In other activities, we have looked at a number of methods for dealing with quadratics. We have learnt to factorise them, how to solve them using the formula and how to use a graph to solve them.

Another way of manipulating quadratics is called **completing the square**.

**The perfect square**

Perfect squares are quadratics that can be written as two brackets that are the same, for example:

x^{2} + 10x + 25 = (x + 5)(x + 5) = (x + 5)^{2}

x^{2} + 12x + 36 = (x + 6)(x + 6) = (x + 6)^{2}

x^{2} - 4x + 4 = (x - 2)(x - 2) = (x - 2)^{2}

However, this is only possible in a very small number of specific quadratics.

**How to complete the square**

When we are dealing with a quadratic that is **not **a perfect square, we have to modify our method by putting it into the form **(x ± a) ^{2} ± b**

**Example:**

Complete the square for x^{2} + 6x - 2

Give your answer in the form (x ± a)^{2} ± b

**Step 1: Find the value of a**

This is the easy part - all we have to do is halve the coefficient of x.

For our question, we have +6x, this means that **a is 3.**

**(x + 3) ^{2} ± b**

**Step 2: Find the value of b**

This takes a bit more thought.

If we look at the expression we have, you will notice that we have a perfect square at the start which we can expand:

(x + 3)^{2 }= (x + 3)(x + 3) = x^{2} + 6x + 9

You will notice that the only difference between this perfect square and our original question is that the last number is -2.

The question we need to ask is how do we get from the **9** we have, to the **-2** we want.

We have to subtract 11.

This means that **b is -11.**

We can now write our quadratic as:

**x ^{2} + 6x – 2 = (x + 3)^{2} - 11**

Let's have a go at some questions now.