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
.
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