Honeyland College Mathematics Scheme of Works for Year 9 - PDF

HONEYLAND COLLEGE MATHEMATICS SCHEME OF WORKS FOR YEAR 9 FIRST TERM S/N TOPICS SUB TOPICS 1. Review of JSS2 Work 2. Number bases Conversion of number bases...

HONEYLAND COLLEGE MATHEMATICS SCHEME OF WORKS FOR YEAR 9 FIRST TERM S/N TOPICS SUB TOPICS 1. Review of JSS2 Work 2. Number bases Conversion of number bases Addition and subtraction of binary numbers Multiplication and division of binary numbers 3. Word problems Sum and difference Product Sum and difference combined with product Word problems involving fractions 4. Proportion Direct proportion Indirect/Inverse proportion Reciprocals 5. Compound Interest Simple interest (revision) Compound interest 6. Factorisation of simple HCF of algebraic expression algebraic expression Factorisation by taken out common factor Simplifying calculations by factorisation Factorisation by grouping like terms 7. Factorization of quadratic Expansion of algebraic expression expression. Coefficients of terms Factorization by quadratic expression Perfect square Difference of two square 8. Simplifying algebraic Simplifying algebraic expressions with expressions involving brackets brackets and fractions Simplifying fractional expressions 9 Equation involving fractions Fractional equations with unknown in the numerators Fractional equations with variable in the denominators. Fractional equations with binomial denominators. Word problems leading to fractional equations. 10. Change of subject of formula 11 Revision 12&13 Examination and Vacation 1  CHAPTER ONE Topic: Number Bases Reference: A.J.S Olawasanmi Essential Mathematics for JSS book 3. Page (1-10) Objectives: Students should be able to: 1. revise basic operations in binary system 2. express numbers in other bases other than 10 3. to convert from a given base to another 4. carry out arithmetical operation in binary system Instructional Material: Chart Showing Number Bases Content Conversion of number bases Addition and subtraction of binary numbers Multiplication and division of binary numbers Conversion of numbers from one base to another. Number base system is a system of counting numbers with the combination of certain digits to which the system uses. The highest digit of any system is one less than the base. In base 10 system of number, the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 are used in the formation of numbers. The numbers in 2 we use only the digit 0 and 1 while numbers in base ten is Known as DECIMAL or DENARY NUMBER. To convert a number from base 10 to a new base, divide the number by the new base. For example; to convert 85ten to base two, divide 85 and the result obtains repeatedly by 2. Example 1 (a) Convert 98ten to base 2 (b) Convert 24810 to octal number (base 8) 2 Solution :. 98ten = 1100111two Example 2: Convert 24810 to octal number (base 8) 8 248 8 31 R 0 8 3 R7 8 0R3 Therefore, 24810 = 3708 Conversion of non-decimal to decimal number (base 10) writing numbers in expanded form The following are expressed in expanded form; 1) 101011102 = (1×27) + (0×26) + (1×25)+ (0×24) +(1×23)+(1×22)+(1×21)+ (0×20) 2) 3054eight=(3×83)+(0×82)+(5×81)+(4×80) 3) 1101.01011two = (1×23)+(1×22)+(0×21)+(1×20) + (0×2-1) + (1×2-2) + (0×2-3) +(1×2-4)+ (1×2-5) conversion of non decimal to denary Example 1 Convert 101011101two to base 10 101011101two = (1×28)+(0×27) +(1×26)+ (0×25) +(1×24)+(1×23) + (1×22) + (0×21)+(1×20) = (1×256)+(0×128)+(1×64) +(0×32)+(1×16)+(1×8)+ (1×4) + (0×2)+(1×1) = 256 + 0 + 64 +0 +16 + 8 + 4 + 0 + 1 = 349ten 3 Example 3: 3054eight = (3×83)+(0×82)+(5×81)+(4×80) = 1536 + 0 + 40 + 4 = 1580ten Example 2: Express 110111.1011two as base ten number 1101.011two= 1 X 23 + 1 X 22 + 0 X 21 + 1 X 20 + 0 X 2-1 + 1 X 2-2 + 1 X 2-3 1 1 = 1 X (2 X 2 X 2) + 1 X (2 X 2) + 0 X 2 + 0 X + 1 X + 2 4 1 1X 8 1 1 = 1X8+1X4+0+0+ + 4 8 1 1 = 8+4+ + 4 8 3 = 12 + 8 = 12 + 0.375 = 12.375 approximately 12.38 Conversion numbers in base 10 to other bases Example 1 Covert 1795 to base four = 1 x 5 2 + 7 x 5 1 + 9 x 50 = 25 + 35 + 9 = 69ten Convert 69ten to base four 1795 = 10112 4 Example 2: Convert 10358 to base 6 = 1 x 83 + 0 x 8 2 + 3 x 8 1 + 5 x 80 = 1 x 512 + 0 + 3 x 8 + 5 x 1 = 512 + 24 + 5 = 541ten 10358 = 10000111012 Converting a decimal number to another bases Example 1: Convert 578.38ten to base 5 578. (38)ten The decimal part (.38), convert to base 5.38 X 5 1. 900 X 5 4. 500 X 5 2. 500 5  If you have 3 d.p, stop and write the numbers = 0.142 Therefore, 578.38ten = 4303.142five Addition and subtraction of binary numbers Examples 1. Add 1112 to 1012 2. Add 11.10 + 111.001 3. Add 304 + 452 in base 6 4. Subtract 11011 from 1101101 5. Subtract 56478 from 72018 Solution A. Add 1112 to 1012 B. Add 11.10 + 111.001 C. Add 3046 + 4526 D. Subtract 11011 from 1101101 in base 6 Note: If the system required you to borrow while subtracting, you are borrowing the base of the system E. Multiplication and division of binary numbers Example: 1. Find the product of 11012 and 1112 2. 1011.1 x 1.1 3. 1110 ÷ 111 Solutions 1. Find the product of 11012 and 1112 Or convert 1 1 0 12 and 111 to base 10 and multiply. Then convert the result to base 2. 2. 1011.1 x 1.1 7  3. 1110 ÷ 111 1110 ÷ 111 102 Other Examples: 1. Evaluate: (10112)2 2. Find the unknown Given that 101p = 5. Solution 1. Evaluate: (10112)2 = 1 0 1 1 x 1 0 1 1 2. Find the unknown Given that 101p = 5 = 1 x p2 + 0 x p 1 + 1 x p 0 = 5 = p2 + 0 + 1 =5 P = 5–1 2 square root both sides. 8  CHAPTER TWO Topic: Word problems Reference: A.J.S Olawasanmi Essential Mathematics for JSS book 3. Page (16-24) Objectives: Students should be able to: 1. solve word problems involving sum, differences and product 2. solve word problems involving combined products with sum and product 3. translate word problems into numerical expressions 4. solve word problems involving fractions Instructional Material: Chart with word problems Content: Sum and difference Product Sum and difference combined with product Word problems involving fractions Meaning of terms related number problems 1. Sum: result of an addition 2. Difference: The result of a subtraction. Positive difference means the result of the subtraction of a smaller number from a larger number while negative difference means the otherwise. 3. Product: The result of a multiplication. 4. Twice (Double): two times 5. Thrice (Treble or triple): three times 6. Half or one-half: ½ , one-third (1/3), a quarter or one – quarter (1/4), three – quarter(3/4), three – fifth(3/5) etc. 7. One and half (1½), two and a quarter (21/4) etc. 8. Consecutive numbers: consecutive numbers are two or more successive numbers of a defined numbers e.g. 3, 4, 5 are consecutive whole number, 8, 10, 12, 14, 16 are 5 consecutive even numbers. 37 and 39 are consecutive odd numbers. NOTE: (1) If x is a whole number, then: x, x + 1, x + 2, x + 3, x + 4…, are consecutive whole numbers. 9  (2) if x is an odd number, then: x, x + 2, x + 4, x + 6, …, are consecutive odd numbers (3) If x is an even number, then: x, x + 2, x + 4, x + 6, …, are consecutive even numbers Sum and Difference Sum: The sum of two or more numbers is the result obtained when they area added together. That is, 10 + 18 + 4 = 32 Difference: This means when two numbers are subtracted from each other. Positive difference: lager number - smaller number Negative difference: smaller number – larger number Example 1: When 25 is added to a number the result is -12. Find the number let the number be x x + 25 = -12 x = -12 – 25 x = - 37 Example 2: Find the product of 2/5 and the positive difference between -1/4 and 0.5 0.5 = ½ 2 1 −1 = ( − ( )) 5 2 4 2 2+ 1 2 3 2 3 3 = ( ) = ( ) 𝑋( ) = 5 4 5 4 5 4 20 Example 3: 10 is subtracted from the product of 4 and a certain number. The result is equal to the sum of 5. and the original number. Find the number Solution the number be x. From the statement; 4x – 10 = 5 + x 10  4x – x = 5 + 10 3x = 15 Divide both sides by 3 x=5 :. The number is 5. 3 1 Examples3: (1) Find the product of 1 , -0.8 and - 2 (2) what number must be multiplied 4 2 3 by 25 to make ? (3) The product of 3 numbers is 3600. If two of the numbers are equal 4 and the third number are 25. Find the two equal numbers. SOLUTION 7 −8 −5 (1). 𝑥( )x( ) (2) Let the number be 𝑥 (3) 25 x 𝑥 x 𝑥 = 3600 4 10 2 7 3 = 25 × 𝑥 = 25𝑥 2 = 3600 4 4 1 =3 100 x = 3 𝑥2 = 3600⁄25 2 3 𝑥= 𝑥 2 = 144 100 𝑥 = 0.03 𝑥 = 12 Example 4: The sum of three consecutive integers is 138. Find the numbers X + x+1 + x + 2 = 138 3x + 3 = 138 3x = 138 – 3 3x = 135 135 X = = 45. 3 Therefore, the numbers are 45, 46 and 47. x = 45 x + 1 = 46 x + 2 = 47 Example 5: Two is added to twice a certain number and the sum is doubled. The result is 12 less than 5 times the original number. Find the original number 11 Solution Let the number be x 2 (2 + 2x) = 5x -12 4 + 4x = 5x -12 4x – 5x = -12 – 4 -x = -16 X = 16 Word Problem Involving age problems Example 1 A father is 24 years older than his son. a. If the son is x years old, how old is the father? b. If the ratio of their ages is 5 : 2. Find the age of the son Example 2: A man is twice as old as his son. 10 years ago, he was three times as old as his son. How old will his son, 5 years’ time? Solution Present age: Son = x, Father = 2x 12 10years ago: Son = x – 10, Father =2x – 10 Thus; 2x – 10 = 3(x – 10) 2x – 10 = 3x – 30 2x – 3x = – 30 + 10 – x = – 20 Dividing both sides by – 1 :. x = 20 The son’s present age = 20 years The son’s present age 5years time = 20 + 5 = 25years More on word problems involving fractions Problems with fractions Examples 1: Find the three-fifth of the sum of 45 and -60. Solution 3/5(45 – 60) = 3/5(-15) = -9 Divide the difference between 25 and 10 by the product of 6 and 5. solution = 25 -10 6x5 15 1 = = = 0.5 30 2 Examples 2: 3 1. When of a number is added to 30. The result is 20 added to the number find the 5 number. 13 2. When the sum of 28 and a certain number is divided by 5. The result is equal to treble the original number. What is the number? SOLUTION (1) Let the number be 𝑥 (2) Let the number be 𝑥 3 28+𝑥 Hence 𝑥 + 30 = 𝑥 + 30 = 3𝑥 5 5 Multiply each term by 5 15𝑥 = 28 + 𝑥 14𝑥 28 3𝑥 + 150 = 5𝑥 + 100 = 14 14 5𝑥 - 3𝑥 = 150 -100 𝑥=2 2𝑥 = 50 × = 25 Problems on real life situations Example The price of a book is thrice that of an exercise book. If a student spends ₦9600 on books and ₦8000 on exercise books, he has 56 items altogether. a) What is the price of an exercise book b) What is the difference between the price of a book and an exercise book? Solution 14  CHAPTER THREE Topic: Proportion Reference: A.J.S Olawasanmi Essential Mathematics for JSS book 3. Page (28-35) Objectives: Students should be able to: 1. solve problems involving direct proportions 2. solve problems involving indirect /inverse proportion 3. apply direct and inverse proportion to real life situation 4. calculate the reciprocal of numbers. Instructional Material: Chart on proportion and reciprocal Content: 1. Direct proportion 2. Indirect/Inverse proportion 3. Reciprocals Proportion. Proportion is an equation in which two ratios are set equal to each other. For example, if there is 1 boy and 3 girls you could write the ratio as: 1:3 (for every one boy there are 3 girls) There are three types of proportion 1. Direct proportion 2. Indirect/inverse proportion 3. Reciprocal Proportion DIRECT PROPORTION: When one quantity increases, the other quantity increase at the same rate and vice versa. Real life examples of the concept of direct proportion are: 1. The distances and time taken by vehicle moving at a uniform (constant or steady) speed. 2. Price and number of articles with no consideration for discount and other factors. 3. Distance covered and fuel consumption. 4. Exchange rate. Etc. Example: Suppose that in 15min a car travels 5km (i) In 30 min it will travel 10 km (ii) In 60 min it will travel 20km 15 If d is directly proportional to t, it is mathematically written as; dαt d = vt where d varies as t and vice versa, then d and t are called variable. v is constant of proportionality usually called velocity or speed. Example 1: 5 litres of petrol cost 100 naira and 8 litres cost 160 naira. (a) Is the cost of the petrol directly proportional to the quantity? (b) Calculate the cost of buying 10litres of petrol (c) Calculate the quantity in litres of petrol if the cost is 400 naira solution a. Ratio of quantity of the petrol = ratio of the corresponding cost 8 𝑙𝑖𝑡𝑟𝑒𝑠 𝑁 160 = 5 𝑙𝑖𝑡𝑟𝑒𝑠 𝑁 100 8 8 = , Since the ratios are the same, the cost is directly proportional to the 5 5 quantity of the petrol b. To calculate the cost of buying 10 litres of petrol Let x represent the cost when the quantity is 10lites X : 100 = 10 : 5 𝑥 10 = 100 5 100 𝑥 10 X= = 200. The cost is 200 naira. 5 c. Let y represents the quantity when the cost is 400 Ratio of the quantity of petrol = ratio of corresponding cost: i.e y : 5 = 400 : 100 𝑦 400 = 5 100 5 𝑋 400 Y= = 20 100 The quantity is 20 litres. Example 2: 20 apples cost ₦300, 16  a) What is the cost of 85 apples? b) how many apples can be bought by ₦1875 unitary method (a) The cost of 20 apples = ₦300 :. The cost of 1 apple = ₦300/20 =₦15 :. The cost of 85 apples = 85 × ₦15 = ₦1275 b) The number of apples of ₦300 worth = 20 The number of apple that can be bought for ₦1 = 20/300 =1/15 The number of apples that can be bought with ₦1875 = 1/15 × 1875 =125 Ratio method (a) Let the cost of 85apples be x :. x/300 = 85/20 300 ×85 x= = 1275 20 :. The cost of 85 apples is ₦1275 b) The number of apple that cost ₦300 = 20 Let the number of apples that costs ₦1875 be y y 1875 /20 = /300 20 × 1875 y= =125 300 y = 125 {the number of apples that will cost ₦1875 is 125} 17  Inverse proportion Two variables are said to be proportional if an increase in one variable leads to a decrease in the other variable or a decrease in one variable leads to an increase in the other variable. It is also known as INDIRECT PROPORTION. 1 Mathematically; t α 𝑛 𝑘 t α , where k is constant 𝑛 k=tXn =tn EXAMPLE 1: 20 men can complete a task in 4 days. How many men will be able to complete the same task in 2 days? Solution Let the number of men needed to finish the task in 2 days be x. The ratio of men in this case is 20 : x and the ratio of the number of days needed is 4 : 2 As the number of men available for work increases, the time needed to complete the task decreases. The number of workers and the number of days needed to finish the task are in inverse proportion. Thus, we have; 20 : x = 2: 4 (inverse of 4 : 2 ) 20 2 = 𝑥 4 20 𝑋 4 X= = 40. 2 It takes 40 men to complete the task in 2 days. Example 2: It takes 8 men 20 days to work on a farm land, how long would it take 32 men if they work at the same rate? 18  Solution Note; As the number of men increases the time taken decreases, as time is inversely to the number of men. let the number of days be x 8 men work on a farm for 20 days 1 man will work on the farm in (20 X 8) days = 160 160 32 men will take = 5 days. (note; 1 man more days, and 32 men less days) 32 Reciprocal proportion 1 Reciprocal of any number is 1 divided by the number. i.e n = 𝑛 Therefore, the reciprocal of the following numbers 2 = ½ = 0.5, 1 1 103 = 3 = = 0.001 10 1000 In general (a) the reciprocal of 10m = 10-m 5 Example: Find the reciprocal of each of the following: (a) 8 (b) 212 (c) (d) 0.4 11 solution (a) The reciprocal of 8 = 1/8 = 0.125 (b) ) 212 = 5/2, so the reciprocal of 5/2 is 2/5 (c) The reciprocal of 5/11 = 11/5 (d) 0.4 = 4/10 = 2/5, the reciprocal of 2/5 = 5/2 19  CHAPTER FOUR Topic: Compound Interest Reference: A.J.S Olawasanmi Essential Mathematics for JSS book 3. Page (39-45) Objectives: Students should be able to: 1. solve problems on simple interest 2. solve problems compound interest 3. apply compound interest to real life situation Instructional Material: Chart on simple interest and newspaper article involving words problems (business involving inflation) Content: Simple interest (revision) Compound interest Simple interest (revision) When the interest paid is not e=reinvested, we say it is simple interest. The principal and the interest remain the same for equal interval of time (e.g annually). Simple interest is calculated using the fomula: 𝑃𝑅𝑇 I= 100 Where; I = interest P = principal R= rate T = Time (in year) Example 1: Mr. Eze saves 50,000 with a bank for 1 year with interest for 512 % per annum. (a) Calculate the interest he will receive after the end of the year. (a) Calculate the simple interest for 412 year Solution 𝑃𝑅𝑇 (a) Using I = , where P = N 50,000 I = ? R = 512 %, T = 1 year 100 50000 𝑋 5.5𝑋 1 I= = N 2,750, we can as well look for the Principal, Rate or Time 100 𝑃𝑅𝑇 (b) Using I = , where P = N 50,000 I = ? R = 512 %, T = 412 years 100 20  50000 𝑋 5.5 𝑋 4.5 1237500 I= = = N12375 100 100 Compound Interest Compound interest is calculated the same way as simple interest except that interest is added to the principal at the end of the year or period. Compound Interest is applicable in banks, ships, building societies credit and credit card companies Example 1: Calculate the compound interest on # 70 000 for 3 years at 5% per annum. Solution 𝑃𝑅𝑇 Using 𝐼= 100 70000 𝑋 5 𝑋 1 1 st year : I = = # 3,500 100 The capital or the amount at the end of 1st year becomes #3500 + # 70000 = # 73500 2nd year: the new principal = # 73500 73500 𝑋 5 𝑋 1 I2 = = # 3675 100 The capital or the amount at the end of 2nd year becomes: # 3675 + # 70 000 = # 77175 3rd year : The new Principal = # 77175 77175 𝑋 5 𝑋 1 I3 = = # 3858.75 100 The capital or the amount at the end of 3rd year becomes: # 3500 + # 3675 + # 3858.75 = #11,033.75 OR The compound interest is calculated using A=P+I I=A–P 21 Therefore, the Compound interest is calculated as = # 81,033.75 - # 70000 = # 11033.75 USING FORMULA FOR CALCULATING COMPOUND INTEREST 𝑅 𝑛 A = P [1 + ] 100 Where P = the principal R = the rate or percentage per annum n = the number of years of period A is the amount. Example 1: using the previous example, calculate the Compound Interest. Solution P = # 70000, R = 5% and n = 3 years 𝑅 𝑛 A= P [1 + ] 100 5 3 = 70000 [1 + ] 100 = 70000 [1 + 0.05]3 = 70000 [1.05]3 = 70000 x 1.1576 = # 81,033.75 A = # 81,033.75 Compound Interest = # 81,033.75 - # 70000 = # 11,033.75 Example 2: A man borrows 600,000 from a cooperative at 8.5% compound interest per annum. If he repays #85000 at the end of each year, find the amount he still owes at the end of 2 years. Solution 1st year: Principal = # 600,000.00 8.5% interest = 8.5% = 0.085 X 600,000 = # 51000.00 (you are to add it first) # 651,000 Repayment (Amount) = - # 85,000 #566,000 22 2nd year: Principal = # 566,000 8.5% interest = 8.5% = 0.085 X 566,000= + # 48,110 # 614,110 Repayment = - # 85,000 # 529, 110 The amount owed after 2 years is # 529, 110 23  CHAPTER FIVE Topic: Factorization of simple algebraic expression Reference: A.J.S Olawasanmi Essential Mathematics for JSS book 3. Page (76-80) Objectives: Students should be able to: 1. simplifying calculation by factorization 2. factorise simple algebraic expressions by taking out the commo factors 3. simplifying calculation 4. solve word problems involving fractions Instructional Material: Chart showing expansion and factorizations Content: HCF of algebraic expression Factorisation by taken out common factor Simplifying calculations by factorization Factorization of simple algebraic expression Example 1: Find the HCF of 4a2b and 6ab Solution 4a2b = 2 X 2 X a x a x b 6ab = 2 X 3 X a X b The HCF = 2 X a X b = 2ab Example 2: Find the HCF of 10pq, pqr, 5p2 10pq = 2 x 5 x p x q Pqr = p x q x r 5p2 = 5 x p x p (Check for the common factor of the three algebraic expressions are p) HCF = p Factorization by taken out common factor Example 1: Simplify the following by factorization (a) 25 x 49 + 64 x 25 25 (49 + 64) 25(113) = 2825 24  Simplifying calculations by factorization Example: Factorize the following expression: 4a (2b + 3c) + 3z(2b + 3c) (2b + 3c) is common Let (2b + 3c) be q Therefore, 4a (2b + 3c) = 4aq + 3zq = q(4a + 3z) Factorization by grouping like terms Example: Factorize this expression a) Xy +az + xz + ay b) 5pq3 – pq2r2 + 5pqr – pr3 Solution a. Xy +az + xz + ay Step 1 (group the term with common factors together) Xy + xz = x (y + z) az + ay = a (z + y) which is equivalent to a ( z + y) the common factor is (z + y) xy + az + xz + ay = (a + x)(z + y) b. 5pq3 – pq2r2 + 5pqr – pr3 = pq2 ( 5q – r2) + pr(5q – r2) = (pq2 + pr) (5q – r2) = p(q2 + r)(5q – r2) 25  CHAPTER SIX Topic: Factorization of quadratic expression Reference: A.J.S Olawasanmi Essential Mathematics for JSS book 3. Page (82-87) Objectives: Students should be able to: 1. multiply out (or expand) two binomial expressions 2. factorize quadratic expressions of the form x2 + bx + c 3. expand and factorise some quadratic expressions known as perfect square Instructional Material: Chart showing expansion and factorizations Content: Expansion of algebraic expression Coefficients of terms Factorization by quadratic expression Perfect square Difference of two square Expansion of algebraic expression Example 1: Expand (x - 1)(3x – 7) = x (3x – 7) -1 (3x - 7) = 3x2 – 7x - 3x + 7 = 3x2 – 10x + 7 OR (x - 1)(3x - 7) = 3x2 – 7x – 3x + 7 = 3x2 – 10x + 7 Example 2: (3x + 4y) (x -2y) + 4x2 = 3x ( x - 2y) + 4y (x – 2y) + 4x2 = 3x2 – 6xy + 4xy – 6y2 + 4x2 = 3x2 - 2xy – 6y2 + 4x2 = 3x2 + 4x2 – 2xy – 6y2 = 7x2 – 2xy – 6y2 1 2 Example 3: Expand ( 𝑥 + 3) ( 𝑥 + 3) 5 5 26  1 2 2 = 𝑥 ( 𝑥 + 3) + 3( 𝑥 + 3) 5 5 5 2 3 6 = 𝑥2 + 𝑥 + 𝑥 + 9 25 5 5 2 3 6 = 𝑥2 + 𝑥 + 𝑥 + 9 25 5 5 2 3 6 = 𝑥2 + x ( + ) +9 { you first factorize the common variable x and find the LCM} 25 5 5 2 9 2 9 = 𝑥2 + x ( ) + 9 = 𝑥 2 + 𝑥+ 9 25 5 25 5 Example 4: Expand (4ab – 3c)2 = (4ab – 3c) (4ab – 3c) = 4ab (4ab – 3c) – 3c (4ab – 3c) = 16a2b2 – 12abc – 12abc + 9c2 = 16a2b2 – 24abc + 9c2 Coefficients of terms Quadratic expression has 2 as its highest power. A quadratic expression is also known as a polynomial of the second degree. It takes the general form as: ax2 + bx + c (a ≠ 0) where a,b and c represent a number. a = the coefficient of x2 b = the coefficient of x c = Is a constant term. Example 1: write down the coefficient of x2, x and the constant term after expanding (3x - 5)(4x + 4). Solution (3x - 5)(4x + 4) = 3x (4x + 4) -5 (4x + 4) = 12x2 + 7x – 20x – 20 = 12x2 – 13x – 20 The coefficient of x2 = 12, x= -13 and the constant = - 20 27  2 9 From example 3 above; 𝑥 2 + 𝑥+ 9 write down the coefficient of x2, x and the constant term. 25 5 2 9 The coefficient of x2 = , x= and the constant = 9 25 5 Factorization by quadratic expression A trinomial is an algebraic expression containing 3 terms. For example: ax2 + bx + c is a trinomial because it has 3 terms which are ax2, bx and c. here, we use sum and Product or factors of the middle term “b” Example 1: Factorize x2 + 3x – 4 (-4x2) = x2 + (-x + 4x) - 4 Method 1 (by sum and product of the middle term) -4x2 = 4x (X) -x = 4x2 (product) -4x = 4x (+) -x = -3x (sum) Method 2: Factors of (-4x2) x2 - x + 4x – 4 (factorize the common factors) x(x – 1) + 4(x – 1) = (x + 4)(x - 1) Example 2: 4x2 –4x – 3 4x2 X -3 = -12x2 using sum and product Sum = -6x + 2x = - 4x, product = -6x X + 2x = -12x2 4x2 – 6x + 2x – 3 2x (2x-3) + 1(2x – 3) = (2x + 1) (2x - 3) Example 3: 16 – 48v + 35v2 First rearrange: 35v2 – 48v + 16 28  The multiplication of 35v2 and 16 = 560v2 Concentrate more on the middle term (the sum); – 48v (-28v -20v) The product; 560v2 (-28v X - 20v) 35v2 – 48v + 16 = 35v2 – 28v – 20v +16 = 7v(5v – 4) – 4 (5v – 4) = (7v – 4)(5v - 4) Difference of two square Difference of two squares means difference of two perfect squares. Perfect squares are numbers or algebraic terms or expressions that can be written in power of two. E.g 1, 4, 9, 16, 25..., x2, a4, (3m)2 , 36y2, 52v2, a2b2, 9x2y2, (2x – y)2 etc. Note: 49v2u2 can be written in power of two as (7vu)2. Examples of difference of two squares are: a2 – 1, x2y2 – 9p2 , (a +2b)2–36 etc. Procedures of factorising difference of two squares Express each of the perfect squares in power of two. E.g x2 – 25 = x2 – 52 Multiply the difference and the sum of the bases obtained. i.e x2 – 52 = (x – 5)(x + 5) Example 1: Factorise p2 – q2 Solution p2 – q2 = (p + q)(p – q) Factorise 81m2 – n2p2 Solution 81m2 – n2p2 = (9m)2 – (np)2 = (9m – np)(9m + np) 29  Example 2 a. Factorise 81m2 – 64n2. Hence, calculate this value when m = 10 and n = 5 Solution 81m2 – 64n2. 92m2 – 82n2 (9m)2 - (8n)2 = (9m – 8n)( 9m + 8n). Hence, (9(10) – 8(5)) (9(10) + 8(5)) = ( 90 – 40) (90 + 40) = (50)(130) = 6,500 Example 3 4𝑥 3𝑥 4𝑥 3𝑥 4𝑥 3𝑥 Factorize: ( )2 - ( )2 = ( − )( + ) 9 9 9 9 9 9 Quadratic expression Quadratic expression is an expression whose variables have highest power of two. The general form of quadratic expression is ax2 + bx + c. Example of quadratic expression are x2 – 9x + 20 , 2p2 – 5p + 1, m2 – 81 etc Solving quadratic expression by factorization Example 1 Solve the equation 4x2 – 4x – 3 Solution Perfect square To make a quadratic expression a perfect square, add square of half of coefficient of linear variable (i.e x or any variable used). Alternatively, the formula b2=4ac can be used, provided that the given equation is written in the form ax2 + bx + c 30 Example 1. What value of k will make the expression the expression x2 – 12x + k a perfect square? Check by Factorising the result. Method 2 By comparing x2 – 12x + k to ax2 + bx + c = 0; a = 1, b = –12 and c = k Using b2 = 4ac :. (–12)2 = 4 × 1 × k 144 = 4k Dividing both sides by 4 k = 36. x2 – 12x + k = x2 – 12x + 36 = x2 – 6x – 6x + 36 = x(x – 6) – 6(x – 6) = (x – 6)(x – 6) = (x – 6)2 (perfect square) 31  CHAPTER SEVEN Topic: Simplifying algebraic expressions involving brackets and fractions Reference: A.J.S Olawasanmi Essential Mathematics for JSS book 3. Page (75-84) New general mathematics by M.F Macrae et al. page 23 and 88 Objectives: Students should be able to: i. perform basic arithmetic operations on expression ii. simplify algebraic expressions iii. expand algebraic expressions by removing bracket Instructional Material: Chart showing algebraic expressions Content: Simplifying algebraic expressions with brackets Simplifying fractional expressions Algebraic Expression Expansion of algebraic expressions is the act of removing brackets by multiplying the immediate term before or after brackets with the terms inside the brackets. Example 1: (a) Expand 2x(x2 y – 5z) (b) Expand – (4a – b + 3a2c) Solution Expand: 2a(a b – 5c) 2 = 2a3b – 10ac note: (bracket means multiplication) Expand: – (4a – b + 3a2c) using the minus sign to open the bracket. = -4a + b – 3a2c Simplification of algebraic expressions with bracket Example 1: Simplify 5(2x – 1) + 4 (x – 3) = 5(2x – 1) + 4 (x – 3) = 10x – 5 + 4x – 12 = 10x + 4x – 5 – 12 Answer = 14x – 17 32 Example 2: Expand and simplify (3a – 2b)(5 – 2a + 3b) = 3a(5 – 2a + 3b) – 2b (5 – 2a + 3b) = 15a – 6a2 + 9ab – 10b + 4ab – 6b2 = - 6a2 – 6b2 + 9ab + 4ab + 15a – 10b Answer = - 6a2 – 6b2 + 13ab + 15a – 10b Example 3: -2ab4c2 (7a2b3 – 3c2) 3 = -42a3b7c2 + 18ab4c4 Simplification of algebraic expressions with fractions To simplify expression involving addition or subtraction of fractions, the LCM of the denominators will need to be determined first. Example 1: Simplify 2(2𝑥−3)+3 (𝑥−6) = 12 4𝑥−6 + 3𝑥−18 = 12 4𝑥+3𝑥−6−18 = 12 7𝑥−24 Answer = 12 1 1 Example 2: Simplify ( 2x + a) - ( 2x – a) 5 5 1(2𝑥+𝑎) 1(2𝑥−𝑎) = - open the bracket and find the LCM of the 5 5 denominator (2𝑥+𝑎)−(2𝑥−𝑎) 2𝑥+𝑎−2𝑥+𝑎 2𝑥−2𝑥+𝑎+𝑎 = = = 5 5 5 2𝑎 Answer = 5 33 Example 3: 3𝑦+5 Simplify 6− 2 6 3𝑦+5 = − find the LCM 1 2 2(6) −1(3𝑦+5) 12−3𝑦−5 = = collect like terms 2 2 12− 5 −3𝑦 = 2 7−3𝑦 Answer = 2 ` 34  CHAPTER EIGHT Topic: Equation involving fractions Reference: A.J.S Olawasanmi Essential Mathematics for JSS book 3. Page (111-112) New general mathematics by M.F Macrae et al. page 96-98 Objectives: Students should be able to: i. perform basic arithmetic operations on fractional equations ii. solve simple equations with fractions Instructional Material: Chart showing fractional expressions and the procedure of clearing fractional Content: Fractional equations with unknown in the numerators Fractional equations with variable in the denominators. Fractional equations with binomial denominators Fractional equations with unknown in the numerators 2𝑥 1 Fracational equations are equations containing fractions e.g = −2 , 3 3 6 3 3 2 1 - = , (𝑥 − 5) = (3x – 1) etc. 𝑥 2𝑥 4 4 3 There are two common methods of solving fractional equations; i. by cross multiplication ii. by clearing fractions (denominations). 2𝑥 8−𝑥 Example 1: solve the equation = 5 10 Solution 35 Method 2 (clearing of fractions) Example 2: BECE 2024, Page 13, question 12 2(4𝑥−1) 9(𝑥+1) Solve for x, = 2 4 8𝑥−2 9𝑥+9 = 2 2 4(8x - 2) = 2(9x + 9) 32x – 8 =18x – 18 CLT 32x – 18x = 18 + 8 14x = 26 x = 26/14 = 13/7 or 167 Example 3: solve the equation 𝑎−5 𝑎 =5+ find the LCM 2 3 𝑎−5 15𝑎+𝑎 = Cross multiply 2 3 3(a -5) = 2(15a + a) 3a – 15 = 30 + 2a 3a-2a = 30 + 15 a = 45 36  Fractional equations with variable in the denominators 8 1 Example 1: BECE 2024, = 4(𝑥−2) 10 Solution 8 1 = (before cross multiplying, first open the bracket at the denominator) 4(𝑥−2) 10 8 1 = 4𝑥−8 10 4x – 8 = 8 x 10 4x – 8 = 80 4x = 80 + 8 4x = 88 x = 88/4 = 22 x = 22. Example 2: Solve for the unknown 5 −7 = 8 4𝑥 5 X 4x = -7 X 8 20x = -56 x = -56/20 x = -14/5 or −245 or -2.8 2 1 3 Example 3: Solve for x in the equation = - and Evaluate; 2y2 – 6y 5 2𝑦 4𝑦 Solution 2 1 3 = - 5 2𝑦 4𝑦 2 1 3 = - 5 2𝑦 4𝑦 LHS RHS { find the LCM of the RHS} and the C.M 2 2−6 = 5 4𝑦 2 x 4y = -4 X 5 8y = -20 37  y = -20/8 y = -5/2 Hence, evaluate 2y2 – 6y. Since y= -5/2 2y2 – 6y −5 −5 = 2( )2 - 6 2 2 25 30 = 2( ) + 4 2 50 = + 15 4 = 12.5 + 15 = 27.5 1 2 2 Example 4: Solve the equation - = 5 7𝑥 35 Solution 1 2 2 - = 5 7𝑥 35 1 2 2 = - = 5 35 7𝑥 7−2 2 = = 35 7𝑥 5 2 = cross multiply 35 7𝑥 5 X 7x = 2 X 35 35x = 70 x = 70/35 x=2 38  Fractional equations with binomial denominators Solution 4 6 Example 2: Solve the equation = 2𝑦+1 𝑦+2 Solution 4 6 = 2𝑦+1 𝑦+2 4(y +2) = 6(2y + 1) 4y + 8 = 12y + 6 4y – 12y = 6 - 8 -8y = -2 −2 1 y= = −8 4 y=¼ 39  5 3 5 Example 3: Solve for x; = - 3𝑥−1 4+𝑥 3𝑥−1 Solution 5 3 5 = - { check the denominator, and collect like terms} 3𝑥−1 4+𝑥 3𝑥−1 5 5 3 + = 3𝑥−1 3𝑥−1 4+𝑥 5+5 3 = (3𝑥−1) 4+𝑥 5+5 3 = (3𝑥−1) 4+𝑥 10 3 = (3𝑥−1) 4+𝑥 = 10(4 + x) = 3 (3x - 1) = 40 + 10x = 9x – 3 = 10x – 9x = -3 – 40 x = - 43 40  CHAPTER NINE Topic: Change of subject formulae Reference: New general mathematics by M.F Macrae et al. page 132-140. Essential mathematics by A.J.S Oluwasanmi page 57- 65. New concept mathematics by Arigbabu et al. page 96-106 Objectives: Students should be able to; i. explain change of subject of formulae ii. solve problems involved in change of subject formulae Instructional Material: Chart displaying problem on subject formulae Content: change of subject formulae Things to know while working on change of subject formulae Don’t rush it. You must go through each stage of your method and not try to skip any. If you are making any variable the subject it means that you want to end up with that varaible say “x” and this is the only x in your answer. Whatever you do to one side of the equals sign you must do to the other. Example 1: Rearrange the formula v = x + at make u the subject. Solution v = x + at v – at = x that is, x = v – at or - x = at – v dividing both sides by “ – ” x = - (at - v) x = -at + v by rearranging x = v – at 41  𝑐𝑥 Example 2: Rearrange the formula a = x + to make x the subject 𝑑 Solution 𝑐𝑥 a=x+ 𝑑 ad = dx + cx multiply both sides by d dx + cx = ad rearrange and factorize then divide by (d + c) x(d + c) = ad 𝒂𝒅 x= 𝒅+𝒄 𝑐 Example 3: Rearrange the formula a = b + to make p the subject. 1+𝑝 𝑐 a=b+ 1+𝑝 a(1 + p) = b(1 + p) + c a + ap = b + bp + c ap – bp = b + c – a p(a – b) = b + c – a 𝒃+𝒄−𝒂 p= 𝒂−𝒃 1 Example 4: Given that 𝑉 = 𝜋𝑟 3 ℎ, make h 3 Solution 1 V = 𝜋𝑟 3 ℎ 3 𝜋𝑟 3 ℎ V= 3 3 X V = 𝜋𝑟 3 ℎ 𝜋𝑟 3 ℎ 3𝑉 3𝑉 = = h= 𝜋𝑟 3 𝜋𝑟 3 𝜋𝑟 3 42