"Citardauq" formula derivation?

Question

I'm trying to understand how we got to the "citardauq" formula   (note: "quadratic", reversed)

I found this question here, first answer by Andre says

Multiply "top" and "bottom" by $-b\mp\sqrt{b^2-4ac}$. After the smoke clears, we obtain $$\frac{2c}{-b \mp \sqrt{b^2-4ac}}.$$

Question is, how does the smoke actually clear? I was able to get to the final result by just distributing and canceling out terms but I wasn't sure I went about it the right way, the $\mp$ confused me a bit. How do you multiply $\pm \sqrt{b^2-4ac}$ with $\mp \sqrt{b^2-4ac}$?

Answer 1

By Vieta's formula, the product of the roots is the ratio of the independent term and the quadratic coefficient.

$$r_0r_1=\frac ca.$$

Then

$$r_1=\frac c{ar_0}.$$

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This formula is useful for the accurate evaluation of the roots, as it trades a difference for a sum, and avoids catastrophic cancellation (https://en.wikipedia.org/wiki/Loss_of_significance).

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You can read the original equation as

$$0=\frac{ax^2+bx+c}{x^2}=a+\frac bx+\frac c{x^2}=a+by+cy^2$$ where $y=1/x$.

The solution of this quadratic equation is

$$y=\frac{-b\pm\sqrt{b^2-4ca}}{2c}$$ or

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

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Regarding the "smoke" method, the computation is

$$\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\frac{-b\mp\sqrt{b^2-4ac}}{-b\mp\sqrt{b^2-4ac}}=\frac{b^2-b^2+4ac}{2a(-b\mp\sqrt{b^2-4ac})}.$$

Answer 2

Basically the thing is we have to multiply $-b \pm \sqrt{b^2-4ac}$ with $-b \mp \sqrt{b^2-4ac}$. Consider the terms $-b = \alpha$ and $\sqrt{b^2-4ac} = \beta$. Then our requirement reduces to multiplying $\alpha + \beta$ and $\alpha -\beta$.

The thing to remember is that we have to multiply $-b + \sqrt{b^2-4ac}$ with $-b -\sqrt{b^2-4ac}$ and not square it. The same argument goes for the term with the negative sign as well. Hope it helps.