扩展：根式法求四次方程

We want to solve the equation $a x^4+b x^3+c x^2+d x+e=0$, assuming that $a\ne0$ so that it really is a quartic equation. First, as with equations of lower degree, we divide through by the leading coefficient: $x^4+\frac{b}{a} x^3+\frac{c}{a} x^2+\frac{d}{a} x+\frac{e}{a}=0$

Now that the equation is monic, we can factor our quartic: If $\alpha$, $\beta$, $\gamma$, and $\delta$ are the roots of our quartic, since it is monic we have: $x^4+\frac{b}{a} x^3+\frac{c}{a} x^2+\frac{d}{a} x+\frac{e}{a}=\left(x-\alpha\right)\left(x-\beta\right)\left(x-\gamma\right)\left(x-\delta\right)$ This holds true for all $x$, so we can equate coefficients of powers of $x$:

$-\alpha-\beta-\gamma-\delta=\frac{b}{a}$ so $\alpha+\beta+\gamma+\delta=-\frac{b}{a}$

$\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=\frac{c}{a}$

$-\alpha\beta\gamma-\alpha\beta\delta-\alpha\gamma\delta-\beta\gamma\delta=\frac{d}{a}$ so $\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta=-\frac{d}{a}$

$\alpha\beta\gamma\delta=\frac{e}{a}$

Now comes the key to this derivation, which is making a convenient substitution. Specifically, since

\begin{pmatrix}

1 & 1 & 1 & 1 \ 1 & 1 & - 1 & -1 \ 1 & -1 & 1 & -1 \ 1 & -1 & -1 & 1

\end{pmatrix}

is an invertible matrix, we can substitute:

$\alpha=\epsilon+\zeta+\eta+\theta$

$\beta=\epsilon+\zeta-\eta-\theta$

$\gamma=\epsilon-\zeta+\eta-\theta$

$\delta=\epsilon-\zeta-\eta+\theta$

Note that the product of the signs on $\zeta$, $\eta$, and $\theta$ is always 1; this will be important at the end.

This substitution allows us to solve for $\epsilon$ immediately: $\alpha+\beta+\gamma+\delta=4\epsilon=-\frac{b}{a}$, so $\epsilon=-\frac{b}{4a}$

$\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=6\epsilon^2-2\zeta^2-2\eta^2-2\theta^2=\frac{c}{a}$ so $\zeta^2+\eta^2+\theta^2=\frac{3b^2-8ac}{16a^2}$

$\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta=4\left(\epsilon\left(\epsilon^2-\zeta^2-\eta^2-\theta^2\right)+2\zeta\eta\theta\right)=-\frac{d}{a}$ so, since $\epsilon=-\frac{b}{4a}$ and $\zeta^2+\eta^2+\theta^2=\frac{3b^2-8ac}{16a^2}$, we obtain $\zeta\eta\theta=\frac{-8a^2d+4abc-b^3}{64a^3}$

$\frac{e}{a}=\alpha\beta\gamma\delta=\epsilon^4+\zeta^4+\eta^4+\theta^4-2\epsilon^2\zeta^2-2\epsilon^2\eta^2-2\epsilon^2\theta^2-2\zeta^2\eta^2-2\zeta^2\theta^2-2\eta^2\theta^2+8\epsilon\zeta\eta\theta$ so, since $\epsilon=-\frac{b}{4a}$ and $\zeta\eta\theta=\frac{-8a^2d+4abc-b^3}{64a^3}$, we obtain $\epsilon^4+\zeta^4+\eta^4+\theta^4-2\epsilon^2\zeta^2-2\epsilon^2\eta^2-2\epsilon^2\theta^2-2\zeta^2\eta^2-2\zeta^2\theta^2-2\eta^2\theta^2=\frac{32a^3e+b\left(-8a^2d+4abc-b^3\right)}{32a^4}$ Since $\zeta^2+\eta^2+\theta^2=\frac{3b^2-8ac}{16a^2}$, we can add $\epsilon^2=\frac{b^2}{16a^2}$ to get $\epsilon^2+\zeta^2+\eta^2+\theta^2=\frac{b^2-2ac}{4a^2}$.

Then squaring both sides gives $\epsilon^4+\zeta^4+\eta^4+\theta^4+2\epsilon^2\zeta^2+2\epsilon^2\eta^2+2\epsilon^2\theta^2+2\zeta^2\eta^2+2\zeta^2\theta^2+2\eta^2\theta^2=\frac{\left(b^2-2ac\right)^2}{16a^4}$

Then subtracting gives us $4\epsilon^2\zeta^2+4\epsilon^2\eta^2+4\epsilon^2\theta^2+4\zeta^2\eta^2+4\zeta^2\theta^2+4\eta^2\theta^2=\left(\epsilon^2+\zeta^2+\eta^2+\theta^2\right)^2-\left(\epsilon^4+\zeta^4+\eta^4+\theta^4-2\epsilon^2\zeta^2-2\epsilon^2\eta^2-2\epsilon^2\theta^2-2\zeta^2\eta^2-2\zeta^2\theta^2-2\eta^2\theta^2\right)=\frac{\left(b^2-2ac\right)^2}{16a^4}-\frac{e}{a}-\frac{b\left(-8a^2d+4abc-b^3\right)}{32a^4} =\frac{2\left(b^2-2ac\right)^2-32a^3e+b\left(8a^2d-4abc+b^3\right)}{32a^4}$ so we can now obtain $\epsilon^2\zeta^2+\epsilon^2\eta^2+\epsilon^2\theta^2+\zeta^2\eta^2+\zeta^2\theta^2+\eta^2\theta^2= \frac{2\left(b^2-2ac\right)^2-32a^3e+b\left(8a^2d-4abc+b^3\right)}{128a^4}$ Since $\epsilon=-\frac{b}{4a}$ and $\zeta^2+\eta^2+\theta^2=\frac{3b^2-8ac}{16a^2}$, $\epsilon^2\zeta^2+\epsilon^2\eta^2+\epsilon^2\theta^2=\epsilon^2\left(\zeta^2+\eta^2+\theta^2\right)=\frac{b^2}{16a^2}\cdot\frac{\left(3b^2-8ac\right)}{16a^2}=\frac{b^2\left(3b^2-8ac\right)}{256a^4}$

Then we can subtract to obtain $\zeta^2\eta^2+\zeta^2\theta^2+\eta^2\theta^2=\frac{4\left(b^2-2ac\right)^2-64a^3e+2b\left(8a^2d-4abc+b^3\right)+b^2\left(8ac-3b^2\right)}{256a^4}$

This means $\zeta^2$, $\eta^2$, and $\theta^2$ are the roots of the cubic $\left(y-\zeta^2\right)\left(y-\eta^2\right)\left(y-\theta^2\right)=y^3-\left(\zeta^2+\eta^2+\theta^2\right)y^2+\left(\zeta^2\eta^2+\zeta^2\theta^2+\eta^2\theta^2\right)y-\left(\zeta\eta\theta\right)^2=y^3+\frac{\left(8ac-3b^2\right)}{16a^2}y^2+\frac{4\left(b^2-2ac\right)^2-64a^3e+2b\left(8a^2d-4abc+b^3\right)+b^2\left(8ac-3b^2\right)}{256a^4}y-\frac{\left(8a^2d-4abc+b^3\right)^2}{4096a^6}$. The importance of the product of signs in the substitutions defining $\epsilon$, $\zeta$, $\eta$, and $\theta$ is that just four sign combinations give roots of our original quartic, and they are the ones making $(\pm\zeta)( \pm\eta)( \pm\theta) = \zeta\eta\theta=\frac{-8a^2d+4abc-b^3}{64a^3}$. So the solutions to our original quartic are given by $x=\frac{-b \pm\sqrt{y_1}\pm\sqrt{y_2}\pm\sqrt{y_3}}{4a}$, where the signs are chosen so that $\left(\pm\sqrt{y_1}\right)\left(\pm\sqrt{y_2}\right)\left(\pm\sqrt{y_3}\right)=-8a^2d+4abc-b^3$

(There's a reason this formula is almost never written explicitly and directly in terms of the coefficients of the original quartic.)