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If you derive the quadratic formula by completing the square on a general quadratic, then should the denominator end up as $2 \ \text{modulus(a)}$ where $a$ is the coefficient on the $x^2$ term which may well be negative?

I know it doesn't matter in the end because of the plus/minus. thanks

callculus42
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Carlos Bacca
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    You've answered your own question: it doesn't matter. I'd prefer just $a$, not $|a|$, which would make most readers think twice to no purpose. – Ethan Bolker Sep 12 '18 at 20:13
  • I know it doesn't matter in the end It does in fact matter in the end, since replacing the denominator with $,2|a|,$ can potentially change the sign of the entire roots, not just the radical. For example, take $,a=-1,$, $b=3,$, $c=-2,$ then the equation is $,-x^2 + 3 x - 2 = 0,$ with roots $,\frac{-3 \pm \sqrt{3^2- 4 \cdot 2}}{2 ,\cdot, (-1)} = {1,2},$. If, instead, you replace the denominator with $,2\cdot|-1|,$, then the formula gives the (wrong) roots $,{-1,-2},$. – dxiv Sep 12 '18 at 21:50
  • Thanks, I don't understand the last part of the derivation then on the denominator it has principal root(4a^2) which I think defines the same function as 2a for a>0 only. Can you explain what I'm not seeing, thanks – Carlos Bacca Sep 12 '18 at 22:34
  • the reason it did not work there was because of the form. It works in the form... -b/2a +/- principalroot(b^2-4ac)/2mod(a) exuse my bad formatting. In the derivation however they seem to just use 2a and then combine the terms and it seems to work ok, don't know why though – Carlos Bacca Sep 12 '18 at 22:52
  • @JamesAnthony don't know why though That's because $,\pm a = \pm |a|,$ in the sense of sets of values. – dxiv Sep 12 '18 at 23:48
  • Think about what happens to the $\pm$ when you negate the denominator. – amd Sep 12 '18 at 23:48
  • thanks for the help isee it now – Carlos Bacca Sep 13 '18 at 09:10

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