In this activity, we will be looking at quadratic equations, which **cannot** be solved by the **'balancing**' method. For example:

x² - 3x - 18 = 0

For these quadratic equations, there are several methods which can be used, but in this activity, we will use the method of **factorisation**. How to factorise a quadratic expression is covered in another activity, and here we assume you have completed that activity, but if you haven't it would be best if you do that before continuing here.

So, the equation **x****² - 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**. Now, remember that each set of brackets stand for a number and these are multiplied together to give the answer zero. So, 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 and this 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 I'll just include the lines of working you would need to show.

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**. But this is unusual, usually there will be two different sets of brackets with two different solutions.

Here's one more example, again slightly different.

2y² - 4y = 0

This one 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 the questions.