 # Descartes' Method for Constructing Roots of Polynomials with 'Simple' Curves - Quadratics

Author(s):
Gary Rubinstein (Stuyvesant High School)

### How Descartes solved quadratic equations by intersecting a circle with a horizontal line

In beginning algebra classes, students learn that they can solve quadratic equations like $x^{2}+8=6x$ by graphing the parabola $y=x^{2}-6x+8$ and locating the $x$-intercepts. The equation $x^{2}+8=6x$ can also be solved using intersections of other parabolas with lines other than the $x$-axis; for example, $y=x^{2}+8$ with $y=6x,$ or $y=x^2$ with $y=6x-8.$ Though this method works well with a graphing calculator, by hand the parabola is difficult to construct accurately enough to determine the $x$-coordinates of the intersection points.

At the beginning of Book III of his Geometry, Descartes explained (Smith 152, 155):

We should always choose with care the simplest curve that can be used in the solution of a problem, but it should be noted that the simplest means not merely the one most easily described, nor the one that leads to the easiest demonstration or construction of the problem, but rather the one of the simplest class that can be used to determine the required quantity.

With this consideration, he showed in Book I that quadratic equations could be solved using just a circle and a horizontal line.

For quadratic equations of the form $x^{2}-px+q=0,$ with $q\ge0,$ Descartes intersected the circle $\left(x-\frac{p}{2}\right)^{2}+y^{2}=\left(\frac{p}{2}\right)^{2}$ with the horizontal line $y=\sqrt{q}$. For example, intersecting $(x-3)^{2}+y^{2}=3^{2}$ with $y=\sqrt{8}$ shows that $x^{2}-6x+8=0$ has roots at $x=2$ and $x=4.$ Notice that if you substitute $\sqrt{8}$ for $y$ in the circle equation, the equation becomes $(x-3)^{2}+8=3^{2},$ which then reduces to $x^{2}-6x+8=0.$

Figure 4. Constructing solutions to quadratic equations. Instructions: Move the sliders to change $p$ and $q$ and thus to find the solutions to the quadratic equation $x^{2}-px+q=0.$ If the equation has a double root, the circle will be tangent to the horizontal line; if the equation has imaginary roots, the circle will not intersect the horizontal line at all.

Descartes actually considered three cases of quadratic equations in his 'Geometry': $x^2 = ax + b^2,$ $x^2 = -ax + b^2,$ and $x^2 = ax - b^2$ (Smith 13-15), writing the constant term as $b^2$ to preserve dimensionality. He assumed implicitly that $a>0$ and $b^2>0,$ and considered only positive roots. I have considered only his third case quadratic in the form $x^2 - ax + b^2 = 0,$ or $x^{2}-px+q=0,$ with just the restriction that $q\ge0.$ Descartes handled the possibility of imaginary roots here as follows (Smith 17):

And if the circle ... neither cuts nor touches the line ..., the equation has no root, so that we may say that the construction of the problem is impossible.

Gary Rubinstein (Stuyvesant High School), "Descartes' Method for Constructing Roots of Polynomials with 'Simple' Curves - Quadratics ," Convergence (May 2016)

## Dummy View - NOT TO BE DELETED

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