A Catalog of Quadratic Equations

It is the time of year when I try to enchant my Pre-calc students with the beauty of complex solutions to polynomials, and particularly quadratics.. A good time to remind them of a beautiful result created by a high school student.....

-----------------------------------
The Mandelbrot set can be thought of as a catalog of closed Julia Sets over the complex plain. Each point on the plain representing a complex number, and the coloring representing whether the Julia set for that point had a closed or open orbit.

The equation y= ¹/₄ x² on the coordinate plain can, in similar fashion, serve as a catalog of all possible quadratic equations. The method was discovered by a high school age British student by the name of Anthony Bayes around 1955 as reported in the Mathematical Gazette in September of that year by H. Gebert, the young man's teacher.

If every point (b,c) on the coordinate plain is representative of a quadratic equation x²+bx+c=0, then the set of all points in the cup of the parabola have only complex solutions [for example, the point (2,5) would represent the equation x²+2x+5=0 which has only complex roots]. Those points beneath the curve have two real distinct solutions, and the points on the curve specify equations which have a double root[ again, (2,1) would represent the perfect square trinomial x²+2x+1=0]. For any point on the curve, the solutions are -1/2 the x-coordinate. In the case of (2,1) the double root of x²+2x+1=0 is at -1/2 of 2 = -1.

Any line tangent to the cuve passes only through points representing equations which share at least one solution. The line y=x-1 which is tanget to the curve at the point (2,1). Since the solutions for x²+2x+1=0 are both -1; all the points on this tangent represent quadratic equations which have x=-1 as one solution. As a case in point, (or a point in this case), (5,4) is on the tangent line, and the solutions of the quadratic equation x²+5x+4=0 are x=-1 and x=-4. If you drew the other tangent to the parabola passing through the point (5,4) it would contain the point (8,16) on the parabola representing the quadratic with a double root at x=-4, that is, the quadratic x²+8x+16=0.

This would allow a simple approximation method to find the solutions of any quadratic with reasonably small values using only a straightedge and a printed graph of the curve y=1/4 x². Simply pick the point (b,c) corresponding to the values of the equation, and then lay a straight edge through the point and tangent to the parabola. The solutions would be -1/2 the x values of the points of tangency, as given above.

Rediscovered, or Did He Cheat?

In a recent post I commented on the "discovery" by a British High School student,Anthony Bayes, of a graphic relationship that allowed every point on the coordinate plane to represent a quadratic equation of the form x²+bx+c=0.

I was impressed that a high school age student would come up with the idea of a transformation from a quadratic to a point on the plane. THEN.... a few days ago I came across a post from the Mathematical Gazette, January of 1913, entitled "A Graphic Solution of the Equation xⁿ-px+q=0, much of which dealt with the idea that the curve 4y=x² (which the author called the Discriminant Curve) would serve as a graphic solution approach for any quadratic of the form x²-px+q=0.
Note the use of -p, a step up from what young Anthony had used, making it a slightly more useful approach, and perhaps testifying to the independence of his discovery. The author of the piece, A. Lodge, (with no other information given), did not claim it as original, nor site another use.

So the solutions to the equation x²-px+q=0 are found by the points where the tangents to 4y=x² intersect the x-axis.

The article also pointed out, and I admit I had never thought of this, that the solutions to x²-px+q=0 are the same as the solutions to $x^2-\frac{p}{k}x+\frac{q}{k^2}=0$. It is easy enough to prove, but for the use of a nomograph like the approximations using the graph of 4y=x², it allows you to work on a much narrower range of x and y values.

The article also referred to solutions of higher powers of trinomials that could be solved in the same way, but my little mind has not grasped it enough to explain it yet, so I save that for another day...

A Mandelbrot-like Set for Quadratics

The Mandelbrot set can be thought of as a catalog of closed Julia Sets over the complex plain. Each point on the plain representing a complex number, and the coloring representing whether the Julia set for that point had a closed or open orbit.

The equation y= ¹/₄ x² on the coordinate plain can, in similar fashion, serve as a catalog of all possible quadratic equations. The method was discovered by a high school age British student by the name of Anthony Bayes around 1955 as reported in the Mathematical Gazette in September of that year by H. Gebert, the young man's teacher.

If every point (b,c) on the coordinate plain is representative of a quadratic equation x²+bx+c=0, then the set of all points in the cup of the parabola have only complex solutions [for example, the point (2,5) would represent the equation x²+2x+5=0 which has only complex roots]. Those points beneath the curve have two real distinct solutions, and the points on the curve specify equations which have a double root[ again, (2,1) would represent the perfect square trinomial x²+2x+1=0]. For any point on the curve, the solutions are -1/2 the x-coordinate. In the case of (2,1) the double root of x²+2x+1=0 is at -1/2 of 2 = -1.

Any line tangent to the curve passes only through points representing equations which share at least one solution. The line y=x-1 which is tangent to the curve at the point (2,1). Since the solutions for x²+2x+1=0 are both -1; all the points on this tangent represent quadratic equations which have x=-1 as one solution. As a case in point, (or a point in this case), (5,4) is on the tangent line, and the solutions of the quadratic equation x²+5x+4=0 are x=-1 and x=-4. If you drew the other tangent to the parabola passing through the point (5,4) it would contain the point (8,16) on the parabola representing the quadratic with a double root at x=-4, that is, the quadratic x²+8x+16=0.
 

Desmos 

The reflection of this tangent in the y-axis gives the tangent through the point representing the equation x^2-2x+1, with the double root at x=+1.  At this point your students should know the point of intersection of the tangent at (8,16) and the tangent at (-2,1).

This would also allow a simple approximation method to find the solutions of any quadratic with reasonably small values using only a straightedge and a printed graph of the curve y=1/4 x². Simply pick the point (b,c) corresponding to the values of the equation, and then lay a straight edge through the point and tangent to the parabola. The solutions would be -1/2 the x values of the points of tangentcy, as given above.