FRACTURE

Moving into the next phase of the challenge: deriving the quadratic formula from a transformation perspective.

Instead of completing the square in the standard way, I’m using the Tschirnhaus transformation $(x = y + k)$ to eliminate the linear term first.

$\displaystyle ax^2 + bx + c = 0 \implies x^2 + \left(\frac{b}{a}\right)x + \left(\frac{c}{a}\right) = 0 \implies x^2 + px + q = 0$

Here, $\ p = \dfrac{b}{a}, \quad q = \dfrac{c}{a}$

$\text{Let, } \ \displaystyle x = t + k \implies (t+k)^2 + p(t+k) + q = 0$

$\phantom{\text{Let, } \ \displaystyle x = t + k} \implies t^2 + (2k + p)t + (k^2 + pk + q)=0$

To remove the $t$ term: $2k + p = 0 \implies k = -\dfrac{p}{2}$

$\displaystyle \therefore \ x = t - \frac{p}{2}$

$\displaystyle \implies \left(t - \frac{p}{2}\right)^2 + p\left(t - \frac{p}{2}\right) + q = 0$

$\displaystyle \implies t^2 - 2t\left(\frac{p}{2}\right) + \left(\frac{p}{2}\right)^2 + pt - p\cdot\frac{p}{2} + q = 0$

$\displaystyle \implies t^2 - \frac{p^2}{4} + q = 0$

$\displaystyle \implies t^2 = \frac{p^2}{4} - q$

$\displaystyle \implies t = \pm\sqrt{\frac{p^2}{4} - q}$

By substituting $k$ and $t$ back, the equation drops beautifully into the classic p-q formula.

$\displaystyle x = t + k = \pm\sqrt{\frac{p^2}{4} - q} - \frac{p}{2} \implies \boxed{x = -\frac{p}{2} \pm \sqrt{\frac{p^2}{4} - q}}$

Substituting $p = \dfrac{b}{a}$ and $q = \dfrac{c}{a}$ brings everything back to the standard quadratic formula we all learn in school.

$\displaystyle x = -\frac{b}{2a} \pm \sqrt{\frac{b^2}{4a^2} - \frac{c}{a}} = -\frac{b}{2a} \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$

$\implies \boxed{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$

Now that the quadratic is down, the stage is officially set for the ultimate challenge: tackling the general cubic.

The Cubic Formula