Descartes's Solution to Quartic Equation

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:

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Proof

First we render the quartic into monic form:

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

Using a Tschirnhaus transformation, $x = y - \dfrac b {4 a}$, we convert $P$ into the reduced quartic:

$y^4 + p y^2 + q y + r = 0$

This is set identically equal to:

$\paren {y^2 + \lambda y + m} \paren {y^2 - \lambda y + n}$

Equating coefficients and eliminating $m$ and $n$ we obtain the following cubic in $\lambda^2$:

$\lambda^6 + 2 p \lambda^4 + \paren {p^2 - 4 r} \lambda^2 - q^2 = 0$

which can be solved using Cardano's Method, for example.

Hence by solving the two quadratic equations:

the solution to the reduced quartic is found.

$\blacksquare$

Also see

• Ferrari's Method

Source of Name

This entry was named for René Descartes.

Historical Note

René Descartes came up with his in $1637$.

Sources

• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): quartic (biquadratic)
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): quartic (biquadratic)