Difference between revisions of "Ferrari method"
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$$x_0 = \frac{q}{4\a_0},$$ | $$x_0 = \frac{q}{4\a_0},$$ | ||
which leads to the equation | which leads to the equation | ||
− | $$(x^2+\frac{p}{2}+\a_0)^2 - 2\a_0(x-x_0)^2 = 0.$$ | + | $$\big(x^2+\frac{p}{2}+\a_0\big)^2 - 2\a_0(x-x_0)^2 = 0.$$ |
This | This | ||
equation of degree 4 splits into two quadratic equations. The roots of | equation of degree 4 splits into two quadratic equations. The roots of |
Revision as of 18:45, 12 December 2014
A method for reducing the solution of an equation of degree 4 to the solution of one cubic and two quadratic equations; it was discovered by L. Ferrari (published in 1545).
The Ferrari method for the equation $$y^4 + a y^3 + by^2 + cy + d = 0$$ consists in the following. By the substitution $y=x-a/4$ the given equation can be reduced to $$x^4+px^2 + qx +r = 0,\label{1}$$ which contains no term in $x^3$. If one introduces an auxiliary parameter $\def\a{\alpha}\a$, the left-hand side of (1) can be written as $$x^4+px^2+qx+r =$$
$$=(x^2+\frac{p}{2}+\a)^2-\big[2\a x^2 - qx +\big(\a^2+p\a+\frac{p^2}{4}-r\big)\big].\label{2}$$ One then chooses a value of $\a$ such that the expression in square brackets is a perfect square. For this the discriminant of the quadratic trinomial must vanish. This gives a cubic equation for $\a$, $$q^2-4\cdot2\a\big(\a^2+p\a+\frac{p^2}{4}-r\big)=0.$$ Let $\a_0$ be one of the roots of this equation. For $\a=\a_0$ the polynomial in square brackets in (2) has one double root, $$x_0 = \frac{q}{4\a_0},$$ which leads to the equation $$\big(x^2+\frac{p}{2}+\a_0\big)^2 - 2\a_0(x-x_0)^2 = 0.$$ This equation of degree 4 splits into two quadratic equations. The roots of these equations are also the roots of (1).
References
[1] | A.G. Kurosh, "Higher algebra" , MIR (1972) (Translated from Russian) |
Ferrari method. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Ferrari_method&oldid=35576