## Friday, January 20, 2012

### A "Backwards" Approach to Completing the Square

One of my students is an intuitive math student, even though she's not so good with memorizing algorithms and often her intuition isn't enough to complete the problem all the way through. She recently encountered a completing-the-square problem on a test, and couldn't remember how to do it. The question said to go from f(x) = 2x^2 - 12x + 5 to the form f(x) = 2(x - k)^2 + h, in order to do further transformational analysis.

Being an intuitive math student, she took the unusual approach of expanding f(x) = 2(x - k)^2 + h into f(x) = 2x^2 - 4kx + 2k^2 + h. And then she set this equal to 2x^2 - 12x + 5 and then just got stuck. I looked at her work and thought it was an interesting alternative to completing the square. Based on her approach, you can simply observe that if the two equations are equal, then it must be true that their x terms are equal: -4kx = -12x, and similarly, their constant terms are equal: 2k^2 + h = 5. So, it follows that k = 3 and h = -13, leading us to the vertex form of the equation as f(x) = 2(x - 3)^2 - 13.

Just thought I'd share an interesting alternative to completing the square "forwards". I see this as an alternative in working "backwards". Funny what the kids can help you see, even when they're not skilled/experienced enough to make it all the way through a problem.

#### 1 comment:

1. Hey, that's brilliant! I'll definitely use it if ever I teach completing the square again.