Completing the Square (faster)

This seems so obvious in hindsight and I have no idea why this never occurred to me.

We have \(x^2+6x\). This becomes \((x+3)^2-(3)^2\). We just put half of \(x\)'s coefficient into the brackets with \(x\) and subtract the square of the rest.

Using this,

If \(x^2\)'s coefficient is not one...

This depends on whether you're dealing with an equation or an expression.

Anyway, it's important to note the main forms of our completed square equation/expression:

I'll show you how to get to the latter two forms.

Let's take \(5x^2-x+14\). What do we need to do to get \(5x^2\) to \(x^2\)? We take 5 as common. You could do this for the entire expression, but it's easier to manage if you do this just for the first two terms.

This gets us \(5(x^2-\frac{1}{5}x)+14\).

Now we complete the square inside the brackets, with the same method as above:



Now we use the distributive property to open up the square brackets, so that we can simplify this expression. Note that we do not distribute the common factor into the \(x^2\) thing.



It's a weird fraction at the end, but I hope it gets the point across. This gets us to \(a(x-b)^2+c\).

Let's take one more example. I'll show you how we get to \((ax-b)^2+c\).

We have \(9x^2-18x+20\). Let's complete the square, as we did before.





Now, let's look at the 9. This is a square of 3. Let's also review this property: \(a^n \times b^n = (ab)^n\).

What we have here is a similar situation: \(3^2(x-1)^2+11\). Both have the same power, so we can distribute 3 into the brackets. Our final form would be \((3x-3)^2+11\).

We now have the form of \((ax-b)^2+c\).

Here, we have been dealing with expressions. If you have an equation that is = to 0, then you can simply divide the entire thing by \(x^2\)'s coefficient.

This article was written on 20/09/2023 and modified the following day. If you have any thoughts, feel free to send me an email with them. Have a nice day!