# Completing the Square

## Problem

Express the equation   in the form   where     and are constants. ## Solution

You’re asked to convert a quadratic equation into vertex format. The original form —  — makes the y-intercept easy to see: It’s   is called vertex format; it makes the vertex’s coordinate easy to see: It’s

## How do you convert it?

By a method called “completing the square.” It’s called that (I think) because you start with something that does not obviously contain a squared binomial and you end with something that does.

These are the steps:

 Ditch the last term. We’ll grab it back later, but now we are working with just the first two terms. Factor out the number that multiplies the first term. Yes, in this case that leaves you with a fraction in the second term. It’s just something you have to deal with. Find the multiplier of x in the second term. Divide that number by 2. Square the result Add that result to   from   above. That should be a perfect square binomial. Express it that way. Multiply   by 3. That’s the 3 we factored out – it’s still there; we just stopped writing it while we focused on  When we added   to , we effectively added   to  . We started with  and added 1/12 to create a perfect square. In adding 1/12, we changed the value of the expression we were given. To change it back to what it was, we need to subtract 1/12. ### Why this works

Consider a perfect square binomial that looks like this: By “like this” I mean starting with x, not with a constant. Note that the multiplier of the second term (that’s ), divided by 2 (that’s ) and squared (that’s ), is the third term. That means that a trinomial that starts with not a constant, if it’s not a perfect square – that is, if it can’t be expressed as – can be made into a perfect square just by the addition or subtraction of a constant equal to the second term’s multiplier divided by 2 and squared.

For example, you can make   into a square binomial – something that can be expressed as – by adding or subtracting some number. What number? Here is how you find it:

• Just for now, ignore that last term. Work with
• The second term is -6x. The multiplier of x is -6.
• Divide that number, -6, by 2. That’s -3.
• Square the result: 9.
• Add that 9 to That gives you which should be a perfect square trinomial.
• Rewrite that in square form:
• By ignoring the last term, you subtract 3. After that you add 9. In total you add You need something that’s equal to what you started with, so subtract 6 from the perfect square you got:

There. That’s how you compile a square.

Oh, Jill, you’re making it too simple. I need to be able to do problems where a number is multiplying

Yes, that’s right. But you can factor that number out so that you can still solve the problem as if it started with

plus something we are temporarily ignoring.

And this is what we did above, where we converted into