# Completing the Square

## Problem

Express the equation $y=3x^2-x+1$  in the form $y=a(x-h)^2+k,$  where $a,$  $h,$  and $k$ are constants.

## Solution

You’re asked to convert a quadratic equation into vertex format. The original form — $y=3x^2-x+1$ — makes the y-intercept easy to see: It’s $(0,&space;1).$ $y=a(x-h)^2+k$ is called vertex format; it makes the vertex’s coordinate easy to see: It’s $(h,&space;k).$

## 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:

 ${\color{Blue}&space;3x^2-x+1}$ ${\color{Blue}&space;3x^2-x}$ Ditch the last term. We’ll grab it back later, but now we are working with just the first two terms. ${\color{Blue}&space;3(x^2-x/3)}$ 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. ${\color{Blue}&space;-1/3}$ Find the multiplier of x in the second term. ${\color{Blue}&space;-1/6}$ Divide that number by 2. ${\color{Blue}&space;1/36}$ Square the result ${\color{Blue}&space;x^2-x/3+1/36}$ Add that result to $x^2-x/3$  from $3(x^2-x/3)$  above. ${\color{Blue}&space;(x-1/6)(x-1/6)}$ ${\color{Blue}&space;=(x-1/6)^2}$ That should be a perfect square binomial. Express it that way. ${\color{Blue}&space;3*(1/36)=1/12}$ Multiply $(x-1/6)^2$  by 3. That’s the 3 we factored out – it’s still there; we just stopped writing it while we focused on $x^2-x/3.$ When we added $1/36$  to $x^2-x/3$, we effectively added $3*1/36=1/12$  to  $3(x^2-x/3)$. ${\color{Blue}&space;3x^2-x+1}$ ${\color{Blue}&space;=3(x-1/6)^2+1-1/12}$ ${\color{Blue}&space;=3(x-1/6)^2+11/12}$ We started with $y=3x^2-x+1$ 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.

That’s the answer: ${\bold{\color{Blue}&space;3(x-1/6)^2+11/12}}$.

### Why this works

Consider a perfect square binomial that looks like this: $(x+d)^2=x^2+2dx+d^2.$ By “like this” I mean starting with x, not with a constant. Note that the multiplier of the second term (that’s $2d$), divided by 2 (that’s $d$) and squared (that’s $d^2$), is the third term. That means that a trinomial that starts with $x,$ not a constant, if it’s not a perfect square – that is, if it can’t be expressed as $(x+d)^2$ – 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 $x^2-6x+3$  into a square binomial – something that can be expressed as $(x+d)^2$ – by adding or subtracting some number. What number? Here is how you find it:

• Just for now, ignore that last term. Work with $x^2-6x.$
• 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 $x^2-6x.$ That gives you $x^2-6x+9,$ which should be a perfect square trinomial.
• Rewrite that in square form: $(x-3)^2.$
• By ignoring the last term, you subtract 3. After that you add 9. In total you add $-3+9=6.$ You need something that’s equal to what you started with, so subtract 6 from the perfect square you got: $x^2-6x+3&space;=&space;(x-3)^2-6.$

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 $x^2.$

Yes, that’s right. But you can factor that number out so that you can still solve the problem as if it started with $x^2.$

$3x^2-x+1=3(x^2-x/3)$ plus something we are temporarily ignoring.

And this is what we did above, where we converted $3x^2-x+1$ into $3(x-1/6)^2+11/12.$