Vieta's Solution of a Cubic Equation

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This Demonstration shows Vieta's solution of the depressed cubic equation , where . To solve it, draw an isosceles triangle with base and unit legs. Let be the angle at the base and . Draw a second isosceles triangle with base angle and unit legs. The base of the second triangle is a root of the equation.



By construction, , . Let . Since and are similar, . The perpendicular projections of and on also give similar right-angled triangles. So . So and . Eliminating gives .

Since , and applying the law of cosines to , we find .


Contributed by: Izidor Hafner (October 2017)
Open content licensed under CC BY-NC-SA




[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.

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