I did a poor job of making my intended point in Irreal’s Quadratic Equations Reconsidered post. The intended point was that Loh had discovered a method of solving quadratic equations without having to remember—or, heaven forefend, rederive—the quadratic formula or to invoke the usual guessing at the factors.
The post showed why the roots of \(x^{2}+bx+c\) sum to \(-b\) and have the product \(c\). The second observation was that since the sum of the roots is \(-b\) their average is \(-b/2\) and that therefore the roots must have the form \((-b/2-u)\) and \((-b/2+u)\) and that since their product is \(c\), we must have \(u^{2}=b^{2}/4 -c\).
I left the technical details there and judging from the comments, readers thought the method involved remembering \(b^{2}/4 -c\) instead of the quadratic formula. They even pointed out that \(b^{2}/4 -c\) is really just the usual discriminant, for the special case of \(a=1\), in the quadratic formula.
But the real point was the method so let me solve a couple of equations to illustrate. First up: \[x^{2}-8x+15=0\]
The roots sum to \(8\) so their average is \(4\) and therefore the roots are \((4-u)\) and \((4+u)\). Since their product is \(15\), we have \(16-u^{2}=15\) so \(u=1\) and the roots are \((4-1)=3\) and \((4+1)=5\).
When the \(x^{2}\) coefficient is not \(1\), we just divide by it first. Thus to solve \[2x^{2}+8x+6=0\] we divide both sides by \(2\) to get \[x^{2}+4x+3=0\]
Now, the roots sum to \(-4\) so their average is \(-2\) and the roots are \((-2-u)\) and \((-2+u)\). Since their product is \(3\) we have \(4-u^{2}=3\) so \(u=1\) and the roots are \((-2-1)=-3\) and \((-2+1)=-1\).
I chose those roots to be simple integers so as not to bog down in more complex arithmetic but the method works fine when the roots involve radicals or are complex. The point is that the only thing you have to remember is that the sum of the roots is \(-b\)—and that therefore the roots are of the form \((-b/2 \pm u)\)—and that their product is \(c\).