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?

  • 1
    $\begingroup$ Just for the sake of being clear, I believe it would help to explicitely state which polynomial we're considering. $\endgroup$
    – gebruiker
    Dec 25, 2015 at 8:53
  • $\begingroup$ I'm asking for a derivation of the quadratic equation... so all of them? $\endgroup$ Dec 25, 2015 at 9:00
  • $\begingroup$ That is what you want: A geometric construction of the roots of a quadratic equation from the coordinates of the max/min point and the value of $a$? $\endgroup$ Dec 25, 2015 at 16:43
  • $\begingroup$ Yup, that would do it. $\endgroup$ Dec 25, 2015 at 18:13

2 Answers 2



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


  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.

enter image description here


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 $$x_H=-\frac{b}{2a}-\frac{\sqrt{|\Delta|}}{2a} .\quad (12)$$

  • $\begingroup$ Wow! This is awesome. Do you know how to transform this construction into a proof of the formula? $\endgroup$ Jan 3, 2016 at 20:22
  • $\begingroup$ @StellaBiderman see my edit, it will give you a sketch of a proof. $\endgroup$ Jan 5, 2016 at 20:20
  • $\begingroup$ In the illustration you wrote $y$$=$$2x^2$-$9x$$+$$4$ rather than $y=2x^2-9x+4.$ It looks like a mixture of something like MathJax or LaTeX with crude pre-Knuth typing. If you used LaTeX embedded in something else, could you just put that whole thing in LaTeX? $\endgroup$ Apr 8, 2017 at 18:05

I may be misunderstanding the specifics of your question, but since the below is a simple geometric solution for $x^2 -ax + b = 0$ that has not been mentioned (it doesn't make use of parabola extreme point as above, so perhaps not what you wanted), thought I'd add it. Note this is only for quadratics with real roots. I don't offhand know a purely geometric construction for real and complex roots: Would be interested to see if anyone knows...

enter image description here


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