Difference between revisions of "Ferrari method"
(Importing text file) |
Ulf Rehmann (talk | contribs) m |
||
| Line 1: | Line 1: | ||
| − | 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). | + | 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 | + | 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|cubic equation]] for $\a$, | |
| − | + | $$q^2-42\a(\a^2+p\a+\frac{p^2}{4}-r)=0.$$ | |
| − | + | 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 | |
| − | + | $$(x^2+\frac{p}{2}+\a_0)^2 - 2\a_0(x-x_0)^2 = 0.$$ | |
| − | One then chooses a value of | + | This |
| − | + | equation of degree 4 splits into two quadratic equations. The roots of | |
| − | + | these equations are also the roots of (1). | |
| − | |||
| − | Let | ||
| − | |||
| − | |||
| − | |||
| − | which leads to the equation | ||
| − | |||
| − | |||
| − | |||
| − | |||
====References==== | ====References==== | ||
| − | <table><TR><TD valign="top">[1]</TD> <TD valign="top"> | + | <table><TR><TD valign="top">[1]</TD> |
| + | <TD valign="top"> A.G. Kurosh, "Higher algebra" , MIR (1972) (Translated from Russian)</TD> | ||
| + | </TR></table> | ||
Revision as of 23:02, 21 November 2011
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 $\a$, $$q^2-42\a(\a^2+p\a+\frac{p^2}{4}-r)=0.$$ 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 $$(x^2+\frac{p}{2}+\a_0)^2 - 2\a_0(x-x_0)^2 = 0.$$ 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=14615