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Revision Quadratic I Solving Quadratic Equations by Factorisation. By I Porter.

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Presentation on theme: "Revision Quadratic I Solving Quadratic Equations by Factorisation. By I Porter."— Presentation transcript:

1 Revision Quadratic I Solving Quadratic Equations by Factorisation. By I Porter

2 Introduction An equation which can be written in the form ax 2 + bx + c = 0, where a ≠0, is called a quadratic equation. Some examples of quadratic equations are: x = 0, x 2 + 5x = 0, x 2 + 6x + 8 = 0, 3x 2 + 2x - 5 = 0 and so on. Quadratic equation can be written in many different forms, such as: x 2 = 16, x 2 = 8x, x 2 = 4x +12. In these case the quadratic equation must be rearranged so that one side is ZERO (0). i.e. x 2 = 16 becomes x = 0. Quadratic equations can be solved by one of 3 methods: *Factorisation *Complete the Square Method *Quadratic Formula

3 Factorisation Method We can solve quadratic equations using a basic law of numbers called the null factor law: If a x b = 0, either a = 0 or b = 0 or both a = 0 and b = 0. If a quadratic is written in the form of two factors and their product is zero, either one of the factors or both of the factors must equal zero (0). For example, if we have (x + a) (x + b) = 0, either x + a = 0or x + b = 0or both (x + a) = 0 and (x + b) = 0.

4 Examples: Factorise and solve each of the following. a) x 2 - 5x = 0 (x - 5) x = 0 So, either x - 5 = 0 orx = 0 x = 5 Factorise fully! The solutions are x = 0 and x = 5. b) x = 0 (x - 4)(x + 4) = 0 Factorise fully! So, either x - 4 = 0orx + 4 = 0 x = 4x = -4 The solutions are x = -4 and x = 4.

5 Examples c) x = 0 Factorise fully! So, either The solutions are and. d) x 2 + 4x - 12 = 0 Factorise fully! (x + 6)(x - 2) = 0 x + 6 = 0 or x - 2 = 0 So, either x = -6x = 2 The solutions are x = -6 and x = 2.

6 Examples e) 3x = 4 - x 2 Rearrange, RHS = 0 x 2 + 3x - 4 = 0 Factorise fully! (x - 1)(x + 4) = 0 So, either x - 1 = 0 or x + 4 = 0 x = 1x = -4 The solutions are x = -4 and x = 1. f) 2x 2 = 19x x x + 24 = 0 Rearrange, RHS = 0 Factorise fully! Sum/product 2 x 24= -1 x -48 sum = -49 = -2 x -24 sum -26 = -3 x -16 sum = -19 2x 2 - 3x - 16x+ 24 = 0 x(2x - 3) - 8(2x - 3) = 0 (x - 8)(2x - 3) = 0 So, either x - 8 = 0 or 2x - 3 = 0 x = 8x = 3 / 2 The solutions are x = 8 and x = 3 / 2.

7 Exercise: Solve each of the following. 1) x 2 + 6x = 02) x = 03) x 2 - 6x = 16 4) 3x = 05) x x + 32 = 06) 2x 2 + x = 3 7) 4x x + 6 = 08) 5x 2 = 6x - 19) 5x - 3x 2 = -28 Sol: x = 0, x = -6Sol: x = -5, x = 5Sol: x = -2, x = 8 Sol: x = -3, x = 3Sol: x = 4, x = 8Sol: x = -1, x = 3 / 2 Sol: x = -2, x = - 3 / 4 Sol: x = 1, x = 1 / 5 Sol: x = 4, x = - 7 / 3


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