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There is some kind of combinatoric approach but that doesn't seem to have a name since apparently every single web article assumes "completing the square" is the only method.

In the alternative method, you start with the assumption $x^2+\tfrac{b}{a}x+\tfrac{c}{a}=0$ by the fundamental theorem of algebra has the form $(x-r_1)(x-r_2)=x^2-(r_1+r_2)x-r_1r_2$ which gives $-r_1-r_2=b/a$ and $r_1r_2=c/a.$

From here, how do you work out which substitution to make? If I solve for either $r_1$ or $r_2$ in terms of the other, it doesn't seem to make things simpler. I think the idea is to leverage that every symmetric polynomial can be simplified in terms of elementary symmetric polynomials.

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A different approach seeks to eliminate the middle term. Note if we make a substitution $u = x- d$ for some constant $d$, we get $x = u + d$ and the new equation becomes $$ 0 = a(u+d)^2 + b(u+d) + c = au^2 + u \left[2ad + b \right] + ad^2+bd +c $$ Since we want the linear term to zero out, we set $$ d = \frac{-b}{2a} $$ and the new equation becomes $$ 0 = au^2 + \frac{b^2}{4a} - \frac{b^2}{2a}+c $$ which you can simplify and solve easily...

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  • $\begingroup$ I've independently derived this approach and I like it because it's analogous to making a depressed cubic when solving a degree 3 equation and that seems elegant somehow. I didn't know this (for a quadratic) was a widely used or referenced method. $\endgroup$
    – Deepak
    Commented Apr 28, 2020 at 3:52
  • $\begingroup$ @Deepak I was absolutely sure that historically this is how they solved it in the first place. In fact, I am pretty positive that (a) the technique for solving cubics was inspired by this idea and (b) the idea of completing the square is in some sense a reversal of this technique... $\endgroup$
    – gt6989b
    Commented Apr 28, 2020 at 13:21
  • $\begingroup$ Interesting, but I'd love to see a reference. $\endgroup$
    – Deepak
    Commented Apr 28, 2020 at 14:06

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