Vieta's Solution of a Cubic Equation

b

0.4

more labels

This Demonstration shows Vieta's solution of the depressed cubic equation

3

x

-3x=b

, where

0<b<2

. To solve it, draw an isosceles triangle

CFG

with base

b

and unit legs. Let

α

be the angle at the base and

ϕ=α/3

. Draw a second isosceles triangle

DCH

with base angle

ϕ

and unit legs. The base of the second triangle is a root of the equation.

Proof

By construction,

DH=HC=CG=FG=1

,

CF=b

. Let

GH=y

. Since

ΔDAH

and

ΔDBG

are similar,

1/(x-1)=(x+1)/(y+1)

. The perpendicular projections of

H

and

G

on

DB

also give similar right-angled triangles. So

1/y=x/(x+b)

. So

y=

2

x

-2

and

xy=x+b

. Eliminating

y

gives

3

x

-3x=b

.

Since

ϕ=

1

3

arccos

b

2

, and applying the law of cosines to

ΔDCH

, we find

x=

2+2cos(2ϕ)

=

2+2cos

2

3

arccos

b

2

.

References

[1] G. E. Martin, Geometric Constructions, New York: Springer, 1998 pp. 126–140.

[2] M. Hladnik, "Some Historical Constructions of the Regular Heptagon" (in Slovenian), Obzornik za matematiko in fiziko, 61(4), 2014 pp. 132–145. www.obzornik.si/61.

External Links

Solving a Cubic via the Trisection of an Angle

Cubic Equation (Wolfram MathWorld)

Archimedes's Neusis Angle-Trisection

Permanent Citation

Izidor Hafner

"Vieta's Solution of a Cubic Equation"
http://demonstrations.wolfram.com/VietasSolutionOfACubicEquation/
Wolfram Demonstrations Project
Published: October 3, 2017