Download presentation

Published byOsvaldo Iredale Modified over 3 years ago

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

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

2
Introduction An equation which can be written in the form ax2 + bx + c = 0 , where a ≠0, is called a quadratic equation. Some examples of quadratic equations are: x2 - 4 = 0, x2 + 5x = 0, x2 + 6x + 8 = 0, 3x2 + 2x - 5 = 0 and so on. Quadratic equation can be written in many different forms, such as: x2 = 16, x2 = 8x, x2 = 4x +12. In these case the quadratic equation must be rearranged so that one side is ZERO (0). i.e. x2 = 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 = 0 or x + b = 0 or both (x + a) = 0 and (x + b) = 0.

4
**Examples: Factorise and solve each of the following.**

a) x2- 5x = 0 Factorise fully! b) x = 0 Factorise fully! (x - 5) x = 0 (x - 4)(x + 4) = 0 So, either So, either x - 5 = 0 or x = 0 x - 4 = 0 or x + 4 = 0 x = 5 x = 4 x = -4 The solutions are x = 0 and x = 5. The solutions are x = -4 and x = 4.

5
**Examples c) x2 - 12 = 0 d) x2 + 4x - 12 = 0 (x + 6)(x - 2) = 0**

Factorise fully! Factorise fully! (x + 6)(x - 2) = 0 So, either So, either x + 6 = 0 or x - 2 = 0 x = -6 x = 2 The solutions are and The solutions are x = -6 and x = 2.

6
**Examples f) 2x2 = 19x - 24 e) 3x = 4 - x2 2x2 - 19x + 24 = 0**

Rearrange, RHS = 0 e) 3x = 4 - x2 Rearrange, RHS = 0 Factorise fully! Sum/product 2x2 - 19x + 24 = 0 x2 + 3x - 4 = 0 Factorise fully! 2 x 24 = -1 x -48 sum = -49 = -2 x -24 sum -26 = -3 x -16 sum = -19 2x2 - 3x - 16x+ 24 = 0 (x - 1)(x + 4) = 0 So, either x(2x - 3) - 8(2x - 3) = 0 x - 1 = or x + 4 = 0 (x - 8)(2x - 3) = 0 So, either x = 1 x = -4 x - 8 = 0 or 2x - 3 = 0 The solutions are x = -4 and x = 1. x = 8 x = 3/2 The solutions are x = 8 and x = 3/2.

7
**Exercise: Solve each of the following.**

1) x2 + 6x = 0 2) x = 0 3) x2 - 6x = 16 Sol: x = 0, x = -6 Sol: x = -5, x = 5 Sol: x = -2, x = 8 4) 3x = 0 5) x2 - 12x + 32 = 0 6) 2x2 + x = 3 Sol: x = -3, x = 3 Sol: x = 4, x = 8 Sol: x = -1, x = 3/2 7) 4x2 + 11x + 6 = 0 8) 5x2 = 6x - 1 9) 5x - 3x2 = -28 Sol: x = -2, x = -3/4 Sol: x = 1, x = 1/5 Sol: x = 4, x = -7/3

Similar presentations

Presentation is loading. Please wait....

OK

Revision Equations I Linear Equations By I Porter.

Revision Equations I Linear Equations By I Porter.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on church leadership Ppt on amending the constitution Ppt on bombay film industry Ppt on history of our national flag Ppt on world bank and imf Free download ppt on unity in diversity Ppt on seven segment display truth Ppt on economic development and growth Ppt on power generation by speed breaker pump Make a ppt on election system in india