# Thread: Why does one have to divide through by "a" when completing the square?

1. ## Why does one have to divide through by "a" when completing the square?

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?

2. Have another go at posting this.

3. We want to write $\displaystyle ax^2+bx$ as a perfect square. The usual procedure says to add (and then subtract) $\displaystyle (b/2)^2$. But $\displaystyle ax^2+bx+(b/2)^2$is only a perfect square when $\displaystyle a=1$.

4. You don't have to divide by a, it just makes the calculations easier.
$\displaystyle (x+ b)^2= x^2+ 2bx+ b^2$
so if you have, say $\displaystyle x^2+ 3x$, you only have to think "2b= 3 so b= 3/2 and then $\displaystyle b^2= 9/4$- I have to add 9/4 to make this a perfect square".

But $\displaystyle (ax+ b)^2= a^2x^2+ 2abx+ b^2$ so if you have, say, $\displaystyle 2x^2+ 3x$, you would have to think "$\displaystyle a^2= 2$ so $\displaystyle a= \sqrt{2}$, then $\displaystyle 2ab= 2\sqrt{2}b= 3$ so $\displaystyle b= \frac{3}{2\sqrt{2}}$ and $\displaystyle b^2= \frac{9}{8}$ so we need to add $\displaystyle \frac{9}{8}$ to make that a perfect square".

It is just simpler to factor out the "a" first.

5. 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.

6. deleted - accidental double post

7. The discriminant of $\displaystyle ax^2+bx+(b/2)^2$ is $\displaystyle b^2-ab^2=b^2(1-a)$. Thus the discriminant is 0 if and only if b=0 or a=1. Thus, if $\displaystyle b\ne 0$ and $\displaystyle a\ne 1$, the quadratic equation $\displaystyle ax^2+bx+(b/2)^2=0$ has 2 distinct roots. Thus, when you factor $\displaystyle ax^2+bx+(b/2)^2$ you get two distinct factors. In order for the expression to be a perfect square, you need the same factor to appear twice.