How to solve a quartic equation

Begin with a general quartic equation

x^4+bx^3+cx^2+dx+e=0,\quad\quad\quad(*)

depress it using the substitution

x = y - \frac{b}{4},

creates a new quartic equation (without the y^3 term) in y:

y^4+qy^2+ry+s=0\quad\quad\quad(**)

where

\begin{cases}q=c-\frac{3b^2}{8}\\r=d-\frac{bc}{2}+\frac{b^3}{8}\\s=e-\frac{bd}{4}+\frac{b^2c}{16}-\frac{3b^4}{256}\end{cases}

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When r=0, i.e., d-\frac{bc}{2}+\frac{b^3}{8}=0, the quartic equation in y becomes a quadratic equation in y^2:

(y^2)^2+qy^2+s=0.

It follows that

y^2 = \frac{-q\pm \sqrt{q^2-4s}}{2} \implies y=\pm\sqrt{\frac{-q\pm\sqrt{q^2-4s}}{2}}\implies x=y-\frac{b}{2}.

For example, to solve quartic equation x^4+2x^3-3x^2-4x+4=0, depress it using the substitution x = y-\frac{1}{2}, we obtain

y^4-\frac{9}{2}y^2+\frac{81}{16}=0.

This is a quadratic equation in y^2:

(y^2)^2-\frac{9}{2}y^2+\frac{81}{16}=0.

Therefore,

y^2 = \frac{\frac{9}{2} \pm \sqrt{(\frac{9}{4})^2-4\cdot 1\cdot \frac{81}{16}}}{2}=\frac{9}{4}\implies y = \pm\sqrt{\frac{9}{4}} = \pm\frac{3}{2}.

Consequently,

x_1 = \frac{3}{2}-\frac{1}{2} = 1, x_2=-\frac{3}{2}-\frac{1}{2} = -2.

In general, we solve the depressed quartic equation as follows:

Substituting u+v+w for y in the left side of (**) yields

y^4+qy^2+ry+s

By \begin {cases} y^2 = (u+v+w)^2=u^2+v^2+w^2+2(uv+uw+vw),\\ y^4 = (y^2)^2 = \left((u^2+v^2+w^2)+2(uv+uw+vw)\right)^2 \\= (u^2+v^2+w^2)^2+4(u^2+v^2+w^2)(uv+uw+vw) + 4(uv+uw+vw)^2 \\= (u^2+v^2+w^2)^2+2(u^2+v^2+w^2)\cdot 2(uv+uw+vw) + 4(uv+uw+vw)^2\end {cases}

= \underbrace{(u^2+v^2+w^2)^2 + 4(u^2+v^2+w^2)(uv+uw+vw)+4(uv+uw+vw)^2}_{y^4}+ q\underbrace{(u^2+v^2+w^2+2(uv+uw+vw))}_{y^2}+ r\underbrace{(u+v+w)}_{y}+s

= (u^2+v^2+w^2)^2+4(u^2+v^2+w^2)(uv+uw+vw)+4(uv+uw+vw)^2 + +q(u^2+v^2+w^2)+2q(uv+vw+uw) + r(u+v+w)+s

= (u^2+v^2+w^2)  +q(u^2+v^2+w^2) + 4(u^2+v^2+w^2)(uv+uw+vw) + 2q(uv+uw+vw) +4(uv+uw+vw)^2 + r(u+v+w)+s

= (u^2+v^2+w^2) + q(u^2+v^2+w^2) + 2(uv+uw+vw)(2(u^2+v^2+w^2) + q) +4(u^2v^2+u^2w^2+v^2w^2 + 2u^2vw + 2v^2uw +2w^2uv) + q(u^2+v^2+w^2)+r(u+v+w)+s

= (u^2+v^2+w^2) + q(u^2+v^2+w^2) + 2(uv+uw+vw)(2(u^2+v^2+w^2) + q) +4(u^2v^2+u^2w^2+v^2w^2) + 8u^2vw + 8v^2uw +8w^2uv) + q(u^2+v^2+w^2)+r(u+v+w)+s

= u^2+v^2+w^2) + q(u^2+v^2+w^2) + 2(uv+uw+vw)(2(u^2+v^2+w^2) + q) +4(u^2v^2+u^2w^2+v^2w^2) + 8u^2vw + 8v^2uw +8w^2uv + r(u+v+w)+s

=(u^2+v^2+w^2) + q(u^2+v^2+w^2) + 2(uv+uw+vw)(2(u^2+v^2+w^2) + q) +4(u^2v^2+u^2w^2+v^2w^2) + 8uvw(u + v + w)  + r(u+v+w)+s

= (u^2+v^2+w^2) + q(u^2+v^2+w^2) + 2(uv+uw+vw)(\underline{2(u^2+v^2+w^2) + q}) +4(u^2v^2+u^2w^2+v^2w^2) + (\underline{8uvw+r})(u + v + w) +s

If \begin{cases} 2(u^2+v^2+w^2) + q=0 \\ 8uvw + r=0\end{cases}\quad\quad\quad(1)

=(u^2+v^2+w^2)^2+q(u^2+v^2+w^2) + 4(u^2v^2+u^2w^2+v^2w^2)+s

Since (1) \implies \begin{cases} u^2+v^2+w^2 = -\frac{q}{2}\\ uvw = -\frac{r}{8}\end{cases}\quad\quad\quad(2)

= (-\frac{q}{2})^2 + q \cdot (-\frac{q}{2}) + 4(u^2v^2+u^2w^2+v^2w^2)+s

= \frac{q^2}{4} - \frac{q^2}{2} + 4(u^2v^2+u^2w^2+v^2w^2)+s

= -\frac{q^2}{4} + s + 4(u^2v^2+u^2w^2+v^2w^2)

= -\frac{q^2-4s}{4} + 4(\underline{u^2v^2+u^2w^2+v^2w^2})

If u^2v^2 + u^2w^2+v^2w^2 = \frac{q^2-4s}{16}\quad\quad\quad(3)

= 0.

It means that u+v+w is a solution of (**) if

\begin{cases} u^2+v^2+w^2=-\frac{q}{2} \\ u^2v^2+u^2v^2+v^2w^2=\frac{q^2-4s}{16}\\ uvw=-\frac{r}{8}\end{cases}\quad\quad\quad(4-1, 4-2, 4-3)

Moreover, squaring (4-3) gives

\begin{cases} u^2+v^2+w^2=-\frac{q}{2} \\ u^2v^2+u^2v^2+v^2w^2=\frac{q^2-4s}{16}\\ u^2v^2w^2=\frac{r^2}{64}\end{cases}\quad\quad\quad(5-1, 5-2, 5-3)

By Vieta’s theorem, u^2, v^2, w^2 satisfying (5-1, 5-2, 5-3) are the three solutions of cubic equation

z^3+\frac{q}{2}z^2+\frac{q^2-4s}{16}z-\frac{r^2}{64}= 0.

(See “How to solve a cubic equation“)

Suppose the three solutions are z_1, z_2, z_3, we have

u^2=z_1 \implies u = \pm\sqrt{z_1}, v^2 = z_2 \implies v = \pm\sqrt{z_2}, w^2=z_3 \implies w=\pm \sqrt{z_3}.

Clearly, there are eight combinations of u,v,w:

\begin{cases} u=\sqrt{z_1} \\  v= \sqrt{z_2} \\w=\sqrt{z_3} \end{cases} \begin{cases} u=\sqrt{z_1} \\  v= \sqrt{z_2} \\w=-\sqrt{z_3} \end{cases} \begin{cases} u=\sqrt{z_1} \\  v= -\sqrt{z_2} \\w=\sqrt{z_3} \end{cases} \begin{cases} u=\sqrt{z_1} \\  v= -\sqrt{z_2} \\w=-\sqrt{z_3} \end{cases}

\begin{cases} u=-\sqrt{z_1} \\  v= \sqrt{z_2} \\w=\sqrt{z_3} \end{cases} \begin{cases} u=-\sqrt{z_1} \\  v= \sqrt{z_2} \\w=-\sqrt{z_3} \end{cases} \begin{cases} u=-\sqrt{z_1} \\  v= -\sqrt{z_2} \\w=\sqrt{z_3} \end{cases} \begin{cases} u=-\sqrt{z_1} \\  v= -\sqrt{z_2} \\w=-\sqrt{z_3} \end{cases}

Among them, only four are valid due to constraint (4-3) placed on the product u\cdot v\cdot w.

From (5-3), we see uvw = \sqrt{z_1}\sqrt{z_2}\sqrt{z_3}= \pm\frac{r}{8}.

If \sqrt{z1}\sqrt{z_2}\sqrt{z_3} = +\frac{r}{8}, the valid ones are:

\begin{cases} u=\sqrt{z_1} \\  v= \sqrt{z_2} \\w=-\sqrt{z_3} \end{cases} since uvw=\sqrt{z_1}\cdot \sqrt{z_2}\cdot -\sqrt{z_3}=-\sqrt{z_1}\sqrt{z_2}\sqrt{z_3} = -\frac{r}{8}

\begin{cases} u=\sqrt{z_1} \\  v= -\sqrt{z_2} \\w=\sqrt{z_3} \end{cases} since uvw=\sqrt{z_1}\cdot -\sqrt{z_2}\cdot \sqrt{z_3}=-\sqrt{z_1}\sqrt{z_2}\sqrt{z_3} = -\frac{r}{8}

\begin{cases} u=-\sqrt{z_1} \\  v= \sqrt{z_2} \\w=\sqrt{z_3} \end{cases} since uvw=-\sqrt{z_1}\cdot \sqrt{z_2}\cdot \sqrt{z_3}=-\sqrt{z_1}\sqrt{z_2}\sqrt{z_3} = -\frac{r}{8}

\begin{cases} u=-\sqrt{z_1} \\  v=-\sqrt{z_2} \\w=-\sqrt{z_3} \end{cases} since uvw=-\sqrt{z_1}\cdot -\sqrt{z_2}\cdot -\sqrt{z_3}=-\sqrt{z_1}\sqrt{z_2}\sqrt{z_3} = -\frac{r}{8}

Otherwise \sqrt{z_1}\sqrt{z_2}\sqrt{z_3}=-\frac{r}{8}, we have

\begin{cases} u=\sqrt{z_1} \\  v= \sqrt{z_2} \\w=\sqrt{z_3} \end{cases} \begin{cases} u=\sqrt{z_1} \\  v= -\sqrt{z_2} \\w=-\sqrt{z_3} \end{cases} \begin{cases} u=-\sqrt{z_1} \\  v= \sqrt{z_2} \\w=-\sqrt{z_3} \end{cases} \begin{cases} u=-\sqrt{z_1} \\  v= -\sqrt{z_2} \\w=\sqrt{z_3} \end{cases}

Consequently, a solution to the general quartic equation (*) is

x = u+v+w - \frac{b}{4}.

There are four such solutions.


Exercise-1 Show that (5-3) \implies uvw = \sqrt{z_1}\sqrt{z_2}\sqrt{z_3}= \pm\frac{r}{8}.

Through the Mind’s Eye

Let’s derive Cardano’s formula for the depressed cubic equation x^3+px=q.

Imaging a large cube with length u. It is comprised of six blocks. Each block has its label on the top.

The large cube’s volume, u^3 is the sum of six smaller pieces:

v^3+v^2(u-v) + v(u-v)u + v(u-v)u+(u-v)^2(u-v)+(u-v)^2v= u^3

That is,

(u-v)^3+\underline{2uv(u-v)}+\underline{v^2(u-v)}+\underline{(u-v)^2v}=u^3-v^3.

Factoring u-v from the underlined terms above gives

(u-v)^3+(2uv + v^2 + (u-v)v)(u-v)=u^3-v^3

or simply,

(u-v)^3+\underbrace{3uv}_{p}(u-v)=\underbrace{u^3-v^3}_{q}.\quad\quad\quad(1)

From the cube and (1), we see that for u>v>0, if 3uv = p and u^3-v^3=q, then u-v, a positive quantity, is a solution of x^3 + px =q.

However, (1) is an algebraic identity regardless of the cube that led us to it. i.e.,

(1) is true for all u, v \in R.

Hence,

\forall u,v \in R, (3uv = q, u^3-v^3=p) \implies u-v, not necessary a positive quantity, is a solution of depressed cubic x^3+px=q.

It follows that to find a solution of x^3+px=q, we solve

\begin{cases} 3uv=p \\ u^3-v^3=q \end{cases}

for u, v, after which the Cardano formula emerges:

x=u-v=\sqrt[3]{\frac{q}{2}+\sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} - \sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}.\quad\quad\quad(\star)


Exercise-1 Show that

u-v=\sqrt[3]{\frac{q}{2}+\sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} - \sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}

after solving

\begin{cases} 3uv=p \\ u^3-v^3=q\end{cases}

for u, v.