Descartes’ Method for Constructing Roots of Polynomials with ‘Simple’ Curves - History of Constructing Solutions Geometrically

Gary Rubinstein (Stuyvesant High School)

The ancient Greeks had a geometric system for solving many of the problems we now solve algebraically or numerically. For example, a line segment of length square root of five can be constructed by creating the mean proportional between a segment of length one and a segment of length five.

In Figure 1 below, segment \({\rm{ABC}}\) is constructed so that \({\rm{AB}}=1\) and \({\rm{BC}}=5.\) Semicircle \({\rm{AC}}\) is then constructed on diameter \({\rm{AC}}.\) If \({\rm{BD}}\) is constructed perpendicular to \({\rm{AC,}}\) then because \({\rm{BD}}\) is the altitude to hypotenuse \({\rm{AC}}\) (of right triangle \({\rm{ADC}}\)), \({\rm{BD}}\) will be the mean proportional between \({\rm{AB}}\) and \({\rm{BC.}}\) In this case \[{\frac{1}{{\rm{BD}}}}={\frac{{\rm{BD}}}{5}}\quad {\rm{or}}\quad {\rm{BD}}^2 = 5\quad {\rm{and}} \quad {\rm{BD}}=\sqrt5.\]

Figure 1. Constructing a segment of length square root of \(n\). Instructions: Drag the point on the slider to change the value of \(n.\)

Along with ‘squaring the circle’ and ‘trisecting the angle,’ the third great problem of antiquity was ‘doubling the cube.’ This problem required constructing a line segment, using only compass and straightedge, so that a cube constructed on that segment would have twice the volume of a cube constructed on a given segment. In modern terms, this requires constructing a segment of length equal to the length of the edge of the original cube multiplied by \(\sqrt[3]{2}.\) The square root construction above is based on the fact that if \({\frac{1}{x}}={\frac{x}{a}}\) (for \(a>0),\) then \(x=\sqrt{a}\). Similarly, Hippocrates (around 450 BCE) realized that finding the cube root of \(a\) is equivalent to finding two mean proportionals between \(1\) and \(a\) (Knorr 23-24). That \({\frac{1}{x}}={\frac{x}{y}}={\frac{y}{a}}\) means that \(y=x^2\) and \(ax=y^2.\) Substituting for \(y,\) we get \(ax=x^4\) so, for \(x>0,\) \(x^3=a.\) Constructing this \(x\) turned out to be impossible if you are limited to compass and straight edge. Menaechmus (around 350 BCE), however, figured out that if you allow the use of other curves, you can construct a segment of this length. Finding the \(x\)-coordinate of the intersection point of the graphs of the two equations \(y=x^2\) and \(ax=y^2\) (both now known to be parabolas) will provide a solution to the problem (Knorr 61).

In Figure 2 below we have graphs of \(y=x^2\) and \(2x=y^2,\) which intersect at a point with an \(x\)-coordinate of \({\sqrt[3]{2}}.\) These parabolas, however, could only be constructed ‘pointwise’ with tools other than compass and straightedge, so the classical ‘doubling the cube’ problem remained unsolved.

Figure 2. Constructing the cube root of the positive integer \(n\) with parabolas \(y=x^2\) and \(nx=y^2,\) which intersect at the origin and at a point with \(x\)-coordinate \(\sqrt[3]{n}.\) Instructions: Drag the point on the slider to change the value of \(n.\)

Thirteen centuries later (around 1100 CE), Omar Khayyam solved cubic equations with different intersections of parabolas, circles, and hyperbolas. Khayyam had 13 different methods depending on which of the coefficients of the equations were positive and the methods were not unified into a coherent general method.

Figure 3. Omar Khayyam's solution to a cubic equation via intersection of a circle with a hyperbola. Instructions: Drag sliders for \(b, c,\) and \(d\) to see Khayyam’s method for graphically solving the equation \(x^3+bx^2+cx=d.\)

The science of solving equations was thoroughly developed by Descartes. A large part of Descartes’ 1637 ‘Geometry’ was devoted to solving polynomial equations of different degrees using the ‘simplest’ and most ‘natural’ curves possible. His solution to quadratic equations did not require parabolas, but simply circles and lines. Cubics and quartics could be solved by intersecting parabolas with circles. Quintics and sextics could be solved by intersecting what are now known as ‘Cartesian parabolas’ with circles.