Graphical method of solving quadratic and cubic equations

Question

I once came across a method for solving quadratic and cubic equations using a graphical method as shown below (where the lengths of the line segments are equal to the coefficients of the equation). However, I can't remember what to do after I have drawn these line segments.

Answer 1

This is Lill's method for general (real) polynomial roots. After drawing the line segments you cast a ray from the isolated endpoint of the line segment corresponding to the highest-power term, turning $90^\circ$ at (the extension of) each segment corresponding to the polynomial's remaining terms, according to the following rules:

• if the ray hits the segment itself, it stays on the same side (is reflected)
• if not, it goes to the opposite side (is refracted)

If this ray eventually hits the other endpoint after all the turns, the slope of the initial ray (before its first turn) corresponds to a root of the polynomial.

Answer 2

Your graph wants to represent the equation

$$x^2+5x+6=0$$

but it begins from the first point to the left instead of to the right. No problem, I will adjust my explanation to it - as you correctly turn clockwise (and not counter-clockwise) for plus signs in coefficients $+5$ and $+6$.)

First which is need to say is, that searching for roots is not a direct geometric construction (in the sense e. g. of an edge and compass) but something as groping around or guessing, but with a quick feedback about a success.

(Something as dividing the segment into $3$ equal parts only by guessing - a quick feedback about a success gives you your eyes.)

Second, it is not a method to find exact values of roots, as the final judge is you, your eyes.

In a summary, the Lill's method (which is its name by its inventor Eduard Lill), is nothing more that a method for a good guessing (and iteratively correcting the guesses) because the quick feedback is very illustrative.

Now, what we are guessing?

The slope of the ray from point A - because it will be a root.

So, lets start with a bad guess. Your equation has roots $-2$ and $-3$, but we will guess $\color{red}{-1}$. Here is the ray $\color{red}k$ with slope $\color{red}{-1}$: It intersect the line $i=BC$ in the point $E$.

Now we construct the perpendicular line $l$ to this ray in point $E$:

It intersects the line $j=CD$ in the point $F$ - and it is different from the point $D$.
So the quick feedback doesn't approve the value $\color{red}{-1}$.

But in spite of it this feedback gives us an idea - we need to move the point $E$ a little upwards for reaching a goal - the coincidence of point $\color{red}F$ with point $\color{blue}D$).

Now we use value (slope) $\color{red}{-2}$, which IS a root of your equation:

Do you see the coincidence of $\color{red}F$ with $\color{blue}D$?
The quick feedback now approves the value $\color{red}{-2}$ as a root of your equation.

Similarly for value (slope) $\color{red}{-3}$:

Answer 3

As an illustration for searching for roots of the equation

$$x^2+5x+6=0$$

(which are $-2$ and $-3$) by Lill's method, here is an animation of it: