Quartic Equations

The quartic formula becomes much less mysterious once one sees what problem it is trying to solve. For a cubic, Cardano found a way to write the unknown as a sum of two quantities, arranged so that the mixed terms combine into the required linear term. For a quartic, the basic aim is different: one tries to factor the polynomial into two quadratic factors. The obstacle is that the correct factorization is not visible at first. Ferrari’s decisive idea is to add a carefully chosen square to the equation until the remaining quadratic expression itself becomes a square. The condition that makes this possible is a cubic equation, called the resolvent cubic. Thus the quartic is solved in two stages: first solve one cubic, and then solve two quadratics.

Start with a monic quartic x^4+Ax^3+Bx^2+Cx+D=0. As with the cubic, the first step is to remove the next-highest term. Put x=y-\frac A4. After expanding, the equation becomes the depressed quartic

\displaystyle y^4+py^2+qy+r=0,

where p=B-\frac{3A^2}{8}, q=C-\frac{AB}{2}+\frac{A^3}{8}, r=D-\frac{AC}{4}+\frac{A^2B}{16}-\frac{3A^4}{256}.

The absence of a y^3 term is crucial. It means that, if the quartic factors into two quadratics, those quadratics must have opposite linear coefficients:y^4+py^2+qy+r =\bigl(y^2-\alpha y+\beta\bigr)\bigl(y^2+\alpha y+\gamma\bigr). Indeed, expanding the right-hand side gives y^4+\bigl(\beta+\gamma-\alpha^2\bigr)y^2 +\alpha(\beta-\gamma)y+\beta\gamma. Thus a factorization of this form must satisfy \beta+\gamma-\alpha^2=p, \quad \alpha(\beta-\gamma)=q, \quad \beta\gamma=r.

The problem is therefore to find the missing quantity \alpha. Once it is known, the sum and difference of \beta and \gamma are known, and the quartic breaks into two ordinary quadratic equations.

Ferrari’s Method

Ferrari’s method reaches this factorization by completing a square. Begin with y^4+py^2+qy+r=0, and add the terms needed to create (y^2+m)^2:

\displaystyle (y^2+m)^2=(2m-p)y^2-qy+(m^2-r).

We now try to choose m so that the right-hand side is also a perfect square. Set z=2m-p,\quad m=\frac{p+z}{2}. If z\ne0, write \alpha=\sqrt z. The coefficient of y forces the desired square to be \big(\alpha y-\frac{q}{2\alpha}\big)^2. For this to have the correct constant term, we require m^2-r=\frac{q^2}{4z}. Substituting m=(p+z)/2 and clearing denominators gives z(p+z)^2-4rz-q^2=0. Equivalently,

\displaystyle z^3+2pz^2+(p^2-4r)z-q^2=0.

This is Ferrari’s resolvent cubic. It is the exact analogue of Cardano’s quadratic resolvent for the cubic: a complicated-looking higher-degree problem has been converted into a lower-degree equation whose roots encode the missing structure.

Let z be any nonzero root of the resolvent cubic, choose \alpha=\sqrt z, and put m=\frac{p+z}{2}. Then the completed-square identity becomes \big (y^2+m\big )^2 =\big(\alpha y-\frac{q}{2\alpha}\big)^2. Taking the difference of the two squares yields the desired factorization:

\displaystyle \big(y^2-\alpha y+m+\frac{q}{2\alpha}\big) \big(y^2+\alpha y+m-\frac{q}{2\alpha}\big)=0.

Thus the four roots come from the two quadratic equations

\displaystyle y^2-\alpha y+m+\frac{q}{2\alpha}=0,

\displaystyle y^2+\alpha y+m-\frac{q}{2\alpha}=0.

Explicitly, the roots of the depressed quartic are

\displaystyle y_{1,2} =\frac{1}{2}\big( \alpha\pm\sqrt{-2p-z-\frac{2q}{\alpha}} \big),

\displaystyle y_{3,4} =\frac{1}{2}\big( -\alpha\pm\sqrt{-2p-z+\frac{2q}{\alpha}} \big).

Finally, returning to the original variable gives the roots x_j=y_j-\frac A4,\quad j=1,2,3,4.

The only genuinely new calculation in the quartic formula is therefore the solution of the resolvent cubic z^3+2pz^2+(p^2-4r)z-q^2=0. Once this cubic has been solved by Cardano’s method, the rest consists only of square roots and quadratic formulas.

There is also a more structural way to understand why a cubic appears. Let y_1,y_2,y_3,y_4 be the roots of the depressed quartic. Since the coefficient of y^3 vanishes, we have y_1+y_2+y_3+y_4=0. There are exactly three ways to divide four roots into two unordered pairs:

\displaystyle \{y_1,y_2\}\mid\{y_3,y_4\}, \quad \{y_1,y_3\}\mid\{y_2,y_4\}, \quad \{y_1,y_4\}\mid\{y_2,y_3\}.

To each pairing attach the square of one pair-sum.

\displaystyle z_1=(y_1+y_2)^2=(y_3+y_4)^2, \\ z_2=(y_1+y_3)^2=(y_2+y_4)^2, \\ z_3=(y_1+y_4)^2=(y_2+y_3)^2.

A symmetric calculation gives z_1+z_2+z_3=-2p, z_1z_2+z_1z_3+z_2z_3=p^2-4r, z_1z_2z_3=q^2. Therefore z_1,z_2,z_3 are exactly the roots of

\displaystyle z^3+2pz^2+(p^2-4r)z-q^2=0.

Thus the resolvent cubic is not an artificial byproduct of algebraic manipulation. Its three roots correspond to the three possible ways of pairing four roots. Solving the resolvent tells us how to divide the four unknown roots into two pairs; once that pairing is known, the quartic factors into the two quadratic equations associated with those pairs. This also explains the choices in Ferrari’s formula. Choosing a particular root z of the resolvent cubic chooses one pairing of the four roots. Choosing the other square root -\sqrt z merely interchanges the two quadratic factors, so it does not create new roots. Choosing a different root of the resolvent cubic produces a different pairing and hence a different factorization, but all such factorizations lead to the same set of four roots.

Euler’s version makes the pairing structure even more transparent. Suppose that z_1,z_2,z_3 are the roots of the resolvent cubic. Choose square roots a=\sqrt{z_1},\qquad b=\sqrt{z_2},\qquad c=\sqrt{z_3}, with signs chosen so that abc=-q. Then the four roots of the depressed quartic are

\displaystyle y_1=\frac{a+b+c}{2}, \\ y_2=\frac{a-b-c}{2}, \\ y_3=\frac{-a+b-c}{2}, \\ y_4=\frac{-a-b+c}{2}.

For example, if a=y_1+y_2, b=y_1+y_3, and c=y_1+y_4, then a+b+c=2y_1, because y_2+y_3+y_4=-y_1. The other three sign patterns recover y_2,y_3,y_4. This is the quartic analogue of Cardano’s use of compatible cube roots: the square roots cannot be selected independently, since their signs must satisfy the relation abc=-q. Different compatible sign choices merely reorder the four roots.

There is one important special case. If q=0, the depressed quartic is even:

\displaystyle y^4+py^2+r=0.

Then no resolvent is needed. Set Y=y^2; one obtains the quadratic equation Y^2+pY+r=0. Thus y^2=\frac{-p\pm\sqrt{p^2-4r}}{2}, and the four roots are obtained by taking the two square roots of each value of y^2. The reason this case is easier is that the roots are automatically paired as y and -y.

Resolvent method

Let \displaystyle y^4+py^2+qy+r=0 be the depressed quartic, with roots y_1,y_2,y_3,y_4. Thus y_1+y_2+y_3+y_4=0. Define the transform

\displaystyle H_0=y_1+y_2+y_3+y_4, \\ H_1=y_1+y_2-y_3-y_4, \\ H_2=y_1-y_2+y_3-y_4, \\ H_3=y_1-y_2-y_3+y_4.

The first quantity is zero, since the quartic is depressed: H_0=0.

The inverse transform is

\displaystyle y_1=\frac{H_0+H_1+H_2+H_3}{4}, \\ y_2=\frac{H_0+H_1-H_2-H_3}{4}, \\ y_3=\frac{H_0-H_1+H_2-H_3}{4}, \\ y_4=\frac{H_0-H_1-H_2+H_3}{4}.

This is related to the three ways to divide four roots into two pairs. Correspondingly, put

\displaystyle \rho=\frac{H_1}{2}=y_1+y_2=-(y_3+y_4), \\ \sigma=\frac{H_2}{2}=y_1+y_3=-(y_2+y_4), \\ \tau=\frac{H_3}{2}=y_1+y_4=-(y_2+y_3).

These are the three nontrivial pair-sums. From them one recovers the roots.

\displaystyle y_1=\frac{\rho+\sigma+\tau}{2},\\ y_2=\frac{\rho-\sigma-\tau}{2}, \\ y_3=\frac{-\rho+\sigma-\tau}{2},\\ y_4=\frac{-\rho-\sigma+\tau}{2}.

Thus \displaystyle \rho,\sigma,\tau are the three possible sums obtained by choosing two roots, with the remaining two roots automatically having the opposite sum. Changing the order of the roots can change signs of \displaystyle \rho,\sigma,\tau, but it only rearranges the three squares

\displaystyle \rho^2,\quad \sigma^2,\quad \tau^2.

Therefore their symmetric combinations depend only on the coefficients of the quartic. Substituting the inverse formulas into the elementary symmetric functions of the roots gives

\displaystyle p=-\frac{\rho^2+\sigma^2+\tau^2}{2}, \\ q=-\rho\sigma\tau, \\ r= \frac{(\rho^2+\sigma^2+\tau^2)^2 -4(\rho^2\sigma^2+\rho^2\tau^2+\sigma^2\tau^2)}{16}.

Hence \displaystyle \rho^2+\sigma^2+\tau^2=-2p, \quad \rho^2\sigma^2+\rho^2\tau^2+\sigma^2\tau^2=p^2-4r, \quad \rho^2\sigma^2\tau^2=q^2. Therefore \displaystyle \rho^2,\sigma^2,\tau^2 are the three roots of

\displaystyle Z^3+2pZ^2+(p^2-4r)Z-q^2=0.

This is Ferrari’s resolvent cubic. Its three roots correspond to the three ways of splitting the four roots into two pairs. After solving the resolvent, choose square roots \rho^2=Z_1,\quad \sigma^2=Z_2,\quad \tau^2=Z_3. Their signs are not independent: they must satisfy \rho\sigma\tau=-q. With this compatibility condition, the inverse formulas recover the four roots.

The Möbius-transformation method

Let f(y)=y^4+py^2+qy+r be a depressed quartic.

The main observation is that a reciprocal quartic, z^4+Az^3+Bz^2+Az+1=0, is easy to solve: dividing by z^2 and setting t=z+\frac1z gives \displaystyle t^2+At+(B-2)=0. Thus we ask whether a change of coordinate can turn f into a reciprocal quartic. Try the simplest projective change, an affine transformation, \displaystyle y=b+az,\quad a\ne0.

Expanding gives f(b+az) =a^4z^4+4a^3bz^3+a^2(6b^2+p)z^2 +a f'(b)z+f(b). For this to be a^4 times a reciprocal polynomial, its constant coefficient must equal its leading coefficient, and its coefficient of z must equal that of z^3. Hence we require f(b)=a^4, f'(b)=4a^2b. Put s=a^2. Then

\displaystyle s^2=f(b),\quad 4sb=4b^3+2pb+q.

Assume q\ne0. Introduce \lambda:=2s-2b^2-p. The second equation becomes q=2b\lambda, so that

\displaystyle b=\frac{q}{2\lambda}, \quad a^2=s=b^2+\frac{p+\lambda}{2} =\frac{q^2}{4\lambda^2}+\frac{p+\lambda}{2}.

Substituting these into s^2=f(b) gives \lambda(p+\lambda)^2-4r\lambda-q^2=0, or equivalently

\displaystyle \lambda^3+2p\lambda^2+(p^2-4r)\lambda-q^2=0.

This is Ferrari’s resolvent cubic. Thus the resolvent appears simply because it is the condition under which a translation and scaling make the quartic reciprocal. Choose a nonzero root \lambda of this cubic, define a,b by the preceding formulas, and put y=b+az. Then z^4+\frac{4b}{a}z^3+ \frac{6b^2+p}{a^2}z^2+ \frac{4b}{a}z+1=0. This is reciprocal. Hence t=z+\frac1z satisfies t^2+\frac{4b}{a}t+(\frac{6b^2+p}{a^2}-2)=0. For each of the two values of t, solve z^2-tz+1=0, and finally return to y=b+az.

Only now does the root geometry become visible. Since the transformed equation is reciprocal, its roots occur in pairs z,\ z^{-1}. In the original y -coordinate, inversion becomes the Möbius involution

\displaystyle \iota_\lambda(y) =b+\frac{a^2}{y-b} =\frac{qy+\lambda(p+\lambda)}{2\lambda y-q}.

Indeed, if z=(y-b)/a, then \frac{\iota_\lambda(y)-b}{a}=\frac1z. Thus \iota_\lambda exchanges the roots in two pairs. There are exactly three ways to split four roots into two pairs, and the three roots of the resolvent cubic encode those three choices. More precisely, if y_1,y_2,y_3,y_4 are the roots, then \displaystyle (y_1+y_2)^2,\quad (y_1+y_3)^2,\quad (y_1+y_4)^2 are the three roots of the resolvent cubic. Choosing one value of \lambda therefore chooses one pairing; the Möbius coordinate turns that selected pairing into the simple symmetry z\leftrightarrow z^{-1}.

Discriminant

Finally, for the depressed quartic y^4+py^2+qy+r, the discriminant \Delta=\prod_{1\le i<j\le4}(y_i-y_j)^2. is

\displaystyle \Delta= 256r^3-128p^2r^2+144pq^2r-27q^4+16p^4r-4p^3q^2.

Hence \Delta=0 precisely when the quartic has a repeated root. For real coefficients, \Delta<0 means that there are two real roots and one non-real conjugate pair. If \Delta>0, the quartic has either four real roots or no real roots; unlike the cubic case, the sign of the discriminant alone does not distinguish these two possibilities. One must also inspect the critical points, which are the roots of the derivative, a cubic equation.

The quartic formula is therefore governed by a simple architecture. First remove the cubic term. Then find a parameter that converts the quartic into a difference of two squares. The condition on that parameter is a cubic because there are three possible pairings of four roots. Solve that resolvent cubic, use it to factor the quartic into two quadratics, and finally solve those quadratics. So each of the methods is really a procedure for discovering the hidden quadratic factorization of a quartic.

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