QUADRATICS EQUATIONS/EXPRESSIONS CONTAINING x2 TERMS

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 - i.e. If the difference of the differences is a 2, the rule contains 1n2 e.g. Write a rule for the following pattern Rule: T = 1n2 + 3 n Term (T) 1 4 2 7 3 12 19 5 28 + 3 1×12 = 1 + 2 + 5 1 = 4 + 3 + 2 + 7 + 2 42 + 3 = 19 + 9 3. Halve the 2nd difference to find the n2 rule 1. Find the difference between terms 2. If difference is not the same, find the difference of the differences! 4. Substitute to find constant 5. Check if rule works

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 3n2 - If the simple trial and error does not work, try this technique: Rule: T = 2n2 + 2n + 1 e.g. Write a rule for the following pattern 2×1= 2 n Term (T) 1 5 2 13 3 25 4 41 61 Term (T) – 2n2 2 = 3 + 1 5 - 2×12 3 + 8 + 2 + 4 13 - 2×22 5 + 12 + 2 + 4 7 + 16 + 2 + 4 9 + 20 + 2 11 3. Halve the 2nd difference to find the n2 rule 1. Find the difference between terms 4. Subtract the n2 rule from the term 2. If difference is not the same, find the difference of the differences! 5. Find the linear part of the rule 6. Check if the rule works 2×42 + 2×4 + 1 = 41

EXPANDING TWO BRACKETS - To expand two brackets, we must multiply each term in one bracket by each in the second Remember integer laws when multiplying e.g. Expand and simplify a) (x + 5)(x + 2) = x2 + 2x + 5x + 10 b) (x - 3)(x + 4) = x2 + 4x - 3x - 12 = x2 + 7x + 10 = x2 + 1x – 12 To simplify, combine like terms c) (x - 1)(x - 3) = x2 - 3x - 1x + 3 c) (2x + 1)(3x - 4) = 6x2 - 8x + 3x - 4 = x2 – 4x + 3 = 6x2 – 5x – 4

PERFECT SQUARES - When both brackets are exactly the same To simplify, combine like terms e.g. Expand and simplify Watch sign change when multiplying a) (x + 8)2 = (x + 8)(x + 8) b) (x - 4)2 = (x – 4)(x – 4) = x2 + 8x + 8x + 64 = x2 - 4x - 4x + 16 = x2 + 16x + 64 = x2 - 8x + 16 Write out brackets twice BEFORE expanding c) (3x - 2)2 = (3x – 2)(3x – 2) = 9x2 - 6x - 6x + 4 = 9x2 – 12 x + 4

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) = x2 + 3x - 3x - 9 b) (x – 6)(x + 6) = x2 + 6x - 6x - 36 = x2 – 9 = x2 – 36 Like terms cancel each other out c) (2x – 5)(2x + 5) = 4x2 + 10x - 10x - 25 = 4x2 – 25

FACTORISING The general equation for a quadratic is ax2 + bx + c When a = 1 - You need to find two numbers that multiply to give c and add to give b e.g. Factorise To check answer, expand and see if you end up with the original question! 1) x2 + 11x + 24 = (x + 3)(x + 8) 1, 24 List pairs of numbers that multiply to give 24 (c) Check which pair adds to give 11 (b) Place numbers into brackets with x 2, 12 3, 8 4, 6 2) x2 + 7x + 6 = (x + 1)(x + 6) 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 2, 3

- Expressions can also contain negatives e.g. Factorise 1) x2 + x – 12 = (x - 3)(x + 4) 2) x2 – 6x – 16 = (x + 2)(x - 8) - 1, 12 1, 16 - As the end number (c) is -12, one of the pair must negative. As the end number (c) is -16, one of the pair must negative. - 2, 6 2, 8 - - 3, 4 4, 4 - Check which pair now adds to give b Make the biggest number of the pair the same sign as b Check which pair now adds to give b Make the biggest number of the pair the same sign as b 3) x2 – 9x + 20 = (x - 4)(x - 5) 4) x2 – 10x + 25 = (x - 5)(x - 5) - 1, 20 - - 1, 25 - = (x - 5)2 As the end number (c) is +20, but b is – 9, both numbers must be negative - 2, 10 - - 5, 5 - As the end number (c) is +25, but b is – 10, both numbers must be negative - 4, 5 - Check which pair now adds to give b

SPECIAL CASES 1. No end number (c) e.g. Factorise a) x2 + 6x + 0 b) x2 – 10x = x( ) x - 10 0, 6 = x(x + 6) Add in a zero and factorise as per normal OR: factorise by taking out a common factor 2. No x term (b) (difference of two squares) e.g. Factorise a) x2 - 25 + 0x = (x - 5)(x + 5) b) x2 – 100 = (x )(x ) - 10 + 10 -5, 5 c) 9x2 – 121 = (3x )(3x ) - 11 + 11 Add in a zero x term and factorise OR: factorise by using A2 – B2 = (A – B)(A + B)

TWO STAGE FACTORISING When a ≠ 1 1. Common factor - Always try to look for a common factor first. e.g. Factorise a) 2x2 + 12x + 16 = 2( ) x2 + 6x + 8 b) 3x2 – 6x – 9 = 3( ) x2 – 2x – 3 1, 8 = 2(x + 2)(x + 4) 1, 3 - = 3(x + 1)(x – 3) 2, 4 c) 3x2 + 24x = 3x( ) x + 8 d) 4x2 – 36 = 4( ) x2 – 9 = 4(x )(x ) - 3 + 3

2. No common factor (HARD) - Use the following technique e.g. Factorise a) 3x2 – 10x – 8 = 3x2 + 2x – 12x – 8 = x( ) 3x + 2 -4( ) 3x + 2 = (x – 4)(3x + 2) 1x, 24x - Multiply first and last terms Find two terms that multiply to -24x2 but add to -10x Replace 10x with the two new terms Factorise 2 terms at a time. 2x, 12x - 3x, 8x - 4x, 6x - 3x2 × - 8 = -24x2 Write in two brackets b) 2x2 + 7x + 3 = 2x2 + x + 6x + 3 = x( ) 2x + 1 + 3( ) 2x + 1 = (x + 3)(2x + 1) 2x2 × 3 = 6x2 1x, 6x 2x, 3x

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 b) (x – 4)(x – 9) = 0 x + 7 = 0 x – 2 = 0 x – 4 = 0 x – 9 = 0 -7 -7 +2 +2 +4 +4 +9 +9 x = - 7 x = 2 x = 4 x = 9 c) x2 + x – 2 = 0 - 1, 2 d) x2 – 5x + 6 = 0 1, 6 - 2, 3 - (x – 1)(x + 2) = 0 (x + 1)(x – 6) = 0 x – 1 = 0 x + 2 = 0 x + 1 = 0 x – 6 = 0 +1 +1 - 2 - 2 -1 -1 +6 +6 x = 1 x = - 2 x = -1 x = 6

e) x2 + 8x = 0 f) x2 – 11x = 0 x( ) = 0 x + 8 x( ) = 0 x – 11 x = 0 x + 8 = 0 x = 0 x – 11 = 0 - 8 - 8 + 11 + 11 x = -8 x = 11 g) x2 - 49 = 0 h) 9x2 - 4 = 0 (x )(x ) = 0 - 7 + 7 (3x )(3x ) = 0 - 2 + 2 x - 7 = 0 x + 7 = 0 3x - 2 = 0 3x + 2 = 0 +7 +7 - 7 - 7 +2 +2 -2 -2 x = 7 x = -7 3x = 2 3x = -2 ÷3 ÷3 ÷3 ÷3 x = 2/3 x = -2/3 i) x2 = 4x + 5 j) x(x + 3) = 180 -4x -5 -4x -5 1, 5 - x2 + 3x = 180 x2 – 4x – 5 = 0 -180 -180 (x + 1)(x – 5) = 0 x2 + 3x – 180 = 0 (x + 15)(x – 12) = 0 x + 1 = 0 x – 5 = 0 x + 15 = 0 x – 12 = 0 -1 -1 +5 +5 -15 -15 +12 +12 x = -1 x = 5 x = -15 x = 12

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 is x + 1 x + 5 x(x + 1) = 20 - 1, 20 x2 + x = 20 - 2, 10 -20 -20 A = 150 m2 x - 4, 5 x2 + x – 20 = 0 (x – 4)(x + 5) = 0 x – 4 = 0 x + 5 = 0 +4 +4 -5 -5 x(x + 5) = 150 x = 4 x = -5 x2 + 5x = 150 The numbers are 4, 5 and -5, -4 -150 -150 x2 + 5x – 150 = 0 (x – 10)(x + 15) = 0 e.g. A paddock of area 150 m2 has a length 5 m longer than its width. Find the dimensions of this paddock. x – 10 = 0 x + 15 = 0 +10 +10 -15 -15 x = 10 x = -15 The dimensions are 10 m by 15 m