Completing the Square
VIDEO COMING SOON…
In this post we will be going over ‘Completing the Square’.
This is used as an additional way to solve quadratic equations alongside the Quadratic Formula and Factorising Quadratics.
Completing the Square is grade 8 (or between A and A*) so you’d assume it’s going to be difficult but if you remember these key steps, you’ll find it really easy.
When \(a = 1\)
I’ll briefly explain how it’s done then we’ll get into a quick example.
- Get your formula into the normal quadratic equation format – \(ax^2+bx+c\)
- divide \(b\) by 2 and put it into a bracket with \(ax^2\) but move the squared sign on the \(ax^2\) to outside the brackets and remove the \(x\) from our \(b\) – it should look like this —> \((ax+\frac{1}{2}b)^2\).
- We now need to deduct \(\frac{1}{2}b^2\), this is because if we were to expand out the \((x+\frac{1}{2}b)^2\), we would have this \(\frac{1}{2}b^2\) that needs to be deducted in order to get back to our \(ax^2+bx+c\).
- So really, our ‘Complete the Square’ formula would look something like \((ax+\frac{1}{2}b)^2-\frac{1}{2}b^2 +c\).
This looks quite complicated on paper but after a few practices, you’ll pick up the steps easily and it will be a breeze.
Let’s go through an example:
\(x^2+6x+2=0\)
Here you can see that \(a=1, b=6, c=2\)
Now we half our \(b\) and put it in brackets with our \(ax^2\) but moving the squared sign to outside of the brackets and removing the \(x\) from our \(b: (x+3)^2\) —> this is our square.
Finally, we minus the \(\frac{1}{2}b^2\) from our equation to ‘complete’ our square. This will now leave us with \((x+3)^2 – 3^2 +2=0\)
After collecting our terms, our final answer is \((x+3)^2-7\).
If you want some comfort, we can expand out the brackets to get back to our \(x^2+6x+2\).
When \(a\) does not \(=1\)
The only difference when our \(a\) doesn’t equal 1 is that we just need to try to factor out our \(a\) so that we can be left with a single \(x^2\).
One thing to watch out for here is that you multiply everything at the end by the number you factored out.
Let’s go through an example:
\(3x^2+18-1\)
Here our \(a = 3\) so the first point of action is to get ride of this by factoring it out. It would look something like this: \(3(x^2+6)-1\)
From our first example above, we know that when we have \(x^2+6x\), the completed square is \((x+3)^2-3^2\) (highlighted above) so let’s sub this into our formula in place of the \((x^2+6x)\).
Now we’ve got an equation that looks a little something like this —> \(3((x+3)^2-9)-1\)
The final step of our question is to expand our bracket by multiplying everything by \(3\): \(3(x+3)^2-27-1\) which when you collect the terms is \(3(x+3)^2-28\).
A key thing to remember is that you need to keep that -9 on the inside of the bracket that is to be multiplied by 3 as this is part of the equation that factored out the 3.
Many people forget this and as a result end up with the wrong answer.
As with most Algebra, to double-check your answer, just expand out your brackets and see if you get back to \(3x^2-18x-1\).
Hopefully, this helped – if you feel confident, go ahead and try some practice questions below and let me know how you get on. If you’ve got any questions, please leave them in the comments and I’ll get back to you as soon as possible.