08Anullfactorlaw

= Solving Quadratic Equations =

Null Factor Law (NFL)
Consider the following equation:

... ... ... ** a × b = 0 **

This is only possible if ... ... ... ** a = 0 ** ... ... or ... ... ** b = 0 **

This is called the ** Null Factor Law (NFL) ** ... .... {Null means nothing or zero}

We use the Null Factor Law to solve quadratic Equations.

** Example 1 **
Solve the following quadratic equations:
 * 1) x(x + 3) = 0
 * 2) (x + 5)(x – 2) = 0
 * 3) 3(2x – 1)(4x + 3) = 0


 * Solution: **

1. Solve x(x + 3) = 0

{this is two things multiplied together and the answer is zero, so apply the Null Factor Law}

... ... x(x + 3) = 0 ... ... {use NFL} ... ... x = 0 .... or .... x + 3 = 0 ... ... x = 0 .... or ....... x = –3

This means the quadratic will equal 0 if x = 0 or if x = –3

{__Both__ answers are correct and __both__ answers must be listed to be a correct solution}

2. Solve (x + 5)(x – 2) = 0

... ... (x + 5)(x – 2) = 0 ... ... {use NFL} ... ... x + 5 = 0 .... or ..... x – 2 = 0 ... ... .. x = –5 ...... or ........ x = 2

3. Solve 3(2x – 1)(4x + 3) = 0

... ... 3(2x – 1)(4x + 3) = 0 ... ... {Use NFL __ 3 <> 0 so ignore it}

math \\ . \qquad 2x - 1 = 0 \quad \text{or} \quad 4x + 3 = 0 \\ . \qquad \quad 2x = 1 \quad \;\; \text{or} \quad \quad 4x = -3 \\ . \qquad \quad \; x = \frac{1}{2} \;\; \quad \text{or} \quad \quad \; x = - \frac{3}{4} \\ math

To solve a quadratic equation:
 * Note: ** Unfortunately, most quadratics are not written in factorised form.
 * 1) If necessary, rearrange to get 0 on the right hand side.
 * 2) Factorise
 * 3) By taking out a a common factor
 * 4) By using the difference of two squares rule
 * 5) By using the shortcut if a = 1
 * 6) By using the full method if a <> 1
 * 7) By completing the square
 * 8) Use the Null Factor Law

** Example 2 **
Solve each of the following quadratic equations
 * 1) 2x² – 8x = 0
 * 2) 4x² – 9 = 0
 * 3) 3x² + 9x – 30 = 0
 * 4) 3x² – 10x = 8
 * 5) x² + 6x + 1 = 0


 * Solution: **

1. Solve 2x² – 8x = 0

math \\ . \qquad 2x^2 - 8x = 0 \qquad \quad \big\{ \text{take out 2x as a common factor} \big\} \\. \\ . \qquad 2x \big(x - 4 \big) = 0 \qquad \quad \big\{ \text{use NFL} \big\} \\. \\ . \qquad 2x = 0 \quad \text{or} \quad x - 4 = 0 \\. \\ . \qquad \; x = 0 \quad \; \text{or} \qquad x = 4 math

2. Solve 4x² – 9 = 0

math \\ . \qquad 4x^2 - 9 = 0 \qquad \qquad \big\{ \text{use difference of two squares} \big\} \\. \\ . \qquad \big( 2x - 3 \big) \big( 2x + 3 \big) = 0 \qquad \;\; \big\{ \text{use NFL} \big\} \\. \\ . \qquad 2x - 3 = 0 \quad \text{or} \quad 2x + 3 = 0 \\. \\ . \qquad \quad \;\; 2x = 3 \; \quad \text{or} \quad \;\; 2x = -3 \\. \\ . \qquad \quad \;\;\; x = \dfrac{3}{2} \quad \text{or} \qquad x = -\dfrac{3}{2} math


 * Note: ** This answer could be written as:

math . \qquad \qquad x = \pm \dfrac{3}{2} math

3. Solve 3x² + 9x – 30 = 0

math \\ . \qquad 3x^2 +9x - 30 = 0 \qquad \qquad \big\{ \text{take out 3 as a common factor} \big\} \\. \\ . \qquad 3 \big( x^2 +3x - 10 \big) = 0 \qquad \quad \big\{ \text{factorise using shortcut} \big\} \\. \\ . \qquad 3 \big( x + 5 \big) \big( x - 2 \big) = 0 \qquad \quad \big\{ \text{Use NFL} \big\} \\. \\ . \qquad x + 5 = 0 \quad \text{or} \quad x - 2 = 0 \\. \\ . \qquad \;\; x = -5 \quad \;\; \text{or} \quad \quad x = 2 math

4. Solve 3x²– 10x = 8

math \\ . \qquad 3x^2 - 10x = 8 \qquad \qquad \big\{ \text{Rearrange to make RHS = 0} \big\} \\. \\ . \qquad 3x^2 - 10x - 8 = 0 \qquad \quad \big\{ \text{Factorise using full method} \big\} \\. \\ . \qquad 3x^2 - 12x + 2x - 8 = 0 \\. \\ . \qquad 3x \big( x - 4 \big) +2 \big( x - 4 \big) = 0 \\. \\ . \qquad \big( x - 4 \big) \big( 3x + 2 \big) = 0 \qquad \quad \big\{ \text{Use NFL} \big\} math . math \\ . \qquad x - 4 = 0 \quad \text{or} \quad 3x + 2 = 0 \\. \\ . \qquad \;\;\; x = 4 \qquad \text{or} \quad \;\; 3x = -2 \\. \\ . \qquad \;\;\; x = 4 \qquad \text{or} \quad \;\; x = - \dfrac{2}{3} math

5. Solve x² + 6x + 1 = 0

math \\ . \qquad x^2 + 6x + 1 = 0 \qquad \quad \big\{ \text{Factorise by completing the square} \big\} \\. \\ . \qquad x^2 + 6x +9 + 1 - 9 = 0 \\. \\ . \qquad \big( x + 3 \big)^2 - 8 = 0 \\. \\ . \qquad \big( x + 3 \big)^2 - \big( \sqrt{8} \big)^2 = 0 \\. \\ . \qquad \big( x + 3 + \sqrt{8} \big) \big( x + 3 - \sqrt{8} \big) = 0 \qquad \big\{ \text{Use NFL} \big\} math . math \\ . \qquad x + 3 + \sqrt{8} = 0 \quad \text{or} \quad x + 3 - \sqrt{8} = 0 \\. \\ . \qquad x = -3 - \sqrt{8} \qquad \text{or} \qquad x = -3 + \sqrt{8} \\. \\ . \qquad x = -3 - 2\sqrt{2} \quad \;\; \text{or} \qquad x = -3 + 2\sqrt{2} \\. \\ . \qquad x = -3 \pm 2\sqrt{2} math

Special Cases

 * If the quadratic can not be factorised, then the equation has ** no real solutions **.


 * If the quadratic is a perfect square, then the equation has ** one solution **.


 * If the quadratic factorises into two different terms, then the equation has ** two solutions **.

** Example 3 **
Solve the following quadratic equations
 * 1) x² + 2x + 6 = 0
 * 2) x² + 4x + 4 = 0


 * Solution: **

1. Solve x² + 2x + 6 = 0

math \\ . \qquad x^2 + 2x + 6 = 0 \qquad \qquad \big\{ \text{Factorise by completing the square} \big\} \\. \\ . \qquad x^2 + 2x + 1 + 6 - 1 = 0 \\. \\ . \qquad \big( x + 1 \big)^2 + 5 = 0 math

... ... ... This is not a difference of two squares because of the plus sign between the two terms ... ... ... Hence the quadratic can not be factorised

... ... ... Hence the quadratic equation has __**no real solutions**__.

2. Solve x² + 4x + 4 = 0

math \\ . \qquad x^2 + 4x + 4 = 0 \qquad \qquad \big\{ \text{Factorise} \big\} \\. \\ . \qquad \big( x + 2 \big) \big( x + 2 \big) = 0 \\. \\ . \qquad \big( x + 2 \big)^2 = 0 \qquad \qquad \big\{ \text{Take the square root of both sides} \big\} \\. \\ . \qquad x + 2 = 0 \\. \\ . \qquad \;\; x = -2 math

... ... ... The quadratic is a perfect square so the equation has __**one solution**__. .