# 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,

• $$x^2+8x$$ becomes $$(x+4)^2-16$$.
• $$x^2+14x+20$$ becomes $$(x+7)^2-29$$.
• $$x^2+9x+11$$ becomes $$(x+\frac{9}{2})^2 -\frac{37}{4}$$.

## 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:

• $$(x-a)^2+b$$
• $$a(x-b)^2+c$$
• $$(ax-b)^2+c$$

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:

$$5[(x-\frac{1}{10})^2-(\frac{1}{10})^2]+14$$

$$5[(x-\frac{1}{10})^2-\frac{1}{100}]+14$$

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.

$$5(x-\frac{1}{10})^2-\frac{1}{20}+14$$

$$5(x-\frac{1}{10})^2+\frac{279}{20}$$

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.

$$9(x^2-2x)+20$$

$$9[(x-1)^2-1]+20$$

$$9(x-1)^2-9+20$$

$$9(x-1)^2+11$$

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!