An algebraic argument for the discriminant
A twitter thread here asked whether there was an explanation for why the discriminant of a quadratic is unchanged when you reverse the order of the coefficients. (I suspect Evariste Galois might have some ideas, but I still haven’t read up on that.)
Instead, here’s my argument:

Suppose $az^2 + bz + c = 0$, with $a\ne 0$ and $c \ne 0$, so that $z \ne 0$.

Then $az + b + \frac{c}{z} = 0$

This can be written as:
 $\left(\sqrt{az} + \sqrt{\frac{c}{z}}\right)^2 = 2 \sqrt{ac}  b$.
 $\left(\sqrt{az}  \sqrt{\frac{c}{z}}\right)^2 = 2 \sqrt{ac}  b$.

Multiplying these together gives $\left(az  \frac{c}{z}\right)^2 = b^2  4ac$.
Swapping $a$ and $c$ on the left doesn’t change the right hand side, so reversing the order of the coefficients of a quadratic doesn’t change the discriminant $\blacksquare$
After a bit of thought, I realised I’d missed a trick, and there was a slightly neater approach that doesn’t implicitly require complex numbers:

If $b = ax  \frac{c}{x}$ then $b^2 = \left(ax + \frac{c}{x}\right)^2$

So $b^2  4ac = \left(ax  \frac cx\right)^2$
Do you have a better argument? I’d love to hear about it!
* Thanks to @kevinhouston for pointing out some minor errors.