telerib | Apr. 12th, 2006

So yesterday was a six hour math problem. Ugh. And it wasn't finished.

This morning, I notice: the denominator is of the form a^2 - b^2. And Sister Myrona's old voice from high school algebra came echoing back over the years:

"The sum and difference of two numbers, the difference of their squa-ares."

Which is to say, a^2 - b^2 = (a+b)*(a-b)

And wouldn't you know it? The numerator was of the form a^2 - 2ab + b^2. Which is trivially factored to (a-b)^2.

So the problem (which was cubic) had terms cancel and is now quadratic, and (also thanks to Sister Myrona) I will never, ever forget the equation to find the roots of a quadratic equation.

"The negative of B plus or minus the nonnegative square root of that same B, squared, minus four-A-C, AAAALL over two A."

(r1,2 = (-b +/- sqrt(b^2 - 4ac)) / 2a, where a, b, and c are the quadratic coefficients of the equation a*x^2 + b*x + c = 0)

You have to chant it in the right singsong voice for the full effect.

Well, it's still a very ugly equation, still, but I now have one variable in terms of the the other variable. HUZZAH!