<s>[INST] What is the connection between the discriminant of a quadratic and the distance formula? The $x$-coordinate of the center of a parabola $ax^2 + bx + c$ is $$-\frac{b}{2a}$$
If we look at the quadratic formula
$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
we can see that it specifies two points at a certain offset from the center
$$-\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$$
This means that $\frac{\sqrt{b^2 - 4ac}}{2a}$ is the (horizontal) distance from the vertex to the roots. If I squint, the two squared-ish quantities being subtracted under a square root sign reminds me of the Euclidean distance formula
$$\sqrt{(x_0 - x_1)^2 + (y_0 - y_1)^2}$$
Is there a connection? If not, is there any intuitive or geometric reason why $\frac{\sqrt{b^2 - 4ac}}{2a}$ should be the horizontal distance from the vertex to the roots?
 [/INST] We have $$f(x)=ax^2+bx+c=a(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a}$$ with $b^2-4ac\geq 0$ and $a\neq 0$. The vertex of the parabola is $$V\left(-\frac{b}{2a},\frac{4ac-b^2}{4a}\right)$$ and the root(s) are at $$R_1\left(-\frac{b}{2a}+\sqrt{\frac{b^2-4ac}{4a^2}},0\right)~\text{and}~R_2\left(-\frac{b}{2a}-\sqrt{\frac{b^2-4ac}{4a^2}},0\right).$$


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*If $b^2-4ac=0$, then we only have one root at $x=-\frac{b}{a}$. Thus we get the point $V\left(-\frac{b}{2a},0\right)$, which simultaneously is the vertex of the parabola, meaning that the horizontal distance from the vertex to the root is $0$.

*If $b^2-4ac>0$, then we have two different roots and the vertex of the parabola does not lie on the x-axis. But as we are only interested in the horizontal distance we again look at $V_x\left(-\frac{b}{a},0\right)$. Now we can apply the formula for the eucledian distance to $V_x$ and $R_1$:
$$d(V_x,R_1)=\sqrt{\left(-\frac{b}{2a}+\sqrt{\frac{b^2-4ac}{4a^2}}-\left(-\frac{b}{2a}\right)\right)^2+\left(0-0\right)^2}=\sqrt{\frac{b^2-4ac}{4a^2}}.$$
For $V_x$ and $R_2$ we get
$$d(V_x,R_2)=\sqrt{\left(-\frac{b}{2a}-\sqrt{\frac{b^2-4ac}{4a^2}}-\left(-\frac{b}{2a}\right)\right)^2+\left(0-0\right)^2}=\sqrt{\left(-\frac{b^2-4ac}{4a^2}\right)^2}=\sqrt{\frac{b^2-4ac}{4a^2}}.$$


So $$d=\sqrt{\frac{b^2-4ac}{4a^2}}$$ is indeed the horizontal distance between the vertex and the roots.
This is also expressed in your formula, as $$-\frac{b}{2a}\pm \sqrt{\frac{b^2-4ac}{4a^2}}$$ only gives you the $x$-coordinate of the roots. As $-\frac{b}{2a}$ is the $x$-coordinate of the vertex, for the roots you only add/subtract $\sqrt{\frac{b^2-4ac}{4a^2}}$, meaning you go "left/right" starting at $V_x$.
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