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 $(0, 1).$ $y=a(x-h)^2+k$ is called vertex format; it makes the vertex’s coordinate easy to see: It’s $(h, k).$

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:

${\color{Blue} 3x^2-x+1}$
${\color{Blue} 3x^2-x}$	Ditch the last term. We’ll grab it back later, but now we are working with just the first two terms.
${\color{Blue} 3(x^2-x/3)}$	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.
${\color{Blue} -1/3}$	Find the multiplier of x in the second term.
${\color{Blue} -1/6}$	Divide that number by 2.
${\color{Blue} 1/36}$	Square the result
${\color{Blue} x^2-x/3+1/36}$	Add that result to $x^2-x/3$ from $3(x^2-x/3)$ above.
${\color{Blue} (x-1/6)(x-1/6)}$ ${\color{Blue} =(x-1/6)^2}$	That should be a perfect square binomial. Express it that way.
${\color{Blue} 3*(1/36)=1/12}$	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 $3(x^2-x/3)$ .
${\color{Blue} 3x^2-x+1}$ ${\color{Blue} =3(x-1/6)^2+1-1/12}$ ${\color{Blue} =3(x-1/6)^2+11/12}$	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: ${\bold{\color{Blue} 3(x-1/6)^2+11/12}}$ .

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: $x^2-6x+3 = (x-3)^2-6.$