Quadratic equations have one variable and the highest exponent is 2.

Quadratic equations can have zero, one or two solutions.

There are four methods for solving quadratic equations.

### By factorising.

Rearrange the equation so that all of the terms are on one side of the equation. Then factorise.

The method depends on the fact that if two factors multiplied together are equal to zero, then either one or both of them must be equal to zero.

 Examples Answers Solve: (a) x2 − 4x − 5 = 0 Factorise to (x − 5)(x + 1) = 0 Either: x − 5 = 0 or x + 1 = 0 x = 5 or x = -1 (b) (2x − 3)(x + 4) = 0 Already factorised Either: 2x − 3 = 0 or x + 4 = 0 2x = 3 or x = − 4 x = 1.5 (c) 3x2 − 6x = 0 Factorise to 3x(x − 2) = 0 Either: 3x = 0 or x − 2 = 0 x = 0 or x = 2 (d) (x − 4)2 = 0 Already factorised (x − 4)(x − 4) = 0 Either: x − 4 = 0 or x − 4 = 0 x = 4

### By taking the square root.

Take the square root of both sides of the equation.

Remember, if x2 = a then

 Examples Answers Solve x2 = 36 Take the square root of both sides. x = ± 6 Solve (x − 4)2 = 9 Take the square root of both sides. x − 4 = ± 3 x = 7 or 1

### By using the Quadratic Formula

When a quadratic equation cannot be factorised at all e.g. x2 + 6x − 1 = 0 or cannot be factorised easily e.g.
then a formula called the Quadratic Formula (proof) can be used:

 For the equation ax2 + bx + c = 0 The Quadratic Formula

 Example 1 Example 2 Solve x2 + 6x − 1 = 0 a = 1b = 6c = -1 (to 4 decimal places) Solve 24x2 + 95x + 50 = 0 a = 24b = 95c = 50

### By graphing.

The solutions for the equation are the x-intercepts of the graph of the equivalent quadratic function which is a parabola.
See
Topic 20

### Word Problems involving Quadratic Equations

There are a range of word problems involving quadratic equations. The basic method is to:

• read the question carefully and assign a variable to the amount that is being found.
• write the solution to the problem

Example

Francis is 10 years older than Michael. The product of their ages is 600. What are their ages?

• Let Michael's age be x
• The equation is x(x + 10) = 600
• Solve the equation:

x(x + 10) = 600

x2 + 10x = 600

x2 + 10x − 600 = 0

(x + 30)(x − 20) = 0

x = -30      OR      x = 20

• Michael cannot be -30 so he must be 20 and Francis is 10 years older at 30.