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Maths educator Po-Shen Loh has discovered a way to solve quadratic equations that is much more intuitive than \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), and is apparently unprecedented in the entire 4000-year history of thought on quadratic equations! See arxiv.org/abs/1910.06709.

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The trick is as follows: Assume, first of all, that \(a=1\), and let \(r\) and \(s\) be the (unknown) roots. Then we can write \(x^2 + bx + c = (x - r)(x - s)\).

Expanding the right-hand side, we get \(x^2 + bx + c = x^2 - (r+s)x + rs\).

So we have \(r + s = -b\), and \(rs = c\).

Now, the only way we can have \(r+s = -b\) is if \(r = -\frac{b}{2} + z\) and \(s = -\frac{b}{2} - z\), for some \(z\).

Since \(rs = c\), this means that \(c = \frac{b^2}{4} - z^2\).

Rearranging, we get \(z = \pm \sqrt{\frac{b^2}{4} - c}\).

And now, substituting into the formula for \(r\) and \(s\), we get \(r, s = -\frac{b}{2} \pm \sqrt{\frac{b^2}{4} - c}\).

But this is just the same as \(\frac{-b \pm \sqrt{b^2 - 4c}}{2}\), which is exactly what we would expect from the standard formula, given that \(a=1\)! The general derivation, for \(a \neq 1\), takes a few more steps, but is fairly straightforward.

The innovation here seems to be in the emphasis placed on the properties of the sum of the roots and the product of the roots.

@skalyan I see, so this person is saying the Vieta equations are unprecedented?

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