Descartes's Geometric Solution of a Quadratic Equation

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This Demonstration shows Descartes's geometric solution of the quadratic equation in the unknown . Consider a circle of radius and let the points and be at and ; the circle meets the negative axis at . Let the vertical line through intersect the circle at and . The solutions are then given by the intersections of the circle and the line. Thus the lengths and are the two roots and of the original quadratic equation. When , the vertical line does not intersect the circle, meaning that the solutions to the quadratic are complex numbers. The slider is therefore stopped at .


The well-known exact solutions to the above quadratic equation are .


Contributed by: Bryan Chen (January 2016)
Open content licensed under CC BY-NC-SA



By Pythagoras's theorem, the and components of both points and are given by , with Thus and . It follows then that and , giving the two roots of the quadratic equation.


[1] P. J. Nahin, An Imaginary Tale: The Story of i, Princeton: Princeton University Press, 1998.

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