Geometric derivation of the quadratic equation

Question

The quadratic equation can be thought of as specifying distances in the Euclidean plane. It tells us that the $x$-intercepts of a function occur at a distance of $\frac{\sqrt{b^2-4ac}}{2a}$ from the $x$ coordinate of the max/min point. Does anyone know a purely geometric derivation of the quadratic formula related to this fact?

Answer 1

Problem:

Let $V(x_V,y_V)$ the vertex of a parable whose leading coefficient is $a$. Find out its roots through geometric construction.

Resolution:

1. Draw line $e$ perpendicular to $x$-axis, such that $V \in e$.
2. Let $A=e\cap x$-axis, mark points $B$ and $C$ on $x$-axis, such that $AB=|a|$ and $AC=1$.
3. Mark point $F$ on $e$, such that $FA=1$, and $V$ and $F$ are in different semi-planes relative to $x$-axis.
4. Draw a line $j$ parallel to $VB$ through $C$.
5. Let $E=j \cap e$. Mark point $G$ on $e$, such that $G$ is the midpoint of $EF$.
6. Draw a circle $\lambda$ centered at $G$ and whose radius is $GF$.
7. Let $\{H,I\}=\lambda \cap x$-axis. The abscissas of $H$ and $I$ are the roots of the quadratic equation that we get when the parabola crosses the $x$-axis.

The picture below shows an example.

Edit/Justification

Recall that: $$x_V= -\frac{b}{2a} \quad (1)$$ and $$y_V=-\frac{\Delta}{4a}. \quad (2)$$ Note that: $$x_A=x_V\quad (3)$$ and $$AV=|y_V|.\quad (4)$$ Note also that $\triangle ABV \sim \triangle ACE$, hence: $$\frac{AB}{AV}=\frac{AC}{AE}. \quad (5)$$ From $(2)$, $(4)$ and $(5)$ we get: $$AE=\frac{|\Delta|}{4a^2}. \quad (6)$$ Recall that an inscribed triangle whose major side is a diameter is right-angled. Therefore $\triangle FHE$ and $\triangle FIE$ are right-angled at $H$ and $I$ respectively. Note that $AH=AI$, and that $AI$ is a height of $\triangle FIE$. As $\triangle FIE$ is a right-angled triangle we have: $$AI^2=AF \cdot AE. \quad (7)$$ Recall that by construction $AF =1$. So we get from $(6)$ and $(7)$: $$AI=AH=\frac{\sqrt{|\Delta|}}{2a}. \quad (8)$$ Note that: $$x_I=x_A+AI \quad(9)$$ and $$x_H=x_A-AH. \quad(10)$$ From $(1)$, $(3)$, $(8)$, $(9)$ and $(10)$, we get: $$x_I=-\frac{b}{2a}+\frac{\sqrt{|\Delta|}}{2a} \quad (11)$$ and $$x_H=-\frac{b}{2a}-\frac{\sqrt{|\Delta|}}{2a} .\quad (12)$$