Have another go at posting this.
Essentially I can't figure out why you should divide out the "a" term in completing the square when you have the non-monic case?
a(x-h)^2+k after all, is just 1(x-h)^2+k and if you don't have to divide when a=1, then why should the case change when a=n>1?
You don't have to divide by a, it just makes the calculations easier.
so if you have, say , you only have to think "2b= 3 so b= 3/2 and then - I have to add 9/4 to make this a perfect square".
But so if you have, say, , you would have to think " so , then so and so we need to add to make that a perfect square".
It is just simpler to factor out the "a" first.
Dr. Steve,
Why wont a quadratic be a perfect square when a doesn't equal 1? I can hypothesize that perhaps, from a geometric POV a square plus a square is more of a rectangle than anything else, and as the ax^2 term is a geometric square 2x^2 must be a rectangle so you have to "square" it, but I'm not sure if I'm right.
The discriminant of is . Thus the discriminant is 0 if and only if b=0 or a=1. Thus, if and , the quadratic equation has 2 distinct roots. Thus, when you factor you get two distinct factors. In order for the expression to be a perfect square, you need the same factor to appear twice.