# A new method to derive the quadratic formula

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...

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.
Nice. But there is a logical flaw: How do you know that $$\alpha, \beta$$ exist?
• You mean real values of $\alpha$ and $\beta$? Apr 18, 2020 at 19:19