Quadratic Equations

A basic quadratic equation contains an x2 term. Quadratic Equations A basic quadratic equation contains an x2 term. Find all the possible values of x from the following: 1. x2 = 16 x2 = 81 x2 + 3 = 12 4. x2 – 8 = 17 x = 4 or x = - 4 x = 9 or x = - 9 x = 3 or x = -3 x = 5 or x = - 5 Note: Nearly all Quadratic equations have two solutions.

To solve any other quadratic equation we need to use the following principle: If A x B = 0 Then either, A = 0 or B = 0 Example: 1. x(x – 6) = 0 Either x = 0 or x – 6 = 0 If x – 6 = 0 x = 6 Therefore x = 6 or x = 0

2. x(x + 3) = 0 Either x = 0 or x + 3 = 0 If x + 3 = 0 x = - 3 Therefore x = - 3 or x = 0 3. 4x(x – 7) = 0 Either 4x = 0 or x – 7 = 0 If 4x = 0 x = 0 x – 7 = 0 x = 7 Therefore x = 7 or x = 0

Quadratic Equations Remember: If A x B = 0 Then either, A = 0 or B = 0 1. (x – 2)(x + 4) = 0 Either x – 2 = 0 or x + 4 = 0 If x – 2 = 0 x = 2 or x + 4 = 0 x = - 4 Therefore x = 2 or x = - 4

2. (x + 5 )(x – 7) = 0 Either x + 5 = 0 or x – 7 = 0 If x + 5 = 0 x = - 5 or x – 7 = 0 x = 7 Therefore x = - 5 or x = 7

QUADRATIC EQUATIONS Most quadratic equations need to be factorised first, in order to solve them using the basic principle. Remember: x2 – 7x = x2 + 5x + 6 = x(x – 7) (x + 3)(x + 2)

Solve the following by factorisation 1. x2 – 7x = 0 Divide by common factor of x x(x – 7) = 0 Either x = 0 or x – 7 = 0 x = 7 So x = 0 or x = 7 2. x2 + 3x = 0 Divide by common factor of x x(x + 3) = 0 Either x = 0 or x + 3 = 0 x = - 3 So x = 0 or x = - 3

3. 5x2 – 20x = 0 5x(x – 4) = 0 Either 5x = 0 or x – 4 = 0 x = 0 x = 4 Divide by common factor of 5x 5x(x – 4) = 0 Either 5x = 0 or x – 4 = 0 x = 0 x = 4 So x = 0 or x = 4

4. x2 + 5x + 6 = 0 (x + 3)(x + 2) = 0 If x + 3 = 0 or x + 2 = 0 Factorise into a double bracket (x + 3)(x + 2) = 0 If x + 3 = 0 or x + 2 = 0 x = - 3 x = - 2 So x = - 3 or x = - 2 5. x2 – 9x + 14 = 0 Factorise into a double bracket (x – 7)(x – 2) = 0 If x – 7 = 0 or x – 2 = 0 x = 7 or x = 2 So x = 7 or x = 2