Procedures for “Completing the Square”
I have seen three approaches to “Completing the Square”, as shown below. Each successfully converts a quadratic equation into vertex form. Which do you prefer, and why?
First Approach
This approach moves all terms that are not part of a perfect square to the other side of the equation to get them out of the way. Therefore, it is only useful if you are working with an equation, or are able (and willing) to turn an expression into an equation:
Second Approach
This approach keeps everything on the same side of the equation, and is therefore suitable for use when working with expressions. It factors the leading coefficient out of all terms before proceeding:
![y=2[x^2+6x\;\;\;\;\;\;\;\;\;+5]](http://s0.wp.com/latex.php?latex=y%3D2%5Bx%5E2%2B6x%5C%3B%5C%3B%5C%3B%5C%3B%5C%3B%5C%3B%5C%3B%5C%3B%5C%3B%2B5%5D&bg=ffffff&fg=333333&s=0 "y=2[x^2+6x\;\;\;\;\;\;\;\;\;+5]")
![y=2[x^2+6x+(\frac{6}{2})^2-(\frac{6}{2})^2+5]](http://s0.wp.com/latex.php?latex=y%3D2%5Bx%5E2%2B6x%2B%28%5Cfrac%7B6%7D%7B2%7D%29%5E2-%28%5Cfrac%7B6%7D%7B2%7D%29%5E2%2B5%5D&bg=ffffff&fg=333333&s=0 "y=2[x^2+6x+(\frac{6}{2})^2-(\frac{6}{2})^2+5]")
![y=2[x^2+6x+3^2-3^2+5]](http://s0.wp.com/latex.php?latex=y%3D2%5Bx%5E2%2B6x%2B3%5E2-3%5E2%2B5%5D&bg=ffffff&fg=333333&s=0 "y=2[x^2+6x+3^2-3^2+5]")
![y=2[(x^2+6x+3^2)-3^2+5]](http://s0.wp.com/latex.php?latex=y%3D2%5B%28x%5E2%2B6x%2B3%5E2%29-3%5E2%2B5%5D&bg=ffffff&fg=333333&s=0 "y=2[(x^2+6x+3^2)-3^2+5]")
![y=2[(x+3)^2-9+5]](http://s0.wp.com/latex.php?latex=y%3D2%5B%28x%2B3%29%5E2-9%2B5%5D&bg=ffffff&fg=333333&s=0 "y=2[(x+3)^2-9+5]")
![y=2[(x+3)^2-4]](http://s0.wp.com/latex.php?latex=y%3D2%5B%28x%2B3%29%5E2-4%5D&bg=ffffff&fg=333333&s=0 "y=2[(x+3)^2-4]")
Third Approach
This approach also keeps everything on the same side of the equation, and is therefore suitable for use when working with expressions or equation. It factors the leading coefficient out of only the two “x” terms:
![y=2[x^2+6x]+10](http://s0.wp.com/latex.php?latex=y%3D2%5Bx%5E2%2B6x%5D%2B10&bg=ffffff&fg=333333&s=0 "y=2[x^2+6x]+10")
![y=2[x^2+6x+(\frac{6}{2})^2-(\frac{6}{2})^2]+10](http://s0.wp.com/latex.php?latex=y%3D2%5Bx%5E2%2B6x%2B%28%5Cfrac%7B6%7D%7B2%7D%29%5E2-%28%5Cfrac%7B6%7D%7B2%7D%29%5E2%5D%2B10&bg=ffffff&fg=333333&s=0 "y=2[x^2+6x+(\frac{6}{2})^2-(\frac{6}{2})^2]+10")
![y=2[x^2+6x+3^2-3^2]+10](http://s0.wp.com/latex.php?latex=y%3D2%5Bx%5E2%2B6x%2B3%5E2-3%5E2%5D%2B10&bg=ffffff&fg=333333&s=0 "y=2[x^2+6x+3^2-3^2]+10")
![y=2[x^2+6x+3^2]-2\cdot 3^2+10](http://s0.wp.com/latex.php?latex=y%3D2%5Bx%5E2%2B6x%2B3%5E2%5D-2%5Ccdot+3%5E2%2B10&bg=ffffff&fg=333333&s=0 "y=2[x^2+6x+3^2]-2\cdot 3^2+10")
![y=2[(x+3)^2]-18+10](http://s0.wp.com/latex.php?latex=y%3D2%5B%28x%2B3%29%5E2%5D-18%2B10&bg=ffffff&fg=333333&s=0 "y=2[(x+3)^2]-18+10")
Discussion
From my perspective, each of the above approaches has advantages and disadvantages. I find two of the above approaches appealing, and one less so. What do you think? Are there other other approaches I should include above?
I have actually used a combination of methods one and three. I will have the students move the constant over to the other side, kind of move it out of the way. We don’t need it just yet like when mixing up cookies, the flour is last, measure it and set aside until the end. Then I have the students factor any leading coefficient out, etc. Then when it comes time to add in the number that will complete the square, that number must also be added to the other side. Of course, we also discuss why that number must be multiplied by the leading coefficient (In your example why it’s 2 times 3 squared, not just three squared). Then, that constant term finally is dealt with by adding it to the term added to the equation and then moved back to the right side of the equation. This also can show the kids why, when the vertex is (h,k) that the h is opposite of what is inside the parenthesis and the k value is the same. Have the students look at the step prior to see, in your example, the y + 8 on the left side. And point out it, if we had written in this form, the vertex (h,k) would be the opposite for both h and k.