# Alternative method for solving quadratic formula?

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.

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...