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I discovered this method of deriving the quadratic formula by taking in account the relationship between zeroes and coefficients.

Let $p(x)$ be a quadratic polynomial of the form : $ax^2+bx+c$
Let $\alpha$ and $\beta$ be its zeroes

So, we have : $\alpha + \beta = \dfrac{-b}{a}$, let's call this the first equation
and $\alpha \beta = \dfrac{c}{a}$
So, by squaring the first equation, we get : $\alpha^2 + \beta^2 + 2\alpha \beta = \dfrac{b^2}{a^2}$
Subtracting $4\alpha \beta$ from both sides, we obtain : $\alpha^2 + \beta^2 - 2\alpha \beta = \dfrac{b^2}{a^2}-4\alpha \beta$ $=$ $\dfrac {b^2}{a^2} - \dfrac{4c}{a}$ $=$ $\dfrac {b^2-4ac}{a^2}$
Square rooting both sides of the above equation, we obtain : $\alpha - \beta = \dfrac {\pm \sqrt{b^2-4ac}}{a}$
So, we now have two equations :
First $\implies$ $\alpha +\beta = \dfrac{-b}{a}$
Second $\implies$ $\alpha - \beta = \dfrac {\pm \sqrt{b^2-4ac}}{a}$
Adding both of these equations, we obtain : $2\alpha = \dfrac {-b \pm \sqrt{b^2-4ac}}{a}$
So, $\alpha = \dfrac {-b \pm \sqrt {b^2-4ac}}{2a}$

Now, we will find the value of $\beta$
For this, let's first multiply our first equation by $-1$
On doing this, we obtain : $-\alpha - \beta = \dfrac {b}{a}$
Now let's add this newly obtained equation and the previously stated second equation
We obtain : $-2\beta = \dfrac {b \pm \sqrt {b^2-4ac}}{a}$
So, $2\beta = \dfrac {-b \pm \sqrt {b^2-4ac}}{a}$
So, $\beta = \dfrac {-b \pm \sqrt {b^2-4ac}}{2a}$
And, we've derived the formula

Let me know what you think about it, thanks...

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2 Answers 2

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It's a valid proof, as are several rival approaches. It can be shortened somewhat: since exchanging $\alpha$ with $\beta$ changes nothing, they can differ only in which sign we take for $\pm$. (In particular, $\sqrt{b^2-4ac}$ is defined as the non-negative root of $y^2=b^2-4ac$, which relies on the fact that $\Bbb R$ is ordered (whereas e.g. $\Bbb C$ is not), but the two roots are algebraically indistinguishable, so one sign gives $\alpha$ while the other gives $\beta$.) So, once you have $\alpha$, you don't need any work to get $\beta$. This is an example of a symmetry argument.

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  • $\begingroup$ I actually thought about that, but did what I did just to make it clearer, thanks for letting me know... $\endgroup$ Apr 18, 2020 at 18:33
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Nice. But there is a logical flaw: How do you know that $\alpha, \beta$ exist?

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  • $\begingroup$ You mean real values of $\alpha$ and $\beta$? $\endgroup$ Apr 18, 2020 at 19:19
  • $\begingroup$ And I think that your answer would have been better as a comment $\endgroup$ Apr 18, 2020 at 19:20

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