# 3 Ways to “Complete 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 can only be used when you are working with an equation. It 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:

$y~=~2x^2+12x+10$

$y-10~=~2x^2+12x$

$\dfrac{y-10}{2}~=~x^2+6x$

$\dfrac{y-10}{2}+(\frac{6}{2})^2~=~x^2+6x+(\frac{6}{2})^2$

$\dfrac{y-10}{2}+9~=~x^2+6x+3^2$

$\dfrac{y-10}{2}+9~=~(x+3)^2$

$\dfrac{y-10}{2}~=~(x+3)^2-9$

$y-10~=~2(x+3)^2-18$

$y~=~2(x+3)^2-8$

### Second Approach

This approach keeps everything on the same side of the equation, and is therefore suitable for use when you either do not have an equation to work with, or do not wish to “mess up” the rest of the equation with your work to complete the square. It factors the leading coefficient out of all terms before proceeding:

$y~=~2x^2+12x+10$

$y~=~2[x^2+6x\;\;\;\;\;\;\;\;\;+5]$

$y~=~2[x^2+6x+(\frac{6}{2})^2-(\frac{6}{2})^2+5]$

$y~=~2[x^2+6x+3^2-3^2+5]$

$y~=~2[(x^2+6x+3^2)-3^2+5]$

$y~=~2[(x+3)^2-9+5]$

$y~=~2[(x+3)^2-4]$

$y~=~2(x+3)^2-8$

### 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 equations. It factors the leading coefficient out of only the two “x” terms:

$y~=~2x^2+12x+10$

$y~=~2[x^2+6x]+10$

$y~=~2[x^2+6x+(\frac{6}{2})^2-(\frac{6}{2})^2]+10$

$y~=~2[x^2+6x+3^2-3^2]+10$

$y~=~2[x^2+6x+3^2]-2\cdot 3^2+10$

$y~=~2[(x+3)^2]-18+10$

$y~=~2(x+3)^2-8$

### 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?