# Descartes's Geometric Solution of a Quadratic Equation

Descartes's Geometric Solution of a Quadratic Equation

This Demonstration shows Descartes's geometric solution of the quadratic equation =az- 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 .

2

z

2

b

z

a/2

M

L

(b,-a/2)

(0,-a/2)

y

L

M

Q

R

MQ

MR

z

1

z

2

b>

a

2

b

b=

a

2

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

z=a±-4

1

2

2

a

2

b