# Solve Quadratic Equations Using Factorisation

In this worksheet, students will learn how to solve a quadratic equation of the form x² + bx + c = 0 by finding common factors and factorising.

Key stage:  KS 4

GCSE Subjects:   Maths

GCSE Boards:   Pearson Edexcel, OCR, Eduqas, AQA

Curriculum topic:   Algebra

Curriculum subtopic:   Solving Equations and Inequalities, Algebraic Equations

Difficulty level:

### QUESTION 1 of 10

In this activity, we will be looking at quadratic equations, which cannot be solved using the balancing method.

For these quadratic equations, there are several methods which can be used.

However, in this activity, we will focus on the method of factorisation.

How we factorise a quadratic expression is covered in another activity.

If you haven't already completed this or are not feeling totally confident with this concept, it would be best to try this before continuing with this activity.

e.g. x² - 3x - 18 = 0

The equation x² - 3x - 18 = 0 can be factorised into two brackets as: (x - 6)(x + 3) = 0

Remember, the numbers in the brackets must multiply to make -18 and add to make -3

Our brackets can be multiplied together to give the answer 0.

If I tell you that I have multiplied together two numbers and the answer is zero, can you tell me anything about my two numbers?

Well, you should be able to tell me that at least one of my numbers is zero itself.

Going back to my equation, that means that either (x - 6) = 0 or (x + 3) = 0 (They cannot both be zero.)

But if x - 6 = 0 then x must be 6 so x = 6 is a solution.

Alternatively, if x + 3 = 0 then x = -3 is another solution.

So there are two possible solutions to this quadratic equation: x = 6 and x = -3

It is a feature of quadratic equations that they nearly always have two solutions.

Here's another example, this time we'll just include the lines of working you would need to show:

e.g. n² - 10n + 25 = 0

(n - 5)(n - 5) = 0

Either n - 5 = 0 or n - 5 = 0

So n = 5 or n = 5

Now in this case, because both sets of brackets are the same, both solutions are the same, so really there is just one solution: n = 5

This is unusual though; usually there will be two different sets of brackets with two different solutions.

Here's one more example, again slightly different:

e.g. 2y² - 4y = 0

This equation will not factorise into two brackets, but we can take out a common factor of 2y from each term on the left:

2y(y - 2) = 0

Either 2y = 0 or y - 2 = 0

So y = 0 or y = 2

Ok, now you should be ready to try some questions.

In this activity, we will solve quadratic equations in the form x² + bx + c = 0 by finding common factors and factorising into two brackets.

x² - 7x + 12 = 0

Below you will see the steps required to solve this equation.

Put these steps into the correct order by matching each to its correct position.

## Column B

1st
Either x - 3 = 0 or x - 4 = 0
2nd
So x = 3 or x = 4
3rd
x² - 7x + 12 = 0
4th
(x - 3)(x -4) = 0

x² + 9x + 20 = 0

The steps for solving this equation are shown below.

Fill in the spaces in each line to complete this process accurately.

## Column B

1st
Either x - 3 = 0 or x - 4 = 0
2nd
So x = 3 or x = 4
3rd
x² - 7x + 12 = 0
4th
(x - 3)(x -4) = 0

Solve this equation:

m² - 5m - 6 = 0

Choose the correct pair of solutions from the options below.

m = 3 or m = -2

m = -3 or m = 2

m = 6 or m = -1

m = -6 or m = 1

Consider this equation:

y² + 9y + 18 = 0

Select the two correct solutions to this equation from the options below.

y = -9

y = -2

y = 3

y = -3

y = 6

y = -6

Look at this equation:

x² - 7x = 0

Which of the solutions below work for this equation?

x = 0

x = 3

x = 7

x = 4

x = -7

p² - 6p + 8 = 0

The steps for solving this equation are shown below.

Fill in the spaces in each line to complete this process accurately.

x = 0

x = 3

x = 7

x = 4

x = -7

The following quadratic equation only has one solution:

a² - 14a + 49 = 0

Choose the solution from the options listed below.

a = -7

a = 7

a = 14

a = -14

One or more of the quadratic equations below has a possible solution of x = 2.

Which equations from the list work with this solution?

x² - 4x + 4 = 0

x² + 4x + 4 = 0

x² + 9x - 22 = 0

x² - 21x + 38

x² - 2x + 2 = 0

Solve this equation:

m² + 7m - 30 = 0

Type your solutions in the spaces below.

x² - 4x + 4 = 0

x² + 4x + 4 = 0

x² + 9x - 22 = 0

x² - 21x + 38

x² - 2x + 2 = 0

Can you solve the following equation?

x² + 6x - 91 = 0

x² - 4x + 4 = 0

x² + 4x + 4 = 0

x² + 9x - 22 = 0

x² - 21x + 38

x² - 2x + 2 = 0

• Question 1

x² - 7x + 12 = 0

Below you will see the steps required to solve this equation.

Put these steps into the correct order by matching each to its correct position.

## Column B

1st
x² - 7x + 12 = 0
2nd
(x - 3)(x -4) = 0
3rd
Either x - 3 = 0 or x - 4 = 0
4th
So x = 3 or x = 4
EDDIE SAYS
1) We need to start with our equation in the form ax² + bx + c = 0 : x² - 7x + 12 = 0 2) We need to factorise this into two brackets: (x - 3)(x - 4) = 0 3) Then we need to equate each bracket to zero: Either x - 3 = 0 or x - 4 = 0 4) Finally, we need to write down the possible values of x: So x = 3 or x = 4 Did you get those steps in the correct order?
• Question 2

x² + 9x + 20 = 0

The steps for solving this equation are shown below.

Fill in the spaces in each line to complete this process accurately.

EDDIE SAYS
These are the steps we need to follow to solve this equation: x² + 9x + 20 = 0 (x + 5)(x + 4) = 0 Either x + 5 = 0 or x + 4 = 0 So x = -5 or x = -4 Did you get those all correct?
• Question 3

Solve this equation:

m² - 5m - 6 = 0

Choose the correct pair of solutions from the options below.

m = 6 or m = -1
EDDIE SAYS
We're getting on a roll now! Here are the steps for solving this equation: m² -5m - 6 = 0 (m - 6)(m + 1) = 0 Either m - 6 = 0 or m + 1 = 0 So m = 6 or m = -1 It is easy to make the mistake of using -3 and 2 or 3 and -2 here because they do multiply to make -6 too. However, when we add these options, they come to -1 or +1 respectively, not -5m which we see in our original equation.
• Question 4

Consider this equation:

y² + 9y + 18 = 0

Select the two correct solutions to this equation from the options below.

y = -3
y = -6
EDDIE SAYS
Did you spot that we needed to select two options from this list? Here are the steps for solving this equation: y² + 9y + 18 = 0 (y + 3)(y + 6) = 0 Either y + 3 = 0 or y + 6 = 0 So y = -3 or y = -6 How are you doing with these? Are you getting the hang of them?
• Question 5

Look at this equation:

x² - 7x = 0

Which of the solutions below work for this equation?

x = 0
x = 7
EDDIE SAYS
Did you spot that this equation factorises into a single bracket? Here are our steps for solving this equation: x² - 7x = 0 x(x - 7) = 0 Either x = 0 or x - 7 = 0 So x = 0 or x = 7 How did you do with this one?
• Question 6

p² - 6p + 8 = 0

The steps for solving this equation are shown below.

Fill in the spaces in each line to complete this process accurately.

EDDIE SAYS
Here is the full working for this one: p² - 8p - 20 = 0 (p - 10)(p + 2) Either p - 10 = 0 or p + 2 = 0 So p = 10 or p = -2 Did you fill the gaps in the method described accurately and in the right order?
• Question 7

The following quadratic equation only has one solution:

a² - 14a + 49 = 0

Choose the solution from the options listed below.

a = 7
EDDIE SAYS
Here is our process: a² - 14a + 49 = 0 (a - 7)(a - 7) = 0 Either a - 7 = 0 or a - 7 = 0 As both the brackets use the same value of a, there is only one solution: So a = 7
• Question 8

One or more of the quadratic equations below has a possible solution of x = 2.

Which equations from the list work with this solution?

x² - 4x + 4 = 0
x² + 9x - 22 = 0
x² - 21x + 38
EDDIE SAYS
We can check each equation by substituting x = 2 into them to see if we reach an answer of 0: x² - 4x + 4 = 0 → 2² - (4 × 2) + 4 = 4 - 8 + 4 = 0 (correct) x² + 4x + 4 = 0 → 2² + (4 × 2) + 4 = 4 + 8 + 4 = 16 (incorrect) x² + 9x - 22 = 0 → 2² + (9 × 2) - 22 = 4 + 18 - 22 = 0 (correct) x² - 21x + 38 → 2² - (21 × 2) + 38 = 4 - 42 + 38 = 0 (correct) x² - 2x + 2 = 0 → 2² - (2 × 2) + 2 = 4 - 4 + 2 = 2 (incorrect) So there are 3 correct equations listed where x = 2 is a viable solution. Did you find all 3 of these?
• Question 9

Solve this equation:

m² + 7m - 30 = 0

Type your solutions in the spaces below.

EDDIE SAYS
Here's the full solution to this equation: m² + 7m -30 = 0 (m + 10)(m -3) = 0 Either m + 10 = 0 or m - 3 = 0 So m = -10 or m = 3 Just one more question left to complete now!
• Question 10

Can you solve the following equation?

x² + 6x - 91 = 0

EDDIE SAYS
This was a tricky one to end with! There are some larger numbers here, but it helps to know that: 91 = 13 × 7 Here's our solution: x² + 6x - 91 = 0 (x - 7)(x + 13) = 0 Either x - 7 = 0 or x + 13 = 0 So x = 7 or -13 Great work - you're all done! Why not try an activity on quadratic equations at the highest level if you are looking for an extra challenge?
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