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

First factorise the quadratic expression

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
(b)5x - 20
(c)x^2 - x
(d)x^2 + 6x
(e)x^3 - x^2
(f)8x + 20x^2
(g)4x - 30x^2
(h)5x + 16x^2
(i)x^4 + x^2
Question 2

Factorise the following:

(a)x^2 + 4x + 3
(b)x^2 - 3x + 2
(c)x^2 - 5x - 14
(d)x^2 - 21 + 20x
(e)x^2 + 12x + 35
(f)x^2 - 10x + 25
(g)x^2 - 11x + 30
(h)x^2 - 2x - 63
(i)x^2 - 14x + 48
Question 3

Solve the following equations:

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

Solve the following equations:

(a)x^2 - 8x + 12 = 0 x =   or   x =
(b)x^2 + 2x - 8 = 0 x =   or   x =
(c)x^2 + x - 6 = 0 x =   or   x =
(d)x^2 +3x - 4 = 0 x =   or   x =
(e)x^2 - 8x + 15 = 0 x =   or   x =
(f)x^2 - 11x + 18 = 0 x =   or   x =
(g)x^2 - 6x - 27 = 0 x =   or   x =
(h)x^2 + 10x + 21 = 0 x =   or   x =
(i)x^2 - 16x - 17 = 0 x =   or   x =
(j)x^2 + 17x + 60 = 0 x =   or   x =
Question 5

Solve the following equations:

(a)x^2 = 8x x =   or   x =
(b)3x^2 = 4x x =   or   x =
(c)x^2 + 5x = 50 x =   or   x =
(d)x^2 + 70 = 17x x =   or   x =
(e)x^2 + x = 56 x =   or   x =
(f)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 =