Thursday 28 September 2023

A few years back I wrote a paper with the tongue-in-cheek title, Twenty Ways to Solve a Quadratic with reference to the song, "Fifty Ways to Leave Your Lover". It included a variety (actually about 20) of approaches, including some graphic approaches, and some history notes on each method.

Recently while looking into the work of Karl George Christian von Staudt (the 147th anniversary of his death is coming up June 1) I found two more graphic methods I had not previously known. I also found an earlier use of one that I had written about in the article above, plus a very early use that I omitted.
Later I changed the name to "Solving Quadratic Equations By analytic and graphic methods; Including several methods you may never have seen." and posted it at Academia.edu.
I will begin with two very early examples and conclude with the von Staudt example.

One of the earliest graphic examples of a solution has to be from Euclid's Elements, in book 2 proposition 11. Euclid's description of the task is to cut a given straight line into two parts so that the rectangle formed by the whole and one of the parts is equal to the square on the second part. If we call the parts of the line b and x, then what we seek is x^2 =(x+b)b or x^2 = bx+b^2.

The construction in the Elements is pretty brief, finding a midpoint and a couple of compass constructions.
To see the image, and Euclid's solution I leave you to the always wonderful web page on the Elements by Professor David Joyce.

When I wrote the paper on solving quadratics, I credited a method (#13 in the list) this way:
13. Real roots by Lill circle. One of the most unusual graphic methods I have ever seen comes from a more general
method of solving algebraic equations first proposed, to my knowledge, by M.E. Lill, in Resolution graphique des équations numériques de tous les degrés..., Nouv. Ann. Math. Ser. 2 6 (1867) 359--362. Lill was supposedly an Artillery Captain, but his method was included in Calcul graphique et nomographie by a more famous French engineer, Maurice d’Ocagne, who called it the “Lill Circle”.
It turns out that before Lill, a young Scottish mathematician named Thomas Carlyle was credited with using the same circle to solve quadratics (To be fair, Lill extended his method to all polynomials, and even complex roots).
In Sir John Leslie's (who died in 1832) "Elements of Geometry" he writes this method "...was suggested to me by Thomas Carlyle, an ingenious young mathematician, and formally my pupil."

Carlyle skipped right to the diameter of the circle in question. Given a quadratic, x2+bx + c=0, plot points at (0,1) and (-b,c) and construct a diameter. The circle with this diameter will intersect the x-axis at the real solutions (if any exist) of the quadratic.
The circle for y=x2 +5x - 6 is shown with the actual parabolic function in red.
Students might be challenged to explain why one end of the diameter will always lie on the function.

In 1908 a little book of less than 100 pages titled Graphic Algebra, written by Arthur Schultze, was published by Macmillan & Co. Schultze was a high school math dept. head at the New York High School of Commerce, and an associate Professor at NYU.
The book is free online in several formats.

On page 47 I found a graphic method of solving quadratic equations I had never seen before. The process uses a standard graph of xy=1. [One of the common approaches when calculators and computers did not exist was to alter an equation to the solution of two equations such as a familiar conic and a straight line.]

I will illustrate his approach with the example he uses in his book. To solve the quadratic equation x2 + 2x - 8 =0 He first makes the simple step of dividing all terms by x to get $x+2 - \frac{8}{x} = 0 ( x\neq 0)$

Now by substitution of y= 1/x (or xy=1) we get x+2-8y=0. So where both of these equations are true, must be a solution to the original equation. Simply picking a couple of convenient points to plot the line x+2-8y = 0 he determines that when y= 0, x = -2; and when y= 1, x = 6. So we graph the equation xy=1 and then plot the points (-2,0) and ((6,1) and hope for an intersection.

The x-coordinates of the two intersections (red) give us the solutions x={-4; 2}.

Schultze's little book also uses the method from my paper (15. Using the graph of y = x^2 and y = -bx – c to find real roots.) as a graphic solution using a simple conic (y=x2)

And for me, the treat of the day was Karl George Christian von Staudt's graphic method of solving quadratics because it is so different from all the others I've ever learned. Staudt was a student at Gottingen and worked with Gauss, who was at the time the director of the observatory, becoming a very good mathematical astronomer in his own right. He began as a high school teacher as well. His book Geometrie der Lage (1847) was a landmark in projective geometry. It was the first work to completely free projective geometry from any metrical basis. In 1857 von Staudt contributed a route to number through geometry called the Algebra of throws, and he made the important discovery that the relation which a conic establishes between poles and polars is really more fundamental than the conic itself, and can be set up independently.
So here is the solution method used by von Staudt. The challenge to teacher (and students) is to see if you can figure out WHY it works (I found this very difficult):
Using the equation x2 - 5x + 6 = 0 we begin by constructing a one unit circle centered at (0,1). Then we construct the points $\frac {c}{-b} , 0)$ and $\frac{4}{-b},2)$ For b= -5 and c= 6 we get the points $\frac {6}{5} , 0)$ and $\frac{4}{5},2)$
Construct the straight line through these two points (in red in the image).
The line intersects the circle in the points I, and J, and it is not essential to know their coordinates, but J is (1,1)
Now use (2,0) to project each of these points onto the x-axis. The result is the solutions to the equation. In this case x= {2, 3}

I remember reading about the poet Omar Khayyam who developed a number of geometric solutions to cubics.  He had a nice (tricky, for me ) way to solve equations like x^3 + a^2 x = b.  One example like this was in the work of Cardano after he learned/stole Tartaglia's method for cubics, x^3 + 6x = 20.  A little mental work might find the solution since it is an integer, but long before (about 400 years)  Cardano, Omar had a graphic method somewhat based on the Greek idea of completing the square, you might call it completing the cube.
Here is how it worked.  He broke the problem down into the graph of a circle and a parabola.  The center of the circle was at (b/2a^2,0) and the circle passed through (0,0).  Then he set up a parabola with x^2 = ay.  Working out how he came up with that is a pretty big challenge, but fun when you get it.
For the Cardano equation x^3 + 6x = 20, 6 is a^2, and b=20, so our circle has a center at (20/(2*6), 0), or just (10/3,0).

x^2 = a y can be simp;ified to y = x^2/a, or in this case, y=x^2/(sqrt(6)).  graphing both on geogebra classic we get:

The intersection at x=2 is our real solution for this problem.

The next thing I thought to try was the famous equation that Rafael Bombelli used to get the whole complex number plain in gear.  It was similar, but with a quirk, x^3 = 15x + 4.  The 15x was on the wrong side.  To use Omar K's method above, the 15 would have to be moved tand become negative, but the poet didn't traffic in no crazy negatives.  I didn't have an example of how he had done this kind of cubic, but a moment of inspiration led be to think of what might well have bee his approach.

Ir we divide all terms by x, we get x^2 = 15 + 4/x.  Since both sides are equal, we can let each of them equal y and so, y= x^2, and y= 15+ 4/x

So I gave them a run on geogebra.
I thought it was interesting that it had solutions at 4 and -4, but I realized it made sense (of course it does, It's math.) when x = +4, 15 + 4/x = 16 and when x = -4, 15 + 4/x = 14.

I found a paper that listed 19 different types of cubics that the Great Poet solved and the method.  I will play with some of these later.