## What you need to know

Completing the square is a method of changing the way that a quadratic is expressed. There are two reasons we might want to do this, and they are

• To help us solve the quadratic equation,
• To find the coordinates of the minimum (or maximum) point of the quadratic graph.

We will be going into number 1 shortly, but to see how to do number 2, click here (https://mathsmadeeasy.co.uk/gcse-maths-revision/turning-points-quadratic-graphs/). Before any of this however, we need to answer the question: what does “completing the square” mean? Well, it involves taking a quadratic, say

$x^2 + 4x - 12$,

and expressing it in the form

$(x + a)^2 + b$

where $a$ and $b$ are constants to be determined. Let’s work through this as an example to see how it’s done.

Example: Write $x^2+4x-12$ in the form $(x+a)^2+b$ where $a$ and $b$ are constants to be determined.

The way we find the two missing constants is as follows. Firstly, the $a$ is found simply by halving the coefficient of $x$, so in this case $a=4\div 2=2$. Then, to find $b$, we take the constant at the end of the quadratic, -12, and subtract $a^2$ from it, so we get $b = -12 - 2^2 = -16$. Therefore, the result of completing the square is

$x^2 + 4x - 12 = (x + 2)^2 - 16$

You might be wondering what’s actually going on here and how we know these two quadratics are actually equal and, well, it’s easiest to see if we try expanding the result and seeing what happens.

$(x + 2)^2 - 16 = x^2 + 2x + 2x + 4 - 16 = x^2 + 4x - 12$

When determining the value of $a$, we had to halve the coefficient of $x$ to compensate for the double appearance of $2x$ when we expand it. On top of this, we can also see that when we expand the bracket we end up with an extra 4 (or $2^2$) at the end, which is why we subtracted 4 from the -12 when determining $b$.

So, this is how you complete the square: halve the coefficient of $x$ to find the number that goes inside the bracket, and then subtract the square of it from the constant term in you quadratic to work out what goes on the end. For the algebra-loving people amongst us, you can think of it like this:

$x^2 + cx + d = \left(x + \dfrac{c}{2}\right)^2 + d - \left(\dfrac{c}{2}\right)^2$

Before getting into how to use this to solve a quadratic equation, we should look at completing the square on a trickier kind of quadratic.

Example: Write $2y^2-16y+6$ in the form $a(x+b)^2+c$, where $a, b,$ and $c$ are constants to be determined.

Firstly, notice how the form that the question asks for is different – there’s now a constant in front of the bracket, and is because of the 2 in front of the $y^2$. To complete the square on an expression like this, we need to start by taking a factor of 2 out of the whole expression. Doing so, we get

$2y^2-16y+6=2(y^2-8y+3)$

Now, all we need to do is complete the square on the bit inside the bracket (as we normally would) and then bring the two back in at the very end. So, the coefficient of $y$ is -8, half of which is -4, therefore we get

$y^2-8y+3 = (y-4)^2+3-(-4)^2=(y-4)^2-13$

Remember, this is just the inside of the bracket, so if we bring the 2 back into the picture, we get

$2\left[(y-4)^2-13\right]$

Finally, expanding the square brackets (and only those), we get the final form of the expression to be

$2(y-4)^2 - 26$

This is precisely the form the question asked for. So, when there’s a number in front of the squared term, it takes that extra step at the beginning and that extra step at the end, but the middle part is still the same process of completing the square.

Using Completing the Square to Solve a Quadratic Equation

To see how this process, let’s jump right into an example.

Example: Use completing the square to solve $x^2 + 4x + 3 = 0$.

The coefficient of $x$ is 4, and half of 4 is 2, so using the completing the square method our equation becomes

$(x + 2)^2 + 3 - 2^2 = (x + 2)^2 - 1 = 0$

Now, to solve this equation for $x$ we will rearrange our equation – not expanding the brackets – to make $x$ the subject. Firstly, add 1 to both sides to get

$(x + 2)^2 = 1$

Then, square root both sides to get

$x + 2 = \pm\sqrt{1} = \pm 1$

Finally, subtracting 2 from both sides makes $x$ the subject and leaves us with

$x = \pm 1 - 2$.

Therefore, the two solutions to our quadratic are $x = -1$ and $x = -3$. Make sure you remember to include both the positive and negative square roots, or you will end up missing one of the solutions.

## Example Questions

The coefficient of $m$ is 5, and half of 5 is $\frac{5}{2}$, so we get

$m^2+5m+6=\left(m+\dfrac{5}{2}\right)^2 + 6 - \left(\dfrac{5}{2}\right)^2$

Now, the bit outside the bracket is more complicated than usual, so we’ll look at that on its own. Firstly, we get

$6-\left(\dfrac{5}{2}\right)^2 = 6 - \dfrac{25}{4}$

Then, by writing 6 as $\frac{6}{1}$ and finding a common denominator, we get

$6-\dfrac{25}{4}=\dfrac{6}{1}-\dfrac{25}{4}=\dfrac{24}{4}-\dfrac{25}{4}=-\dfrac{1}{4}$

Now we know the constant that goes outside the bracket, the final result of completing the square is

$\left(m+\dfrac{5}{2}\right)^2-\dfrac{1}{4}$

In order to be able to apply our normal process of completing the square, we need to take a factor of 2 out of this whole expression:

$2x^2 - 8x + 10 = 2(x^2 - 4x + 5)$

Now the bit inside the bracket looks much more like something we can complete square on. The coefficient of $x$ is -4, half of which is -2, so we get:

$2(x^2 - 4x + 5) = 2\left[(x - 2)^2 + 5 - (-2)^2\right]) = 2\left[(x - 2)^2 + 1\right]$

Now, if we expand the outside bracket (but not the inside one, careful now), we get

$2\left[(x -2)^2 + 1\right] = 2(x - 2)^2 + 2$,

which is in the form asked for in the question.

The coefficient of $z$ is 14, half of which is 7, so we get:

$z^2 + 14z - 1 = (z + 7)^2 - 1 - 7^2 = (z + 7)^2 - 50$

Meaning our equation is now

$(z + 7)^2 - 50 = 0$

Now we must rearrange this equation (without expanding the bracket) to make $z$ the subject

$(z + 7)^2 = 50$

$z + 7 = \pm \sqrt{50}$

$z = -7 \pm \sqrt{50}$

So, the two solutions are: $x = -7 + \sqrt{50}$ and $x = -7 - \sqrt{50}$.

## Completing the Square Revision and Worksheets

Completing the Square
Level 7-8
Completing the Square (2)
Level 8-9
Completing the Square (3)
Level 8-9
Completing the Square Hard
Level 8-9

## Completing the Square Teaching Resources

Completing the square revision materials and resources can be accessed via this page. You may be a York Maths Tutor or you may be a maths teacher in London, wherever you are, whether you are teaching or tutoring Maths, the completing he square resources will be a useful addition to your bank of GCSE Maths questions that you use with your pupils.