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1. 1. MATHEMATICS FORM 4 Quadratic Expressions & Equations
2. 2. 2.1 QUADRATIC EXPRESSIONS (QE) # Is an expression in the form of ax 2 + bx + c, where a, b and c are constants, a ≠ 0 and x is an unknown. # Has : only one unknown : the highest power of the unknown is 2 Eg : 3x 2 + 2x - 1 m 2 + 2m 5 - y 2 9p 2
3. 3. 2.2 FACTORISATION OF QUADRATIC EXPRESSIONS  Factorisation of a QE is the process of writing the expresion as a product of two linear expressions.  Four types of factorise QE of the form : 1.1) ax 2 + c i) take out the HCF of a and c. ii) write “what remains” as the second factor. eg: a) 6x 2 + 8 = 2 ( 3x 2 + 4) b) 9 + 6m 2 = 3 (3 + 2m 2 ) (HCF )
4. 4. 1.2) ax 2 + bx i) take out the unknown (x) and HCF of a and b. ii) write “what remains” as the second factor. eg: a) 12x 2 + 9x = 3x ( 4x + 3) b) 7y + 14y 2 = 7y (1 + 2y ) unknown y (HCF )
5. 5. 2) px 2 − q, where p and q are perfect square i) write p = a 2 and q = b 2 . ii) write the answer as (ax + b)(ax − b). eg: a) x 2 − 9 = x 2 − 3 2 = (x + 3)(x−3) b) 49 + 16y 2 = 7 2 − 4 2 y 2 = 7 2 − (4y) 2 = (7 + 4y)(7 − 4y)
6. 6. 3) ax 2 + bx + c, where a, b and c are not equal to zero. i) list down the pair of numbers p and q such that pq = c . ii) select the pair of p and q such that p + q = b iii) write the answer as (x + p)(x + q). eg: a) x 2 + 7x + 10 = (x + 2)(x + 5) pq 10 p 1 2 q 10 5 p+ q 11 7 Since b and c are positive the values of p and q must also positive
7. 7. eg: b) x 2 − 8x + 12 = (x − 2)(x − 6) pq 12 p −1 −2 q −12 −6 p+ q −13 −8 Since c is positive and b is negative, that is, pq is positive and p + q is negative,then the values of p and q must be negative.
8. 8. eg: c) x 2 + 5x − 6 = (x − 1)(x + 6) pq − 6 p 1 −1 q −6 6 p+ q −5 5 Since c is negative, that is, pq is negative, then the values of p and q must be of different signs, where one is positive and the other is negative.
9. 9. eg: d) x 2 − 3x − 18 = (x + 3)(x − 6) e) 16 − 8x + x 2 = x 2 − 8x + 16 = (x − 4)(x − 4) or = (x − 4) 2 pq − 18 p 1 −1 2 −2 3 q −18 18 −9 9 −6 p+ q −17 17 −7 7 −3 pq 16 p −1 −2 −4 q −16 −8 −4 p+ q −17 −10 −8
10. 10.  Other method that can be used to factorise ax 2 + bx + c is called cross method. The steps to be followed are: i) Factorise a as m n, then factorise c as p q such that mq + np = b. mx +p +npx (nx)(p) nx +q +mqx (mx)(q) mnx 2 +pq +(mq + np)x ii) Write the answer as (mx + p)(nx + q)
11. 11. eg: 2x 2 − 13x + 15 = (x − 3)(2x − 5) eg: 8 + 2x − x 2 = −x 2 + 2x + 8 = (−x + 4)(x + 2) x − 5 −10x 2x −3 −3x 2x 2 15 −13x −x 2 4x x 4 −2x −x 2 8 +2x
12. 12. 4) Factorise QE containing coefficients with common factors.  To factorise completely a QE containing coefficients with a common factor, take out the HCF first before finding the other two factors. eg: 2x 2 + 16x + 24 = 2(x 2 + 8x + 12) = 2(x + 2)(x + 6)
13. 13. 2.3 QUADRATIC EQUATIONS  A quadratic equation (QEq) with one unknown has an equal sign and the highest power of the unknown is 2. eg: x 2 + 4x + 3 = 0 m 2 = 4m  General form : ax 2 + bx + c = 0 eg: Write 4/x = 5 − x in general form. x 2 − 5x + 4 = 0
14. 14. 2.4 ROOTS OF QUADRATIC EQUATIONS  The value of the unknown which satisfy a QEq are called the roots of the QEq.  Verifying roots : - Substituting a given value for the unknown in QEq to determine whether it is a root. eg: 2y 2 = 4 − 7y; y = 4, y = 1/2 Value of y Left hand side (LHS) Right hand side (RHS) conclusion 4 2(4 2 )= 32 4 − 7(4) = −24 LHS ≠ RHS 4 is not the root. 1/2 2(1/2 ) 2 = 1/2 4 − 7(1/2) = 1/2 LHS = RHS 1/2 is not the root.
15. 15.  Determining roots by factorisation. Steps : 1) express the equation in general form : ax 2 + bx + c = 0 2) Factorise ax 2 + bx + c to express the equation in factor form (mx +p)(nx + q) = 0. 3) Write mx + p = 0 and nx + q = 0 4) Solve for x, x = − p/m and x = −q/n
16. 16. Eg: w(w + 3) = 9(w − 1) w 2 + 3w = 9w – 9 Step 1: w 2 – 6w − 9 = 0 Step 2: (w – 3)(w − 3) = 0 Step 3: w − 3 = 0 and w − 3 = 0 Step 4: w = 3 [repeated root]
17. 17. Eg: 3p 2 + 4p = 2 p + 4 3p 2 + 4p = 2(p + 4) Step 1: 3p 2 + 2p − 8 = 0 Step 2: (3p – 4)(p + 2) = 0 Step 3: 3p − 4 = 0 and p + 2 = 0 Step 4: p = 4/3, p = − 2