Quartic function

This article is about the univariate quartic. For the bivariate quartic, see Quartic plane curve.

Graph of a polynomial of degree 4, with 3 critical points and four real roots (crossings of the x axis) (and thus no complex roots). If one or the other of the local minima were above the x axis, or if the local maximum were below it, or if there were no local maximum and one minimum below the x axis, there would only be two real roots (and two complex roots). If all three local extrema were above the x axis, or if there were no local maximum and one minimum above the x axis, there would be no real root (and four complex roots).

In algebra, a quartic function, is a function of the form

$f(x)=ax^4+bx^3+cx^2+dx+e,$

where a is nonzero, which is defined by a polynomial of degree four, called quartic polynomial.

Sometimes the term biquadratic is used instead of quartic, but, usually, biquadratic function refers to a quadratic function of a square (or, equivalently, to the function defined by a quartic polynomial without terms of odd degree), having the form

$f(x)=ax^4+cx^2+e.$

A quartic equation, or equation of the fourth degree, is an equation that equates a quartic polynomial to zero, of the form

$ax^4+bx^3+cx^2+dx+e=0 ,$

where a ≠ 0.

The derivative of a quartic function is a cubic function.

Since a quartic function is defined by a polynomial of even degree, it has the same infinite limit when the argument goes to positive or negative infinity. If a is positive, then the function increases to positive infinity at both ends; and thus the function has a global minimum. Likewise, if a is negative, it decreases to negative infinity and has a global maximum. In both cases it may or may not have another local maximum and another local minimum.

The degree four (quartic case) is the highest degree such that every polynomial equation can be solved by radicals.

History

Lodovico Ferrari is credited with the discovery of the solution to the quartic in 1540, but since this solution, like all algebraic solutions of the quartic, requires the solution of a cubic to be found, it could not be published immediately. The solution of the quartic was published together with that of the cubic by Ferrari's mentor Gerolamo Cardano in the book Ars Magna (1545).

The Soviet historian I. Y. Depman claimed that even earlier, in 1486, Spanish mathematician Valmes was burned at the stake for claiming to have solved the quartic equation. Inquisitor General Tomás de Torquemada allegedly told Valmes that it was the will of God that such a solution be inaccessible to human understanding. However Beckmann, who popularized this story of Depman in the West, said that it was unreliable and hinted that it may have been invented as Soviet antireligious propaganda. Beckmann's version of this story has been widely copied in several books and internet sites, usually without his reservations and sometimes with fanciful embellishments. Several attempts to find corroborating evidence for this story, or even for the existence of Valmes, have failed.

The proof that four is the highest degree of a general polynomial for which such solutions can be found was first given in the Abel–Ruffini theorem in 1824, proving that all attempts at solving the higher order polynomials would be futile. The notes left by Évariste Galois prior to dying in a duel in 1832 later led to an elegant complete theory of the roots of polynomials, of which this theorem was one result.

Applications

Each coordinate of the intersection points of two conic sections is a solution of a quartic equation. The same is true for the intersection of a line and a torus. It follows that quartic equations often arise in computational geometry and all related fields such as computer graphics, computer-aided design, computer-aided manufacturing and optics. Here are example of other geometric problems whose solution amounts of solving a quartic equation.

In computer-aided manufacturing, the torus is a shape that is commonly associated with the endmill cutter. To calculate its location relative to a triangulated surface, the position of a horizontal torus on the Z-axis must be found where it is tangent to a fixed line, and this requires the solution of a general quartic equation to be calculated.

A quartic equation arises also in the process of solving the crossed ladders problem, in which the lengths of two crossed ladders, each based against one wall and leaning against another, are given along with the height at which they cross, and the distance between the walls is to be found.

In optics, Alhazen's problem is "Given a light source and a spherical mirror, find the point on the mirror where the light will be reflected to the eye of an observer." This leads to a quartic equation.

Finding the distance of closest approach of two ellipses involves solving a quartic equation.

The eigenvalues of a 4×4 matrix are the roots of a quartic polynomial which is the characteristic polynomial of the matrix.

The characteristic equation of a fourth-order linear difference equation or differential equation is a quartic equation. An example arises in the Timoshenko-Rayleigh theory of beam bending.

Intersections between spheres, cylinders, or other quadrics can be found using quartic equations.

Inflection points and golden ratio

Letting F and G be the distinct inflection points of a quartic, and letting H be the intersection of the inflection secant line FG and the quartic, nearer to G than to F, then G divides FH into the golden section:

$\frac{FG}{GH}=\frac{1+\sqrt{5}}{2}= \text{the golden ratio}.$

Moreover, the area of the region between the secant line and the quartic below the secant line equals the area of the region between the secant line and the quartic above the secant line. One of those regions is disjointed into sub-regions of equal area.

Solving a quartic equation

Nature of the roots

Given the general quartic equation

$ax^4 + bx^3 + cx^2 + dx + e = 0$

with real coefficients and $a \ne 0,$ the nature of its roots is mainly determined by the sign of its discriminant

$\begin{align} \Delta\ =\ &256 a^3 e^3 - 192 a^2 b d e^2 - 128 a^2 c^2 e^2 + 144 a^2 c d^2 e - 27 a^2 d^4 \\ &+ 144 a b^2 c e^2 - 6 a b^2 d^2 e - 80 a b c^2 d e + 18 a b c d^3 + 16 a c^4 e \\ &- 4 a c^3 d^2 - 27 b^4 e^2 + 18 b^3 c d e - 4 b^3 d^3 - 4 b^2 c^3 e + b^2 c^2 d^2 \end{align}$

This may be refined by considering the signs of four other polynomials:

$P = 8ac - 3b^2$

such that $\frac{P}{8a^2}$ is the second degree coefficient of the associated depressed quartic (see below);

$Q= b^3+8da^2-4abc,$

such that $\frac{Q}{8a^3}$ is the first degree coefficient of the associated depressed quartic;

$\Delta_0 = c^2 - 3bd + 12ae,$

which is 0 if the quartic has a triple root; and

$D = 64 a^3 e - 16 a^2 c^2 + 16 a b^2 c - 16 a^2 bd - 3 b^4$

which is 0 if the quartic has two double roots.

The possible cases for the nature of the roots are as follows:

• If $\Delta < 0$ then the equation has two real roots and two complex conjugate roots.
• If $\Delta > 0$ then the equation's four roots are either all real or all complex.
  • If P < 0 and D < 0 then all four roots are real and distinct.
  • If P > 0 or D > 0 then there are two pairs of complex conjugate roots.
• If $\Delta = 0$ then either the polynomial has a multiple root, or it is the square of a quadratic polynomial. Here are the different cases that can occur:
  • If P < 0 and D < 0 and $\Delta_0\not =0$, there is a real double root and two real simple roots.
  • If D > 0 or (P > 0 and (D ≠ 0 or Q ≠ 0)), there is a real double root and two complex conjugate roots.
  • If $\Delta_0 = 0$ and D ≠ 0, there is a triple root and a simple root, all real.
  • If D = 0, then:
    • If P < 0, there are two real double roots.
    • If P > 0 and Q = 0, there are two complex conjugate double roots.
    • If $\Delta_0 = 0$, all four roots are equal to $-\frac{b}{4a}$

There are some cases that do not seem to be covered, but they cannot occur. For example $\Delta$ > 0, P = 0 and D ≤ 0 is not one of the cases. However if $\Delta$ > 0 and P = 0 then D > 0 so this combination is not possible.

General formula for roots

File:Quartic Formula.svg

The four roots $x_1, x_2, x_3, x_4$ for the general quartic equation

$ax^4+bx^3+cx^2+dx+e=0 \,$

with a ≠ 0 are given in the following formula, which is deduced from the one in the section Solving by factoring into quadratics by back changing the variables (see section Converting to a depressed quartic) and using the formulas for the quadratic and cubic equations.

$\begin{align} x_{1,2}\ &= -\frac{b}{4a} - S \pm \frac12\sqrt{-4S^2 - 2p + \frac{q}{S}}\\ x_{3,4}\ &= -\frac{b}{4a} + S \pm \frac12\sqrt{-4S^2 - 2p - \frac{q}{S}} \end{align}$

where p and q are the coefficients of the second and of the first degree respectively in the associated depressed quartic

$\begin{align} p &= \frac{8ac-3b^2}{8a^2}\\ q &= \frac{b^3 - 4abc + 8a^2d}{8a^3} \end{align}$

and where

$\begin{align} S &= \frac{1}{2}\sqrt{-\frac23\ p+\frac{1}{3a}\left(Q + \frac{\Delta_0}{Q}\right)} \quad\qquad\ &\text{(if } S= 0 \text{, see Special cases of the formula, below)}\\ Q\ &=\ \sqrt[3]{\frac{\Delta_1 + \sqrt{\Delta_1^2 - 4\Delta_0^3}}{2}} &\text{(if } Q= 0 \text{, see Special cases of the formula, below)} \end{align}$

with

$\begin{align} \Delta_0 &= c^2 - 3bd + 12ae\\ \Delta_1 &= 2c^3 - 9bcd + 27b^2 e + 27ad^2 - 72ace \end{align}$

and

$\Delta_1^2-4\Delta_0^3 = - 27 \Delta\ ,$ where $\Delta$ is the aforementioned discriminant. The mathematical expressions of these last four terms are very similar to those of their cubic counterparts.

Special cases of the formula

• If $\Delta > 0,$ the value of $Q$ is a non-real complex number. In this case, either all roots are non-real or they are all real. In the latter case, the value of $S$ is also real, and one may prefer to express it in a purely real way, by using trigonometric functions, as follows:

$S = \frac{1}{2}\sqrt{-\frac23\ p+\frac{2}{3a}\sqrt{\Delta_0}\cos\frac{\phi}{3}}$

where

$\phi = \arccos\left(\frac{\Delta_1}{2\sqrt{\Delta_0^3}}\right).$

• If $\Delta \neq 0$ and $\Delta_0 = 0,$ the sign of $\sqrt{\Delta_1^2 - 4 \Delta_0^3}=\sqrt{\Delta_1^2}$ has to be chosen to have $Q \neq 0,$ that is one should define $\sqrt{\Delta_1^2}$ as $\Delta_1,$ maintaining the sign of $\Delta_1.$

• If $S = 0,$ then one must change the choice of the cube root in $Q$ in order to have $S \neq 0.$ This is always possible except if the quartic may be factored into $\left(x+\tfrac{b}{4a}\right)^4.$ The result is then correct, but misleading because it hides the fact that no cube root is needed in this case. In fact this case may occur only if the numerator of $q$ is zero, in which case the associated depressed quartic is biquadratic; it may thus be solved by the method described below.

• If $\Delta = 0$ and $\Delta_0 = 0,$ and thus also $\Delta_1 = 0,$ at least three roots are equal, and the roots are rational functions of the coefficients.

• If $\Delta=0$ and $\Delta_0 \neq 0,$ the above expression for the roots is correct but misleading, hiding the fact that the polynomial is reducible and no cube root is needed to represent the roots.

Simpler cases

Reducible quartics

Consider the general quartic

$Q(x) = a_4x^4+a_3x^3+a_2x^2+a_1x+a_0.$

It is reducible if Q=RS, where R and S are non-constant polynomials with rational coefficients (or more generally with coefficients in the same field as the coefficients of Q). There are two ways to write such a factorization: either

$Q(x) = (x-x_1)(b_3x^3+b_2x^2+b_1x+b_0)$

or

$Q(x) = (c_2x^2+c_1x+c_0)(d_2x^2+d_1x+d_0).$

In either case, the roots of Q are the roots of the factors, which may be computed by solving quadratic or cubic equations.

Detecting such factorizations can be done by using the factor function of every computer algebra system. But, in many cases, it may be done by hand-written computation. In the preceding section, we have already seen that the polynomial is always reducible if its discriminant $\Delta$ is zero (this is true for polynomials of every degree).

A very special case of the first case of factorization is when a₀=0. This implies that x₁=0 is a first root, b₃=a₄, b₂=a₃, b₁=a₂, b₀=a₁, and the other roots may be computed by solving a cubic equation.

If $a_4+a_3+a_2+a_1+a_0=0,$ then $Q(1)=0$ and we have a factorization of the first kind with x₁=1. Similarly, if $a_4-a_3+a_2-a_1+a_0=0,$ then $Q(-1)=0$ and we have a factorization of the first kind with x₁=-1.

Once a root x₁ is known, the second factor of the factorization of the first kind is the quotient of the Euclidean division of Q by x-x₁. It is

$a_4x^3+(a_4x_1+a_3)x^2+(a_4x_1^2+a_3x_1+a_2)x+a_4x_1^3+a_3x_1^2+a_2x_1+a_1.$

If $a_0, a_1, a_2, a_3, a_4$ are small integers a factorization of the first kind is easy to detect: if $x_1=\frac{p}{q}\ ,$ with p and q coprime integers, then q divides evenly a₄, and p divides evenly a₀. Thus, computing $Q\left(\frac{p}{q}\right)$ for every possible values of p and q allows to find the rational roots, if any.

In the case of two quadratic factors or of large integer coefficients, the factorization is harder to compute, and, in general, it is better to use the factor function of a computer algebra system (see polynomial factorization for a description of the algorithms that are involved).

Biquadratic equations

If $a_3=a_1=0,\,$ then the biquadratic function

$Q(x) = a_4x^4+a_2x^2+a_0\,\!$

defines a biquadratic equation, which is easy to solve.

Let $z=x^2.\,$ Then Q becomes a quadratic q in $z,\,$

$q(z) = a_4z^2+a_2z+a_0.\,\!$

Let $z_+\,$ and $z_-\,$ be the roots of q. Then the roots of our quartic Q are

$\begin{align} x_1&=+\sqrt{z_+}, \\ x_2&=-\sqrt{z_+}, \\ x_3&=+\sqrt{z_-}, \\ x_4&=-\sqrt{z_-}. \end{align}$

Quasi-palindromic equation

The polynomial

$P(x)=a_0x^4+a_1x^3+a_2x^2+a_1 m x+a_0 m^2$

is almost palindromic, as

$P(mx)=\frac{x^4}{m^2}P(\frac{m}{x})$

(it is palindromic if m = 1).

The change of variables $z = x + \frac{m}{x}$ in $\frac{P(x)}{x^2}=0$ produces the quadratic equation $a_0z^2+a_1z +(a_2-2ma_0) =0.$ As $x^2-xz+m=0,$ the quartic equation

$P(x)=0$

may be solved by applying the quadratic formula twice.

Solution methods

Converting to a depressed quartic

For solving purposes, it is generally better to convert the quartic into a depressed quartic by the following simple change of variable. All formulas are simpler and some methods work only in this case. The roots of the original quartic are easily recovered from that of the depressed quartic by the reverse change of variable.

Let

$a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0 = 0$

be the general quartic equation we want to solve.

Dividing by a₄, provides the equivalent equation

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

with

$a=\frac{a_3}{a_4},\quad b=\frac{a_2}{a_4},\quad c=\frac{a_1}{a_4},\quad d=\frac{a_0}{a_4}.$

Substituting y - a/4 for x gives, after regrouping the terms, the equation

$y^4+py^2+qy+r=0,$

where

$\begin{align} p=&\frac{8b-3a^2}{8} &=&\frac{8a_2a_4-3a_3^2}{8a_4^2}\\ q=&\frac{a^3-4ab+8c}{8} &=&\frac{a_3^3-4a_2a_3a_4+8a_1a_4^2}{8a_4^3}\\ r=&\frac{-3a^4+256d-64ca+16a^2b}{256}&=&\frac{-3a_3^4+256a_0a_4^3-64a_1a_3a_4^2+16a_2a_3^2a_4}{256a_4^4}. \end{align}$

If y₁, y₂, y₃, y₄ are the roots of this depressed quartic, then the roots of the original quartic, x₁, x₂, x₃, x₄ are

$x_i = y_i-\frac{a}{4} =y_i-\frac{a_3}{4a_4}, \text{ for } 1 \leq i \leq 4.$

Ferrari's solution

As explained in the preceding section, we may start with a depressed quartic equation

$u^4 + \alpha u^2 + \beta u + \gamma = 0.$

(1)

This depressed quartic can be solved by means of a method discovered by Lodovico Ferrari. The depressed equation may be rewritten (this is easily verified by expanding the square and regrouping all terms in the left-hand side)

$(u^2 + \alpha)^2 = \alpha u^2 + \alpha^2- \beta u - \gamma.$

(2)

Then, we introduce a variable y into the factor on the left-hand side by adding $2u^2y+2\alpha y + y^2$ to both sides. After regrouping the coefficients of the power of u in the right-hand side, this gives the equation

$(u^2 + \alpha + y)^2 = (\alpha + 2 y) u^2 - \beta u + (y^2 + 2 y \alpha + \alpha^2 - \gamma),$

(3)

which is equivalent to the original equation, whichever value is given to y.

As the value of y may be arbitrarily chosen, we will choose it in order to get a perfect square in the right-hand side. This implies that the discriminant in u of this quadratic equation is zero, that is y is a root of the equation

$(-\beta)^2 - 4 (2 y + \alpha) (y^2 + 2 y \alpha + \alpha^2 - \gamma) = 0.\,$

which may be rewritten

$y^3 + {5 \over 2} \alpha y^2 + (2 \alpha^2 - \gamma) y + \left( {\alpha^3 \over 2} - {\alpha \gamma \over 2} - {\beta^2 \over 8} \right) = 0.$

(4)

This is the resolvent cubic of the quartic equation. The value of y may thus be obtained from the formulas provided in the article Cubic equation.

When y is a root of equation (4), the right-hand side of equation (3) the square of

$\left(u\sqrt{\alpha + 2 y}-\frac{\beta}{2\sqrt{\alpha + 2 y}}\right)^2.$

However, this induces a division by zero if $\alpha + 2 y=0.$ This implies $\beta=0,$ and thus that the depressed equation is bi-quadratic, and may be solved by an easier method (see above). This was not a problem at the time of Ferrari, when one solved only explicitly given equations with numeric coefficients. For a general formula that is always true, one thus need to choose a root of the cubic equation such that $\alpha + 2 y \not=0.$ This is always possible unless for the depressed equation x⁴=0.

Now, if y is a root of the cubic equation such that $\alpha + 2 y \not=0,$ equation (3) may be rewritten

$\left(u^2 + \alpha + y+u\sqrt{\alpha + 2 y}-\frac{\beta}{2\sqrt{\alpha + 2 y}}\right) \left(u^2 + \alpha + y - u\sqrt{\alpha + 2 y}+\frac{\beta}{2\sqrt{\alpha + 2 y}}\right)=0,$

and the equation is easily solved by applying to each factor the formula for quadratic equations. Solving them we may write the four roots as

$u={\pm_1\sqrt{\alpha + 2 y} \pm_2 \sqrt{-\left(3\alpha + 2y \pm_1 {2\beta \over \sqrt{\alpha + 2 y}} \right)} \over 2},$

where $\pm_1$ and $\pm_2$ denote either + or -. As the two occurrences of $\pm_1$ must denote the same sign, this leave four possibilities, one for each root.

Therefore the solutions of the original quartic equation are

$x=-{a_3 \over 4a_4} + {\pm_1\sqrt{\alpha + 2 y} \pm_2 \sqrt{-\left(3\alpha + 2y \pm_1 {2\beta \over \sqrt{\alpha + 2 y}} \right)} \over 2}.$

Solving by factoring into quadratics

One can solve a quartic by factoring it into a product of two quadratics, which do not need to have rational coefficients. Let

$\begin{array}{lcl} 0 = x^4 + ax^3 + bx^2 + cx + d & = & (x^2 + px + q)(x^2 + rx + s) \\ & = & x^4 + (p + r)x^3 + (q + s + pr)x^2 + (ps + qr)x + qs \end{array}$

By equating coefficients, this results in the following set of simultaneous equations:

$\begin{array}{lcl} a & = & p + r \\ b & = & q + s + pr \\ c & = & ps + qr \\ d & = & qs \end{array}$

This can be simplified by starting again with a depressed quartic where $a = 0$, which can be obtained by substituting $(x - a/4)$ for $x$, then $r = -p$, and:

$\begin{array}{lcl} b + p^2 & = & s + q \\ c & = & (s - q) p \\ d & = & sq \end{array}$

One can now eliminate both $s$ and $q$ by doing the following:

$\begin{array}{lcl} p^2(b + p^2)^2 - c^2 & = & p^2(s + q)^2 - p^2(s - q)^2 \\ & = & 4p^2sq \\ & = & 4p^2d \end{array}$

If we set $P = p^2$, then this equation turns into the resolvent cubic equation

$P^3 + 2bP^2 + (b^2 - 4d)P - c^2 = 0\,$

which is solved elsewhere. Then, if p is a square root of a non-zero root of this resolvent (such a non-zero root exists except for the quartic x⁴, which is trivially factored),

$\begin{array}{lcl} r & = & -p \\ 2s & = & b + p^2 + c/p \\ 2q & = & b + p^2 - c/p \end{array}$

The symmetries in this solution are as follows. There are three roots of the cubic, corresponding to the three ways that a quartic can be factored into two quadratics, and choosing positive or negative values of $p$ for the square root of $P$ merely exchanges the two quadratics with one another.

The above solution shows that the quartic polynomial with a zero coefficient on the cubic term is factorable into quadratics with rational coefficients if and only if either the resolvent cubic $P^3 + 2bP^2 + (b^2 - 4d)P - c^2\,$ has a non-zero root which is the square of a rational, or $b^2-4d$ is the square of rational and c=0; this can readily be checked using the rational root test.

Solving by Lagrange resolvent

The symmetric group S₄ on four elements has the Klein four-group as a normal subgroup. This suggests using a resolvent cubic whose roots may be variously described as a discrete Fourier transform or a Hadamard matrix transform of the roots; see Lagrange resolvents for the general method. Denote by x_i, for i from 0 to 3, the four roots of

$x^4 + ax^3 + bx^2 + cx + d = 0$

If we set

$\begin{align} s_0 &= \tfrac12(x_0 + x_1 + x_2 + x_3), \\ s_1 &= \tfrac12(x_0 - x_1 + x_2 - x_3), \\ s_2 &= \tfrac12(x_0 + x_1 - x_2 - x_3), \\ s_3 &= \tfrac12(x_0 - x_1 - x_2 + x_3), \end{align}$

then since the transformation is an involution we may express the roots in terms of the four s_i in exactly the same way. Since we know the value s₀ = -a/2, we only need the values for s₁, s₂ and s₃. These are the roots of the polynomial

$(s^2 - s_1^2)(s^2-s_2^2)(s^2-s_3^2).$

Substituting the s_i by their values in term of the x_i, this polynomial may be expanded in a polynomial in s whose coefficients are symmetric polynomials in the x_i. By the fundamental theorem of symmetric polynomials, these coefficients may be expressed as polynomials in the coefficients of the monic quartic. If, for simplification, we suppose that the quartic is depressed, that is a=0, this results in the polynomial

${s}^{6}+2b\,{s}^{4}+({b}^{2}-4d)\,{s}^{2}-{c}^{2}$

(5)

This polynomial is of degree six, but only of degree three in s², and so the corresponding equation is solvable by the method described in the article Cubic function. By substituting the roots in the expression of the x_i in terms of the s_i, we obtain expression for the roots. In fact we obtain, apparently, several expressions, depending on the numbering of the roots of the cubic polynomial and of the signs given to their square roots. All these different expressions may be deduced from one of them by simply changing the numbering of the x_i.

These expressions are unnecessarily complicated, involving the cubic roots of unity, which can be avoided as follows. If s is any non-zero root of (5), and if we set

$F_1 = {x}^{2}+sx+ \frac{b}{2} + \frac{s^2}{2} - \frac{c}{2s}$

$F_2 ={x}^{2}-sx+ \frac{b}{2} + \frac{s^2}{2} + \frac{c}{2s}$

then

$F_1 F_2 = x^4 + bx^2 + cx + d$

We therefore can solve the quartic by solving for s and then solving for the roots of the two factors using the quadratic formula.

Note that this gives exactly the same formula for the roots as the preceding section.

Solving with algebraic geometry

An alternative solution using algebraic geometry is given in (Faucette 1996), and proceeds as follows (more detailed discussion in reference). In brief, one interprets the roots as the intersection of two quadratic curves, then finds the three reducible quadratic curves (pairs of lines) that pass through these points (this corresponds to the resolvent cubic, the pairs of lines being the Lagrange resolvents), and then use these linear equations to solve the quadratic.

The four roots of the depressed quartic $x^4+px^2+qx+r=0$ may also be expressed as the x coordinates of the intersections of the two quadratic equations $y^2+py+qx+r=0,$ $y-x^2=0;$ i.e., using the substitution $y=x^2;$ that two quadratics intersect in four points is an instance of Bézout's theorem. Explicitly, the four points are $P_i := (x_i,x_i^2)$ for the four roots $x_i$ of the quartic.

These four points are not collinear because they lie on the irreducible quadratic $y=x^2,$ and thus there is a 1-parameter family of quadratics (a pencil of curves) passing through these points. Writing the projectivization of the two quadratics as quadratic forms in three variables:

$\begin{align} F_1(X,Y,Z) &:= Y^2 + pYZ + qXZ + rZ^2,\\ F_2(X,Y,Z) &:= YZ - X^2 \end{align}$

the pencil is given by the forms $\lambda F_1 + \mu F_2$ for any point $[\lambda,\mu]$ in the projective line – in other words, where $\lambda$ and $\mu$ are not both zero, and multiplying a quadratic form by a constant does not change its quadratic curve of zeros.

This pencil contains three reducible quadratics, each corresponding to a pair of lines, each passing through two of the four points, which can be done $\textstyle{\binom{4}{2} = 6}$ different ways. Denote these $Q_1 = L_{12} + L_{34},$ $Q_2 = L_{13} + L_{24},$ $Q_3 = L_{14} + L_{23}.$ Given any two of these, their intersection is exactly the four points.

The reducible quadratics, in turn, may be determined by expressing the quadratic form $\lambda F_1 + \mu F_2$ as a 3×3 matrix: reducible quadratics correspond to this matrix being singular, which is equivalent to its determinant being zero, and the determinant is a homogeneous degree three polynomial in $\lambda$ and $\mu,$ and corresponds to the resolvent cubic.

See also

• Linear function
• Quadratic function
• Cubic function
• Quintic function
• Polynomial
• Newton's method

External links

• Quartic formula as four single equations at PlanetMath.org.
• Ferrari's achievement