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AS Mathematics Algebra – Quadratic equations
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Objectives Be confident in the use of brackets Be able to factorise quadratic expressions Be able to solve quadratic equations by factorisation
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Factorising quadratic expressions The general form of a quadratic expression is: where a, b, c are constants and a ≠ 0 ax 2 + bx +c
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(i) x 2 + bx + c (a=1) You should be able to spot most of these by inspection
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Example 1 Factorisex 2 + 5x + 6 All positive, so (x + )(x + ) coefficent of x 2 is 1, so (x )(x ) Need 2 numbers with a product of +6... 1, 6 2, 3.. that sum to +5 ____ (x + 2)(x + 3) Check your answer by expanding the brackets.
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Example 2 Factorise x 2 - 5x + 6 (x )(x ) (x - )(x - ) (x - 2)(x - 3) coefficent of x 2 is 1, so (x )(x ) need 2 numbers with a +ve product but a –ve sum… both numbers must be -ve. Need 2 numbers with a product of +6... -1, -6 -2, -3.. that sum to -5 _____ Check your answer by expanding the brackets.
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(ii) ax 2 + bx + c, a≠1,0 The inspection method is fine for quadratics where a=1, but for complex quadratics a more rigorous method is required.
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Example 3 Factorise 3x 2 + 6x - 24 3x 2 + 12x - 6x -24 3x(x + 4) - 6(x + 4) Find two numbers so that sum = b AND product = ac sum = +6 product = 3(-24) = -72 } +12 -6 split the middle term factorise by grouping (x + 4)(3x – 6) Check your answer by expanding the brackets.
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Example 4 Factorise 6y 2 + 19y + 15 6y 2 + 10y + 9y + 15 2y(3y + 5) + 3(3y + 5) Find two numbers so that sum = b AND product = ac sum = +19 product = 6(15) = +90 } +10 +9 split the middle term factorise by grouping (3y + 5)(2y + 3) Check your answer by expanding the brackets.
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Example 5 Factorise 8d 2 + 10d - 12 8d 2 + 16d - 6d - 12 4d(2d + 4) - 3(2d + 4) sum = +10 product = 8(-12) = -96 } +16 -6 (2d + 4)(4d - 3) Check your answer by expanding the brackets.
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Solving quadratic equations 4 methods: (iv) graphically (iii) completing the square (ii) quadratic formula (i) factorisation We start with solving quadratic equations by factorisation
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Solving quadratic equations by factorisation Example 1 Solve the equation 3x 2 – 1 = 2x 3x 2 - 3x + x – 1 = 0 3x(x - 1) + 1(x - 1) = 0 sum = -2 product = 3(-1) = -3 } -3 +1 (x - 1)(3x + 1) = 0 Have you checked your answer? Rearrange to get... = 0 3x 2 – 2x – 1 = 0 Either x - 1 = 0 or 3x + 1 = 0 x = 1 or x = -⅓ The roots of this equation are x = 1 and x = - 1 / 3 ___________
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Example 2 Solve the equation (3x + 1)(2x-1) – (x + 2) 2 = 5 5x 2 - 5x – 10 = 0 5[x 2 – x – 2] = 0 sum = -1 product = 1(-2) = -2 } -2 +1 5(x + 1)(x – 2) = 0 Have you checked your answer? Expand and rearrange 6x 2 – 3x + 2x – 1 – [x 2 + 4x + 4] = 5 Either x + 1 = 0 or x - 2 = 0 x = -1 or x = 2 ______ The roots of this equation are x = -1 and x = 2
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