Completing the square

Suppose we have $a x^{2} + b x + c = 0$ .

It’d be easier to complete the square if the $x$ term were even, so let’s double:

$2 a x^{2} + 2 b x + 2 c = 0$

It’s also be nicer if the $x^{2}$ term were a square, so let’s multiply by $2 a$ :

$4 a^{2} x^{2} + 4 a b x + 4 a c = 0$

The first two terms are $(2 a x) (2 a x + 2 b)$ , which by difference of two squares, is $(2 a x + b)^{2} - b^{2}$ :

$(2 a x + b)^{2} - b^{2} + 4 a c = 0$

Or better:

$(2 a x + b)^{2} = b^{2} - 4 a c$

Unsquare everything:

$2 a x + b = \pm \sqrt{b^{2} - 4 a c}$

Rearrange:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

Now, I know that won’t come as a surprise. However, there are two nuggets in doing it this way that I especially like:

Making the first term a square and the second term even hugely simplifies the process of completing the square (who among us has not got confused by the “uh… it’s $a {(x - \frac{b}{2 a})}^{2} - \frac{b^{2}}{4 a^{2}}$ , I think…?” process?).
Dealing with the first two terms as a difference of two squares is probably unnecessary, but a link I hadn’t drawn before.

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