QUICK STEPS TO COMPLETE THE SQUARE

This article is going to be a “quickie” because I believe Completing the Square should be just that....quick, easy.  So let’s dispose of all the middle steps and just move swiftly from the General Quadratic Equation to the Vertex Form.

We’ll use a sample equation.......12X^2 + 8X - 10 = Y

1.  Start with the pattern
     of the Vertex Form.................................A(x + h)^2 + K = Y

2.  From the original equation,
     think of the X-variable
     terms as a separate entity......(12X^2 + 8X) - 10 = Y

3.  Factor out the leading
     coefficient, so that the
     coefficient on the X^2
     is just 1..............................12(X^2 + 8/12 X) - 10 = Y

     I’d reduce that improper
     fraction...............................12(X^2 + 2/3 X) - 10 = Y

4.  Fill in “A” in the
     Vertex Form..........................................12(X + h)^2 + K = Y

5.  Fill in the “h” value
     of the Vertex Form
     with 1/2 the new
     coefficient on X in
     Step 3..............................................12(X + 2/6)^2 + K = Y

     I like smaller numbers, so I’m reducing
     that pesky improper fraction................12(X + 1/3)^2 + K = Y

6.  Think for a second about what we’ve
     done.  By creating the (X + h) binomial
     and squaring it, we’ve actually added
     more to the equation.  (FOIL it through
     if you need proof.)  So we need to
     remove it again.

     Square “h”, multiply it by “A”,
     and subtract it from the
     constant in the General
     Form equation...................12(X + 1/3)^2 + (- 10 - 12/9) = Y

     I feel I’m doing more reducing that any
     real work here.....................12(X + 1/3)^2 + (- 10 - 4/3) = Y

7.  A little arithmetic and,
     TAH-DAH, I’m ready
     to solve for X and label
     the vertex on a graph......................12(X + 1/3)^2 - 34/3 = Y


Many texts add the step of finding the perfect square trinomial before factoring it into the binomial squared.  I think it’s just a way to insure that you subtract out the extraneous constant value but I prefer "the elegant solution."