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.