Solving a quadratic equation

[Fred_Kohlhepp]

New and old from Po Shen Loh

[LucMeekes]

6. Where u=sqrt(B^2/4-C).

Simply:

This is the famous quadratic root formula, but with A=1.

Luc

[Fred_Kohlhepp]

This is the famous quadratic root formula, but with A=1.

Well, duh!  You didn't expect a different answer, did you?   😉

What I found noteworthy, that maybe didn't come across in my post, was the development of the solution in his paper, https://arxiv.org/abs/1910.06709, was the logic behind the derivation. (I like starting from basics and developing the solution rather than memorizing an equation.  Probably because my memory is so poor.)

[LucMeekes]

First off: I did not mean to be offensive in any way. Please accept my apologies if my reaction came across that way.

No, I did not expect a different answer.

Unfortunately the link you provide does not work (for me), but I was able to follow the gist of the 'derivation' by the 5 steps. I felt that the calculation to find the value of u should be explicit, rather than just hinted at in step 4. Hence my step 6; in hindsight this should be added to step 4.

Note that the method also works for Ax^2+Bx+C=0 as follows: Divide the entire quadratic by A, gives you x^2 + B'x+C'=0, with B'=B/A and C'=C/A.

Luc

[Fred_Kohlhepp]

I took no offense, and hope I didn't give any.

Try a Google search on "po shen loh quadratic equation" which should point you to https://www.nytimes.com/2020/02/05/science/quadratic-equations-algebra.html eventually.

[LucMeekes]

That link works immediately. Thanks!

Luc

[Werner_E]

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@LucMeekes wrote:
.....

Unfortunately the link you provide does not work (for me),

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The comma after the address was mistakenly still part of the link. So it should only be

https://arxiv.org/abs/1910.06709

A nice find.