3 3 - 2x (2x 3)(3 - 2x) (2x 3)(3 - 2x) (2x 3)(3 - 2x) (ii) 3 1 3(3 2x) 1(2x 3) 9 6x 2x 3 6 8x 2x 3 3 - 2x (2x 3)(3 - 2x) (2x 3)(3 - 2x) (2x 3)(3 - 2x) 4.27 Seniority of Multiplication/Division over addition/subtraction You are aware now that the rules for multiplication are different from the rules for add/subtract. E4.27 A4.27 For example: -2 + -3 gives a negative result but -2(-3) gives a positive result. Evaluate: -2 + 5 and -2(5). -2 + 5 = 3 and -2(5) = -10 The square of a sum versus the square of a product. Compare (a+b)2 with (ab)2. 4.28 The first is the square of a sum and the second is the square of a product. We want so see if it is correct to distribute the exponent of 2 to each of members a and b in the bracket. i.e. Does (a+b)2 = a2 + b2 and does (ab)2 = a2b2 ? We will use verification to test these with a = 3 and b = 5. Test if (a+b)2 = a2 + b2: LHS = (3+5)2 = 82 = 64 RHS = 32 + 52 = 9 + 25 = 34 so LHS ≠ RHS. (a+b)2 ≠ a2 + b2 Now test if (ab)2 = a2b2: LHS = (3(5))2 = 152 = 225 RHS = (32)(52)=9(25) = 225 so LHS = RHS. (ab)2 = a2b2 Strictly speaking this is not a proof as we have just used a specific case. Seeing one black sheep is enough to disprove the claim that: all sheep are white but seeing one white sheep does not prove that all sheep are white. E4.28 (i) (ii) A4.28 Use x = 2 and y = 3 to test if the square of a quotient is the quotient of the squares Use x = 2 and y = 3 to test if the square of a difference is the difference of the squares (i) 2 x x2 Test : 2 y y 2 9 LHS = 32 9 3 92 81 RHS 2 9 3 9 So LHS RHS This supportsthe rule : 2 x x2 2 y y (ii) Test : x - y x 2 - y 2 2 LHS = 9 - 3 6 36 2 2 RHS 92 32 81 9 72 So LHS RHS So x - y x 2 - y 2 2 4.29 The important point is that multiplication ‘wins’ over addition. Here are some examples: The cube of a sum is the sum of the cubes: (a+b)3 = a3 + b3 False The cube of a product is the product of the cubes: (ab)3 = a3b3 (a+b)n = an + bn (ab)n = anbn The modulus of a sum is the sum of the moduli. |a+b| = |a| + |b| The modulus of a product is the product of the moduli. |ab| = |a||b| E4.29 (i) (ii) (iii) (iv) (v) A4.29 True False True False True Is the reciprocal of a sum of a and b the sum of the reciprocals of a and b? Is the reciprocal of the product of a and b the product of the reciprocals? Is the root of the sum of and b the sum of the roots of a and b? Is the root of the product of a and b the product of the roots? verify (ii) and (iv) above using x = 2 and y = 3. (i) Is the the sum of the reciprocals of a and b equal to the reciprocal of a sum of a and b? No 1 1 1 a b ab (ii) Is the reciprocal of the product of a and b the product of the reciprocals? Yes 1 1 1 a b ab (iii)Is the root of the sum of and b the sum of the roots of a and b? No. a b a b (iv) Is the root of the product of a and b the product of the roots? Yes. ab a b 4.30 Expansion of a perfect square Finally in this chapter we want to determine the expansion of (a + b)2. We know that: (a + b)2 = a2 + b2 is a FALSE statement. To find a true statement let us take a numerical example. Look at the outer, large square below. It is 8 by 8 units i.e (5 + 3) by (5 + 3). The area of the whole square is therefore (5 + 3)2: 5 3 5 3 It were look at the inner shapes we can identify a large 5x5 square, two 5x3 rectangles and a smaller 3x3 square. If we add the areas of the four rectangles we get 52 + 32 + 2(5)(3) Hence we get (5 + 3)2 = 52 + 32+ 2(5)(3). This is called the rule of the perfect square. E4.30 Complete the Table below. The first row has been done for you. Outer Square 102 = 100 = = = 144 Inner large square 62 = 36 82 = = = Rectangles 2(6)(4) = 48 = 2(3)(4) = Inner smaller square 42 = 16 = 36 = 42 = 16 A4.30 Outer Square 4.31 Inner large square 2 10 = 100 62 = 36 142 = 196 82 = 64 2 7 = 49 42 = 16 12 = 144 82 = 64 So now we state the algebraic result: Rectangles 2(6)(4) = 48 2(8)(6) = 96 2(3)(4) =24 2(8)(4) = 64 Inner smaller square 42 = 16 62 = 36 32 =9 42 = 16 (a + b)2 = a2 + 2ab + b2 Note the first and last terms of the RHS are the areas of the two squares and the middle term is the area of the two rectangles. We read the RHS as: the square of the first term in bracket, the square of the last term in the bracket and twice the cross of the two terms in the middle. E4.31 Write out the full expansion of the perfect square when (i) the square of the first term in the bracket is x2 and the square of the last term in the bracket is 100. (ii) A4.31 (i) The square of the first term in the bracket is 25 and twice the cross of the two terms in the bracket is 10y. the square of the first term in the bracket is x2 and the square of the last term in the bracket is 100. (x + 10)2 = (x2 + 2(x)(10) + 102) = (x2 + 20x + 100) (ii) The square of the first term in the bracket is 25 and twice the cross of the two terms in the bracket is 10y. (5 + y)2 = (52 + 2(5)(y) + y2) = (25 + 10y + y2) 4.32 Mixing numbers with letters we get: (x + 3)2 = x2 + 2(3)(x) + 32 = x2 + 6x + 9 Also (2x + 3y)2 = (2x)2 + 2(2x)(3y) + (3y)2 = 4x2 + 12xy + 9y2 E4.32 Complete the following statements: (i) (x + 1)2 = (ii) (x + )2 = x2 + 6x + 9 (iii) ( + 5)2 = x2 + + 25 (iv)(2x + )2 = 4x2 + 12x + 9 (v) (3x + 2y)2 = (vi) ( )2 = + 2xy + (vii) ( ) = 9x2 + + 100 2 2 (i) (x + 1) = x + 2x + 1 (ii) (x + 3)2 = x2 + 6x + 9 (iii) (x + 5)2 = x2 + 10x + 25 (iv)(2x + 3)2 = 4x2 + 12x + 9 (v)(3x + 2y)2 = 9x2 + 12xy + 4y2 (vi) (x + y)2 = x2 + 2xy + y2 (vii) (3x + 10)2 = 9x2 + 60x + 100 Note when we write: (x + y)2 = x2 + 2xy + y2 , the LHS has just 1 term but the LHS has 3 terms. A4.32 4.33 So we expand (x + y)2 to get x2 + 2xy + y2 and We factorise x2 + 2xy + y2 to get (x + y)2 E4.33 Complete the Table below. Factorised form (x + 2)2 Expanded Form 4x2 + 4xy + y2 4m2 + 12m + 9 (√x + 1)2 A4.33 4.34 Factorised form Expanded Form 2 (x + 2) x2 + 4x + 4 2 (2x + y) 4x2 + 4xy + y2 (2m + 3)2 4m2 + 12m + 9 2 (√x + 1) x + 2√x + 1 Now if we have the square of a difference rather than the square of a sum then we get the results: (x - 3)2 = x2 + 2(-3)(x) + (-3)2 = x2 - 6x + 9 Also (2x - 3y)2 = (2x)2 + 2(2x)(-3y) + (-3y)2 = 4x2 - 12xy + 9y2 With practice you should be able to directly to the final form without the intermediate stage. But do this only when you are confident enough. Remember ‘the square of the 1st, the square of the last and twice the cross in the middle’ For example we write down directly: (3x - 2y)2 = 9x2 - 12xy + 4y2 E4.34 Complete the Table below. Factorised form (x - 2)2 Expanded Form 4x2 - 4xy + y2 4m2 - 12m + 9 (√x - 1)2 A4.34 Factorised form (x – 2)2 (2x – y)2 (2m – 3)2 (√x - 1)2 Expanded Form x2 – 4x + 4 4x2 - 4xy + y2 4m2 - 12m + 9 x – 2√x + 1 Exercise 4 Q1. Write expressions for (i) the sum of a, b and c. (ii) the product of m and n. (iii) the reciprocal of x (iv) the sum of the reciprocals of x and y. (v) the cube of the product of x and y. (vi) m taken away from n. (viii) the modulus of the sum of x and y. (viii) the sum of absolute values of x and y. (xii) the arithmetic mean of p, q and r. (xiii) the root of the mean of the squares of p, q and r. Q2 Using the seniority of multiply/divide over add/subtract identify at sight whether the following statements are TRUE or FALSE: (i) (a+b)2 = a2 + b2 (ii) (ab)2 = a2b2 (iii) √(x-y) = √x - √y 3 3 1 a b a b 1 1 (iv) √(xy) = √x√y (v) (vi) 3 ab a b c c Where possible correct the incorrect statements. 3 Q3. The Present Value, PV of an annuity of £S per annum for n years when the interest rate is r is given by the formula: 1 n 1 S 1 r PV 1 r 1 1 1 r Q4 Simplify each of the following: (i) 3(3 – 2x) – (-2 + x) (ii) (x + 4y)2 – xy(x + y) 3x 2 y3z (iii) (iv) 4x2 + y2 – (x – 2y)2 9(xyz)3 Q5. Simplify 3x x (i) 5y 10y 2 (ii) 3 1 2 5x y 10xy 2 Q6 Simplify and write the following expressions with the exponents in the numerator: i. 3 2x 2 8x 2 y 2 x ii 2x 4 y 3 2 1 iii. 5 x2 y iv. x3 y 3 xy 2 v. 3 x 1

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