Ferrari's Method

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Theorem

Let $P$ be the quartic equation:

$a x^4 + b x^3 + c x^2 + d x + e = 0$

such that $a \ne 0$.


Then $P$ has solutions:

$x = \dfrac {-p \pm \sqrt {p^2 - 8 q}} 4$

where:

$p = \dfrac b a \pm \sqrt {\dfrac {b^2} {a^2} - \dfrac {4 c} a + 4 y_1}$
$q = y_1 \mp \sqrt {y_1^2 - \dfrac {4 e} a}$

where $y_1$ is a real solution to the cubic:

$y^3 - \dfrac c a y^2 + \left({\dfrac {b d} {a^2} - \dfrac {4 e} a}\right) y + \left({\dfrac {4 c e} {a^2} - \dfrac {b^2 e} {a^3} - \dfrac {d^2} {a^2}}\right) = 0$


Ferrari's method is a technique for solving this quartic.


Proof

First we render the quartic into monic form:

$x^4 + \dfrac b a x^3 + \dfrac c a x^2 + \dfrac d a x + \dfrac e a = 0$


Completing the square in $x^2$:

$\left({x^2 + \dfrac b {2a} x}\right)^2 + \left({\dfrac c a - \dfrac {b^2} {4 a^2}}\right) x^2 + \dfrac d a x + \dfrac e a = 0$

Then we introduce a new variable $y$:

$\left({x^2 + \dfrac b {2a} x + \dfrac y 2}\right)^2 + \left({\dfrac c a - \dfrac {b^2} {4 a^2} - y}\right) x^2 + \left({\dfrac d a - \dfrac b {2a} y}\right) x + \left({\dfrac e a - \dfrac {y^2} 4}\right) = 0$

This equation is valid for any $y$, so let us pick a value of $y$ so as to make:

$\left({\dfrac c a - \dfrac {b^2} {4 a^2} - y}\right) x^2 + \left({\dfrac d a - \dfrac b {2a} y}\right) x + \left({\dfrac e a - \dfrac {y^2} 4}\right)$

have a zero discriminant.

That is:

$\left({\dfrac d a - \dfrac b {2a} y}\right)^2 = 4 \left({\dfrac c a - \dfrac {b^2} {4 a^2} - y}\right) \left({\dfrac e a - \dfrac {y^2} 4}\right)$

After some algebra, this can be expressed as a cubic in $y$:

$y^3 - \dfrac c a y^2 + \left({\dfrac {b d} {a^2} - \dfrac {4 e} a}\right) y + \left({\dfrac {4 c e} {a^2} - \dfrac {b^2 e} {a^3} - \dfrac {d^2} {a^2}}\right) = 0$

Using (for example) Cardano's Formula, we can find a real solution of this: call it $y_1$.


Now a quadratic equation $p x^2 + q x + r$ can be expressed as:

$p \left({\left({x + \dfrac q {2p}}\right)^2 - \dfrac {q^2 - 4 p r} {4 p^2}}\right)$

If that quadratic has a zero discriminant, i.e. $q^2 = 4 p r$, then this reduces to:

$p \left({\left({x + \dfrac q {2p}}\right)^2}\right)$

... which in turn becomes:

$p \left({\left({x + \pm \sqrt{\dfrac r p}}\right)^2}\right)$

as $q^2 = 4 p r \implies \dfrac {q^2} {4 p^2} = \dfrac r p$.

So, as:

$\left({\dfrac c a - \dfrac {b^2} {4 a^2} - y_1}\right) x^2 + \left({\dfrac d a - \dfrac b {2a} y_1}\right) x + \left({\dfrac e a - \dfrac {y_1^2} 4}\right)$

has a zero discriminant (we picked $y_1$ to make that happen), we can write it as:

$\left({\dfrac c a - \dfrac {b^2} {4 a^2} - y_1}\right)\left({x \pm \dfrac {\sqrt{\left({\dfrac e a - \dfrac {y_1^2} 4}\right)}} {\sqrt{\left({\dfrac c a - \dfrac {b^2} {4 a^2} - y_1}\right)}}}\right)^2$


Now we return to the equation:

$\left({x^2 + \dfrac b {2a} x + \dfrac {y_1} 2}\right)^2 + \left({\dfrac c a - \dfrac {b^2} {4 a^2} - y_1}\right) x^2 + \left({\dfrac d a - \dfrac b {2a} y_1}\right) x + \left({\dfrac e a - \dfrac {y_1^2} 4}\right) = 0$

which can now be written:

$\left({x^2 + \dfrac b {2a} x + \dfrac {y_1} 2}\right)^2 = \left({\dfrac {b^2} {4 a^2} - \dfrac c a + y_1}\right) \left({x \mp \dfrac {\sqrt{\left({\dfrac {y_1^2} 4 - \dfrac e a}\right)}} {\sqrt{\left({\dfrac {b^2} {4 a^2} - \dfrac c a + y_1}\right)}}}\right)^2$

Taking square roots of both sides:

$x^2 + \dfrac b {2a} x + \dfrac {y_1} 2 = \pm x \sqrt {\left({\dfrac {b^2} {4 a^2} - \dfrac c a + y_1}\right)} \mp \sqrt{\dfrac {y_1^2} 4 - \dfrac e a}$

Arranging into canonical quadratic form:

$(1): \quad x^2 + \left({\dfrac b {2a} \pm \dfrac 1 2 \sqrt {\dfrac {b^2} {a^2} - \dfrac {4 c} a + 4 y_1}}\right) x + \dfrac 1 2 \left({y_1 \mp \sqrt{y_1^2 - \dfrac {4 e} a}}\right) = 0$


Let:

$p = \dfrac b a \pm \sqrt {\dfrac {b^2} {a^2} - \dfrac {4 c} a + 4 y_1}$
$q = y_1 \mp \sqrt {y_1^2 - \dfrac {4 e} a}$


Then equation $(1)$ can be written as:

$x^2 + \dfrac p 2 x + \dfrac q 2 = 0$


Using the Quadratic Formula, putting $a = 1, b = \dfrac p 2, c = \dfrac q 2$:

\(\displaystyle x\) \(=\) \(\displaystyle \dfrac {- \dfrac p 2 \pm \sqrt {\dfrac {p^2} 4 - 4 \dfrac q 2} } 2\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {- \dfrac p 2 \pm \sqrt {\dfrac 1 4} \sqrt {p^2 - 8 q} } 2\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {- p \pm \sqrt {p^2 - 8 q} } 4\)


Hence the result.

$\blacksquare$


Source of Name

This entry was named for Lodovico Ferrari.


Historical Note

Ferrari's Method was published by Gerolamo Cardano in $1545$, in his Artis Magnae, Sive de Regulis Algebraicis, on which Lodovico Ferrari collaborated.


Sources