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Descartes's Geometric Solution of a Quadratic Equation

​
a
25
b
8
z
1
= MR = 22.1047
z
2
= MQ = 2.89531
This Demonstration shows Descartes's geometric solution of the quadratic equation
2
z
=az-
2
b
in the unknown
z
. Consider a circle of radius
a/2
and let the points
M
and
L
be at
(b,-a/2)
and
(0,-a/2)
; the circle meets the negative
y
axis at
L
. Let the vertical line through
M
intersect the circle at
Q
and
R
. The solutions are then given by the intersections of the circle and the line. Thus the lengths
MQ
and
MR
are the two roots
z
1
and
z
2
of the original quadratic equation. When
b>
a
2
, the vertical line does not intersect the circle, meaning that the solutions to the quadratic are complex numbers. The
b
slider is therefore stopped at
b=
a
2
.
The well-known exact solutions to the above quadratic equation are
z=
1
2
a±
2
a
-4
2
b
.
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