1. Quadratic Equations

2. 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:
x2 = 16
x2 = 81
x2 + 3 = 12
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.

3. 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

4. 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

5. 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

6. 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

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 =
x(x – 7)
(x + 3)(x + 2)

8. 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

9. 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 = x = 4
So x = 0 or x = 4

10. 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 = 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