# 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 .

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Contributed by: Bryan Chen (January 2016)

Open content licensed under CC BY-NC-SA

## Snapshots

## Details

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.

Reference

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

## Permanent Citation