Tag Archive for: vertex format for a quadratic

Completing the Square

Completing the Square


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


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.

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.