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QUADRATICS EQUATIONS/EXPRESSIONS CONTAINING x 2 TERMS

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SIMPLE QUADRATIC PATTERNS - Quadratic number patterns are sequences of numbers where the difference between terms is not the same and the rule contains a squared term - You need to halve the difference of the differences to find the squared term e.g. Write a rule for the following pattern nTerm (T) Find the difference between terms If difference is not the same, find the difference of the differences! + 2 Rule: T = 3. Halve the 2 nd difference to find the n 2 rule 1n21n2 4. Substitute to find constant 1×1 2 = 1 1 = = Check if rule works i.e. If the difference of the differences is a 2, the rule contains 1n 2

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HARDER QUADRATIC PATTERNS - The squared term of the rule is found by halving the difference of the differences - i.e. If the difference of the differences is a 6, the rule contains 3n 2 - If the simple trial and error does not work, try this technique: e.g. Write a rule for the following pattern nTerm (T) Find the difference between terms If difference is not the same, find the difference of the differences! + 4 Rule: T = 3. Halve the 2 nd difference to find the n 2 rule 2n22n2 4. Subtract the n 2 rule from the term 2×1= 2 2 = × ×4 + 1 = Find the linear part of the rule + 1 Term (T) – 2n 2 6. Check if the rule works 5 - 2× × n

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EXPANDING TWO BRACKETS - To expand two brackets, we must multiply each term in one bracket by each in the second e.g. Expand and simplify a) (x + 5)(x + 2)b) (x - 3)(x + 4) c) (x - 1)(x - 3)c) (2x + 1)(3x - 4) = x 2 + 2x+ 5x+ 10 To simplify, combine like terms = x 2 + 7x + 10 = x 2 + 4x- 3x- 12 Remember integer laws when multiplying = x 2 + 1x – 12 = x 2 - 3x- 1x+ 3 = x 2 – 4x + 3 = 6x 2 - 8x+ 3x- 4 = 6x 2 – 5x – 4

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PERFECT SQUARES - When both brackets are exactly the same e.g. Expand and simplify a) (x + 8) 2 = x 2 + 8x + 64 To simplify, combine like terms = x x + 64 = (x + 8)(x + 8)b) (x - 4) 2 = x 2 - 4x + 16 = x 2 - 8x + 16 = (x – 4)(x – 4) c) (3x - 2) 2 = 9x 2 - 6x + 4 = 9x 2 – 12 x + 4 = (3x – 2)(3x – 2) Watch sign change when multiplying Write out brackets twice BEFORE expanding

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DIFFERENCE OF TWO SQUARES - When both brackets are the same except for signs (i.e. – and +) e.g. Expand and simplify a) (x – 3)(x + 3)b) (x – 6)(x + 6) c) (2x – 5)(2x + 5) = x 2 + 3x- 3x- 9 Like terms cancel each other out = x 2 – 9 = x 2 + 6x- 6x- 36 = x 2 – 36 = 4x x- 10x- 25 = 4x 2 – 25

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FACTORISING The general equation for a quadratic is ax 2 + bx + c When a = 1 - You need to find two numbers that multiply to give c and add to give b e.g. Factorise 1) x x + 24 List pairs of numbers that multiply to give 24 (c) 1, 24 2, 12 3, 8 4, 6 Check which pair adds to give 11 (b) Place numbers into brackets with x = (x + 3)(x + 8) 2) x 2 + 7x + 6 2, 3 1, 6 List pairs of numbers that multiply to give 6 (c) Check which pair adds to give 7 (b) Place numbers into brackets with x = (x + 1)(x + 6) To check answer, expand and see if you end up with the original question!

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- Expressions can also contain negatives e.g. Factorise 1) x 2 + x – 12 1, 12 2, 6 3, 4 As the end number (c) is -12, one of the pair must negative. Make the biggest number of the pair the same sign as b Check which pair now adds to give b = (x - 3)(x + 4) 2) x 2 – 6x – 16 1, 16 2, 8 4, 4 As the end number (c) is -16, one of the pair must negative. Make the biggest number of the pair the same sign as b Check which pair now adds to give b = (x + 2)(x - 8) 3) x 2 – 9x , 20 2, 10 4, 5 As the end number (c) is +20, but b is – 9, both numbers must be negative Check which pair now adds to give b = (x - 4)(x - 5) 4) x 2 – 10x , 25 5, = (x - 5)(x - 5) = (x - 5) 2 As the end number (c) is +25, but b is – 10, both numbers must be negative

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SPECIAL CASES 1. No end number (c) e.g. Factorise a) x 2 + 6xb) x 2 – 10x Add in a zero and factorise as per normal + 0 0, 6 = (x + 0)(x + 6) = x(x + 6) = x( ) 2. No x term (b) (difference of two squares) e.g. Factorise a) x OR: factorise by taking out a common factor Add in a zero x term and factorise + 0x -5, 5 = (x - 5)(x + 5) OR: factorise by using A 2 – B 2 = (A – B)(A + B) b) x 2 – = (x )(x ) x - 10 c) 9x 2 – 121 = (3x )(3x )

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TWO STAGE FACTORISING When a ≠ 1 1. Common factor - Always try to look for a common factor first. e.g. Factorise a) 2x x + 16= 2( ) 1, 8 2, 4 = 2(x + 2)(x + 4) b) 3x 2 – 6x – 9= 3( ) 1, 3-= 3(x + 1)(x – 3) c) 3x x= 3x( )x + 8d) 4x 2 – 36= 4( ) = 4(x )(x ) x 2 + 6x + 8x 2 – 2x – 3 x 2 – 9

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2. No common factor (HARD) - Use the following technique e.g. Factorise a) 3x 2 – 10x – 8 Multiply first and last terms 3x 2 × - 8 = -24x 2 Find two terms that multiply to -24x 2 but add to -10x 1x, 24x 2x, 12x 3x, 8x 4x, 6x Replace 10x with the two new terms = 3x 2 + 2x – 12x – 8 Factorise 2 terms at a time. = x( )3x + 2-4( )3x + 2= (x – 4)(3x + 2) Write in two brackets b) 2x 2 + 7x + 3 2x 2 × 3 = 6x 2 1x, 6x 2x, 3x = 2x 2 + x + 6x + 3= x( )2x ( ) 2x + 1= (x + 3)(2x + 1)

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SOLVING QUADRATICS To solve use the following steps: 1. Move all of the terms to one side, leaving zero on the other 2. Factorise the equation 3. Set each factor to zero and solve. e.g. Solve a) (x + 7)(x – 2) = 0 x + 7 = 0x – 2 = x = - 7x = 2 b) (x – 4)(x – 9) = 0 x – 4 = 0x – 9 = x = 4x = 9 c) x 2 + x – 2 = 0 (x – 1)(x + 2) = 0 1, 2- x – 1 = 0x + 2 = x = 1x = - 2 d) x 2 – 5x + 6 = 0 (x + 1)(x – 6) = 0 x + 1 = 0x – 6 = 0 +6 x = -1x = 6 1, 6 2, 3 - -

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e) x 2 + 8x = 0 x( ) = 0 x = 0x + 8 = x = -8 x + 8 f) x 2 – 11x = 0 x( ) = 0 x = 0x – 11 = x = 11 x – 11 g) x = 0 (x )(x ) = 0 x + 7 = x = x - 7 = 0 +7 x = 7 h) 9x = 0 (3x )(3x ) = x - 2 = x = 2 ÷3 x = 2/3 3x + 2 = x = -2 ÷3 x = -2/3 i) x 2 = 4x + 5 (x + 1)(x – 5) = 0 x + 1 = 0x – 5 = 0 +5 x = -1x = 5 1, 5--4x -5 x 2 – 4x – 5 = 0 j) x(x + 3) = 180 x 2 + 3x = x 2 + 3x – 180 = 0 (x + 15)(x – 12) = 0 x + 15 = 0x – 12 = x = -15x = 12

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WRITING EQUATIONS - Involves writing an equation from the information then solving e.g. The product of two consecutive numbers is 20. What are they? If x = a number, then the next consecutive number isx + 1 x(x + 1) = 20 x 2 + x = x 2 + x – 20 = 0 1, 20 2, 10 4, (x – 4)(x + 5) = 0 x – 4 = 0x + 5 = x = 4x = -5 The numbers are 4, 5 and -5, -4 e.g. A paddock of area 150 m 2 has a length 5 m longer than its width. Find the dimensions of this paddock. A = 150 m 2 x x + 5 x(x + 5) = 150 x 2 + 5x = x 2 + 5x – 150 = 0 (x – 10)(x + 15) = 0 x – 10 = 0x + 15 = x = 10x = -15 The dimensions are10 m by 15 m

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