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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).
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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
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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
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$$x^4+px^2 + qx +r = 0,\label{1}$$
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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
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$$x^4+px^2+qx+r =$$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/f/f038/f038460/f0384601.png" /></td> </tr></table>
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$$=(x^2+\frac{p}{2}+\a)^2-\big[2\a x^2 - qx +\big(\a^2+p\a+\frac{p^2}{4}-r\big)\big].\label{2}$$
 
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One then chooses a value of $\a$ such that the expression in
consists in the following. By the substitution <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/f/f038/f038460/f0384602.png" /> the given equation can be reduced to
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square brackets is a perfect square. For this the discriminant of the
 
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quadratic trinomial must vanish. This gives a
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/f/f038/f038460/f0384603.png" /></td> <td valign="top" style="width:5%;text-align:right;">(1)</td></tr></table>
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[[Cubic equation|cubic equation]] for $\a$,  
 
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$$q^2-42\a(\a^2+p\a+\frac{p^2}{4}-r)=0.$$
which contains no term in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/f/f038/f038460/f0384604.png" />. If one introduces an auxiliary parameter <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/f/f038/f038460/f0384605.png" />, the left-hand side of (1) can be written as
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Let $\a_0$ be one of the
 
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roots of this equation. For $\a=\a_0$ the polynomial in square brackets in
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/f/f038/f038460/f0384606.png" /></td> <td valign="top" style="width:5%;text-align:right;">(2)</td></tr></table>
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(2) has one double root,  
 
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$$x_0 = \frac{q}{4\a_0},$$
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/f/f038/f038460/f0384607.png" /></td> </tr></table>
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which leads to the equation  
 
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$$(x^2+\frac{p}{2}+\a_0)^2 - 2\a_0(x-x_0)^2 = 0.$$
One then chooses a value of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/f/f038/f038460/f0384608.png" /> 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 <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/f/f038/f038460/f0384609.png" />,
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This
 
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equation of degree 4 splits into two quadratic equations. The roots of
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/f/f038/f038460/f03846010.png" /></td> </tr></table>
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these equations are also the roots of (1).
 
 
Let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/f/f038/f038460/f03846011.png" /> be one of the roots of this equation. For <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/f/f038/f038460/f03846012.png" /> the polynomial in square brackets in (2) has one double root,
 
 
 
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/f/f038/f038460/f03846013.png" /></td> </tr></table>
 
 
 
which leads to the equation
 
 
 
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/f/f038/f038460/f03846014.png" /></td> </tr></table>
 
 
 
This equation of degree 4 splits into two quadratic equations. The roots of these equations are also the roots of (1).
 
  
 
====References====
 
====References====
<table><TR><TD valign="top">[1]</TD> <TD valign="top"> A.G. Kurosh,   "Higher algebra" , MIR (1972) (Translated from Russian)</TD></TR></table>
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<table><TR><TD valign="top">[1]</TD>
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<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)
How to Cite This Entry:
Ferrari method. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Ferrari_method&oldid=19674
This article was adapted from an original article by I.V. Proskuryakov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article