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The concept of Vieta's formula can be found in the work of the 12th century Arabic mathematician Sharaf al-Din al-Tusi. It is plausible that the algebraic advancements made by Arabic mathematicians such as al-Khayyam, al-Tusi, and al-Kashi influenced 16th-century algebraists, with Vieta being the most prominent among them. [2]
The formula can be derived as a telescoping product of either the areas or perimeters of nested polygons converging to a circle. Alternatively, repeated use of the half-angle formula from trigonometry leads to a generalized formula, discovered by Leonhard Euler, that has Viète's formula as a special case. Many similar formulas involving nested ...
Since α, β, and γ are the roots of , it is a consequence of Vieta's formulas that their product is equal to q 2 and therefore that √ α √ β √ γ = ±q. But a straightforward computation shows that √ α √ β √ γ = r 1 r 2 r 3 + r 1 r 2 r 4 + r 1 r 3 r 4 + r 2 r 3 r 4.
In mathematics, a quartic equation is one which can be expressed as a quartic function equaling zero. The general form of a quartic equation is The general form of a quartic equation is Graph of a polynomial function of degree 4, with its 4 roots and 3 critical points .
Quadratic formula; Quartic equation; Quartic function; R. Rationalisation (mathematics) S. ... Vieta's formulas; Z. Zero-product property This page was ...
In mathematics, especially in algebraic geometry, a quartic surface is a surface defined by an equation of degree 4. More specifically there are two closely related types of quartic surface: affine and projective.
Vieta may refer to: François Viète (1540–1603), commonly known by the Latin form of his name Franciscus Vieta, a French mathematician; Vieta (crater), a crater on the Moon, named after him; Vieta's formulas, expressing the coefficients of a polynomial as signed sums and products of its roots. Artūras Vieta (born 1961), Lithuanian sprint canoer
Replace some a i by a variable x in the formulas, and obtain an equation for which a i is a solution. Using Vieta's formulas, show that this implies the existence of a smaller solution, hence a contradiction. Example. Problem #6 at IMO 1988: Let a and b be positive integers such that ab + 1 divides a 2 + b 2. Prove that a 2 + b 2 / ab + 1 ...