﻿ Unit 17 Section 2 : Quadratic Equations: Factorisation

# Unit 17 Section 2 : Quadratic Equations: Factorisation

In this section we look at factorisation and how this can be used to solve quadratic equations. In Unit 11 you factorised expressions; we now take this one stage further to solve equations. In Unit 11 you looked at factorizing expressions with common factors. We now develop this to solving equations with common factors.

## Example 1

Factorise:

(a)

12x + 8

12x + 8 = 4(3x + 2)

(b)

x^3 + x^2

x^3 + x^2 = x^2 (x + 1)

(c)

3x^2 + 15x

3x^2 + 15x = 3x(x + 5)

## Example 2

Factorise:

As both expressions contain x^2, they will factorise in the form:

(x ± )(x ± )

We must determine the missing numbers, and whether a ' +' or a '–' sign is required in each bracket.

(a)

x^2 + 6x + 8

For x^2 + 6x + 8 we require two numbers that multiply together to give 8 and add together to give 6.

So the numbers are 2 and 4.

Hence x^2 + 6x + 8 = (x + 2)(x + 4).

(b)

x^2 - 5x + 6

For x^2 - 5x + 6 we require two numbers that multiply together to give 6 and add together to give –5.

So the numbers are –2 and –3.

Hence x^2 - 5x + 6 = (x - 2)(x - 3).

## Example 3

Use factorisation to solve the following equations:

(a)

x^2 + 6x = 0

x^2 + 6x = 0

x(x + 6) = 0

For the left-hand-side to be zero, either:

x = 0   or   x + 6 = 0

The second equation has solution x = -6, so the equation x^2 + 6x = 0 has solution x = 0   or   x = -6.

(b)

x^2 + 3x + 2 = 0

First factorise the left-hand-side of the equation:

x^2 + 3x + 2 = 0

(x + 2)(x + 1) = 0

x + 2 = 0   or   x + 1 = 0

so the equation x^2 + 3x + 2 = 0 has solution

x = -2   or   x = -1

(c)

x^2 - 8x + 16 = 0

Factorise the left-hand-side of the equation:

x^2 - 8x + 16 = 0

(x - 4)(x - 4) = 0

(x - 4)^2 = 0

So x - 4 = 0

which means that the equation x^2 - 8x + 16 = 0 has just one solution, namely

x = 4

Note that in this example the equation has only one solution.

## Example 4

Solve the following equations:

(a)

2x^2 = 3x

First, we rearrange the equation so that it has 0 on the right-hand-side.

This first step is essential.

2x^2 = 3x

2x^2 - 3x = 0 Subtracting 3x from both sides

Now we factorise the left-hand-side and solve the equation as we did in Example 2.

x(2x - 3) = 0

so   x = 0   or   2x - 3 = 0.

The second equation has solution x = 1.

Therefore the equation 2x^2 = 3x has solution x = 0   or   x = 1.

(b)

x^2 = 7x - 6

Again, the first step is to rearrange the equation so that it has 0 on the righthand-side.

x^2 = 7x - 6

x^2 - 7x + 6 = 0 Subtracting 7x from both sides and adding 6 to both sides

Now we factorise the left-hand-side and solve the equation.

(x - 1)(x - 6) = 0

so   x - 1 = 0   or   x - 6 = 0.

Therefore the equation x^2 = 7x - 6 has solution x = 1   or   x = 6.

## Exercises

Note: To type indices on this page use ^ sign. e.g.   n2:
Question 1

Factorise the following:

(a) 3x + 21 5x - 20 x^2 - x x^2 + 6x x^3 - x^2 8x + 20x^2 4x - 30x^2 5x + 16x^2 x^4 + x^2
Question 2

Factorise the following:

(a) x^2 + 4x + 3 x^2 - 3x + 2 x^2 - 5x - 14 x^2 - 21 + 20x x^2 + 12x + 35 x^2 - 10x + 25 x^2 - 11x + 30 x^2 - 2x - 63 x^2 - 14x + 48
Question 3

Solve the following equations:

(a) (b) x^2 - 4x = 0 x =   or   x = x^2 + 3x = 0 x =   or   x = x^2 - 7x = 0 x =   or   x = x - 4x^2 = 0 x =   or   x = 7x - 3x^2 = 0 x =   or   x = 2x^2 - 5x = 0 x =   or   x =
Question 4

Solve the following equations:

(a) (b) x^2 - 8x + 12 = 0 x =   or   x = x^2 + 2x - 8 = 0 x =   or   x = x^2 + x - 6 = 0 x =   or   x = x^2 +3x - 4 = 0 x =   or   x = x^2 - 8x + 15 = 0 x =   or   x = x^2 - 11x + 18 = 0 x =   or   x = x^2 - 6x - 27 = 0 x =   or   x = x^2 + 10x + 21 = 0 x =   or   x = x^2 - 16x - 17 = 0 x =   or   x = x^2 + 17x + 60 = 0 x =   or   x =
Question 5

Solve the following equations:

(a) (b) x^2 = 8x x =   or   x = 3x^2 = 4x x =   or   x = x^2 + 5x = 50 x =   or   x = x^2 + 70 = 17x x =   or   x = x^2 + x = 56 x =   or   x = x^2 = 14x + 51 x =   or   x =
Question 6

Use the following graph to tell the number of the solutions of the equation x^2 + 2x - 3 = 0.

The equation has solutions.

Graph crosses x-axis at two points, which are the x-values that give y = 0.

Question 7

The following graph shows the curve with equation y = x^2 + 2x + 1

(a)

How many solutions will the equation x^2 + 2x + 1 = 0 have?

The equation has solutions.
(b)

Check your answer to (a) by factorising x^2 + 2x + 1.

x^2 + 2x + 1 =
Question 8

Use the following graph to tell the number of the solutions of the equation x^2 + 2x + 2 = 0.

Number of solutions =
Question 9

A rectangle is 3 cm longer than it is wide. The width of the rectangle is w cm and the area is 10 cm² .

(a)

The area of the rectangle: = 10

(b)

Rewrite the equation in part (a) in the form ax^2 + bx + c = 0

(c)

Solve the equation in part (b).

x =   or   x =
(d)

State the dimensions of the rectangle.

cm by cm
Question 10

The surface area of this cuboid is 18 cm².
Determine the volume of the cuboid.

Volume = cm³
Question 11

Solve for y,

= y + 2

y =   or   y =