Inalgebra, aquartic function is afunction of the form^α

${\displaystyle f(x)=a{x}^4+b{x}^3+c{x}^2+dx+e,}$

wherea is nonzero,which is defined by apolynomial ofdegree four, called aquartic polynomial.

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

${\displaystyle a{x}^4+b{x}^3+c{x}^2+dx+e=0,}$

wherea ≠ 0.^[1]Thederivative of a quartic function is acubic function.

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

${\displaystyle f(x)=a{x}^4+c{x}^2+e.}$

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 negativeinfinity. Ifa is positive, then the function increases to positive infinity at both ends; and thus the function has aglobal minimum. Likewise, ifa 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 byradicals, according to theAbel–Ruffini theorem.

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since this solution, like all algebraic solutions of the quartic, requires the solution of acubic to be found, it could not be published immediately.^[2] The solution of the quartic was published together with that of the cubic by Ferrari's mentorGerolamo Cardano in the bookArs Magna.^[3]

The proof that four is the highest degree of a general polynomial for which such solutions can be found was first given in theAbel–Ruffini theorem in 1824, proving that all attempts at solving the higher order polynomials would be futile. The notes left byÉvariste Galois prior to dying in a duel in 1832 later led to an elegantcomplete theory of the roots of polynomials, of which this theorem was one result.^[4]

Eachcoordinate of the intersection points of twoconic sections is a solution of a quartic equation. The same is true for the intersection of a line and atorus. It follows that quartic equations often arise incomputational geometry and all related fields such ascomputer graphics,computer-aided design,computer-aided manufacturing andoptics. Here are examples of other geometric problems whose solution involves solving a quartic equation.

Incomputer-aided manufacturing, the torus is a shape that is commonly associated with theendmill cutter. To calculate its location relative to a triangulated surface, the position of a horizontal torus on thez-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.^[5]

A quartic equation arises also in the process of solving thecrossed 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.^[6]

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.^[7][8][9]

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

Theeigenvalues of a 4×4matrix are the roots of a quartic polynomial which is thecharacteristic polynomial of the matrix.

The characteristic equation of a fourth-order lineardifference equation ordifferential equation is a quartic equation. An example arises in theTimoshenko-Rayleigh theory of beam bending.^[10]

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

LettingF andG be the distinctinflection points of the graph of a quartic function, and lettingH be the intersection of the inflectionsecant lineFG and the quartic, nearer toG than toF, thenG dividesFH into thegolden section:^[11]

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

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.

Given the general quartic equation

${\displaystyle a{x}^4+b{x}^3+c{x}^2+dx+e=0}$

with real coefficients anda ≠ 0 the nature of its roots is mainly determined by the sign of itsdiscriminant

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

${\displaystyle P=8ac-3b^2}$

such that⁠P/8a²⁠ is the second degree coefficient of the associated depressed quartic (seebelow);

${\displaystyle R=b^3+8d{a}^2-4abc,}$

such that⁠R/8a³⁠ is the first degree coefficient of the associated depressed quartic;

${\displaystyle {\mathrm{\Delta }}_0=c^2-3bd+12ae,}$

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

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

which is 0 if the quartic has two double roots.

The possible cases for the nature of the roots are as follows:^[12]

• If∆ < 0 then the equation has two distinct real roots and twocomplex conjugate non-real roots.
• If∆ > 0 then either the equation's four roots are all real or none is.
  • IfP < 0 andD < 0 then all four roots are real and distinct.
  • IfP > 0 orD > 0 then there are two pairs of non-real complex conjugate roots.^[13]
• If∆ = 0 then (and only then) the polynomial has amultiple root. Here are the different cases that can occur:
  • IfP < 0 andD < 0 and∆₀ ≠ 0, there are a real double root and two real simple roots.
  • IfD > 0 or (P > 0 and (D ≠ 0 orR ≠ 0)), there are a real double root and two complex conjugate roots.
  • If∆₀ = 0 andD ≠ 0, there are a triple root and a simple root, all real.
  • IfD = 0, then:
    • IfP < 0, there are two real double roots.
    • IfP > 0 andR = 0, there are two complex conjugate double roots.
    • If∆₀ = 0, all four roots are equal to−⁠b/4a⁠

There are some cases that do not seem to be covered, but in fact they cannot occur. For example,∆₀ > 0,P = 0 andD ≤ 0 is not a possible case. In fact, if∆₀ > 0 andP = 0 thenD > 0, since${\displaystyle 16a^2{\mathrm{\Delta }}_0=3D+P^2;}$ so this combination is not possible.

The four rootsx₁,x₂,x₃, andx₄ for the general quartic equation

${\displaystyle a{x}^4+b{x}^3+c{x}^2+dx+e=0}$

witha ≠ 0 are given in the following formula, which is deduced from the one in the section onFerrari's method by back changing the variables (see§ Converting to a depressed quartic) and using the formulas for thequadratic andcubic equations.

wherep andq are the coefficients of the second and of the first degree respectively in theassociated depressed quartic

and where

(ifS = 0 orQ = 0, see§ Special cases of the formula, below)

with

and

${\displaystyle {\mathrm{\Delta }}_1^2-4{\mathrm{\Delta }}_0^3=-27\mathrm{\Delta },}$ where${\displaystyle \mathrm{\Delta }}$ is the aforementioneddiscriminant. For the cube root expression forQ, any of the three cube roots in the complex plane can be used, although if one of them is real that is the natural and simplest one to choose. The mathematical expressions of these last four terms are very similar to those of theircubic counterparts.

• If${\displaystyle \mathrm{\Delta }>0,}$ the value of${\displaystyle Q}$ 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${\displaystyle S}$ is also real, despite being expressed in terms of${\displaystyle Q;}$ this iscasus irreducibilis of the cubic function extended to the present context of the quartic. One may prefer to express it in a purely real way, by usingtrigonometric functions, as follows:

${\displaystyle S=\frac{1}{2}\sqrt{-\frac{2}{3}p+\frac{2}{3a}\sqrt{{\mathrm{\Delta }}_0}\cos \frac{\phi }{3}}}$

where

${\displaystyle \phi =\arccos \left(\frac{{\mathrm{\Delta }}_1}{2\sqrt{{\mathrm{\Delta }}_0^3}}\right).}$

• If${\displaystyle \mathrm{\Delta }\ne 0}$ and${\displaystyle {\mathrm{\Delta }}_0=0,}$ the sign of${\displaystyle \sqrt{{\mathrm{\Delta }}_1^2-4{\mathrm{\Delta }}_0^3}=\sqrt{{\mathrm{\Delta }}_1^2}}$ has to be chosen to have${\displaystyle Q\ne 0,}$ that is one should define${\displaystyle \sqrt{{\mathrm{\Delta }}_1^2}}$ as${\displaystyle {\mathrm{\Delta }}_1,}$ maintaining the sign of${\displaystyle {\mathrm{\Delta }}_1.}$
• If${\displaystyle S=0,}$ then one must change the choice of the cube root in${\displaystyle Q}$ in order to have${\displaystyle S\ne 0.}$ This is always possible except if the quartic may be factored into${\displaystyle {\left(x+\frac{b}{4a}\right)}^4.}$ 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^{[clarification needed]} may occur only if thenumerator of${\displaystyle q}$ is zero, in which case the associateddepressed quartic is biquadratic; it may thus be solved by the method describedbelow.
• If${\displaystyle \mathrm{\Delta }=0}$ and${\displaystyle {\mathrm{\Delta }}_0=0,}$ and thus also${\displaystyle {\mathrm{\Delta }}_1=0,}$ at least three roots are equal to each other, and the roots arerational functions of the coefficients. The triple root${\displaystyle x_0}$ is a common root of the quartic and its second derivative${\displaystyle 2(6a{x}^2+3bx+c);}$ it is thus also the unique root of the remainder of theEuclidean division of the quartic by its second derivative, which is a linear polynomial. The simple root${\displaystyle x_1}$ can be deduced from${\displaystyle x_1+3x_0=-b/a.}$
• If${\displaystyle \mathrm{\Delta }=0}$ and${\displaystyle {\mathrm{\Delta }}_0\ne 0,}$ the above expression for the roots is correct but misleading, hiding the fact that the polynomial isreducible and no cube root is needed to represent the roots.

Consider the general quartic

${\displaystyle Q(x)=a_4{x}^4+a_3{x}^3+a_2{x}^2+a_{1}x+a_0.}$

It isreducible ifQ(x) =R(x)×S(x), whereR(x) andS(x) are non-constant polynomials withrational coefficients (or more generally with coefficients in the samefield as the coefficients ofQ(x)). Such a factorization will take one of two forms:

${\displaystyle Q(x)=(x-x_1)(b_3{x}^3+b_2{x}^2+b_{1}x+b_0)}$

or

${\displaystyle Q(x)=(c_2{x}^2+c_{1}x+c_0)(d_2{x}^2+d_{1}x+d_0).}$

In either case, the roots ofQ(x) are the roots of the factors, which may be computed using the formulas for the roots of aquadratic function orcubic function.

Detecting the existence of such factorizations can be doneusing the resolvent cubic ofQ(x). It turns out that:

• if we are working overR (that is, if coefficients are restricted to be real numbers) (or, more generally, over somereal closed field) then there is always such a factorization;
• if we are working overQ (that is, if coefficients are restricted to be rational numbers) then there is an algorithm to determine whether or notQ(x) is reducible and, if it is, how to express it as a product of polynomials of smaller degree.

In fact, several methods of solving quartic equations (Ferrari's method,Descartes' method, and, to a lesser extent,Euler's method) are based upon finding such factorizations.

Ifa₃ =a₁ = 0 then the function

${\displaystyle Q(x)=a_4{x}^4+a_2{x}^2+a_0}$

is called abiquadratic function; equating it to zero defines abiquadratic equation, which is easy to solve as follows

Let the auxiliary variablez =x².ThenQ(x) becomes aquadraticq inz:q(z) =a₄z² +a₂z +a₀. Letz₊ andz₋ be the roots ofq(z). Then the roots of the quarticQ(x) are

The polynomial

${\displaystyle P(x)=a_0{x}^4+a_1{x}^3+a_2{x}^2+a_{1}mx+a_0{m}^2}$

is almostpalindromic, asP(mx) =⁠x⁴/m²⁠P(⁠m/x⁠) (it is palindromic ifm = 1). The change of variablesz =x +⁠m/x⁠ in⁠P(x)/x²⁠ = 0 produces thequadratic equationa₀z² +a₁z +a₂ − 2ma₀ = 0. Sincex² −xz +m = 0, the quartic equationP(x) = 0 may be solved by applying thequadratic formula twice.

For solving purposes, it is generally better to convert the quartic into adepressed 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

${\displaystyle a_4{x}^4+a_3{x}^3+a_2{x}^2+a_{1}x+a_0=0}$

be the general quartic equation we want to solve.

Dividing bya₄, provides the equivalent equationx⁴ +bx³ +cx² +dx +e = 0, withb =⁠a₃/a₄⁠,c =⁠a₂/a₄⁠,d =⁠a₁/a₄⁠, ande =⁠a₀/a₄⁠.Substitutingy −⁠b/4⁠ forx gives, after regrouping the terms, the equationy⁴ +py² +qy +r = 0,where

Ify₀ is a root of this depressed quartic, theny₀ −⁠b/4⁠ (that isy₀ −⁠a₃/4a₄⁠) is a root of the original quartic and every root of the original quartic can be obtained by this process.

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

${\displaystyle y^4+p{y}^2+qy+r=0.}$

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

${\displaystyle {\left(y^2+\frac{p}{2}\right)}^2=-qy-r+\frac{p^2}{4}.}$

Then, we introduce a variablem into the factor on the left-hand side by adding2y²m +pm +m² to both sides. After regrouping the coefficients of the power ofy on the right-hand side, this gives the equation

${\displaystyle {\left(y^2+\frac{p}{2}+m\right)}^2=2m{y}^2-qy+m^2+mp+\frac{p^2}{4}-r,}$

1

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

As the value ofm may be arbitrarily chosen, we will choose it in order tocomplete the square on the right-hand side. This implies that thediscriminant iny of thisquadratic equation is zero, that ism is a root of the equation

${\displaystyle (-q{)}^2-4(2m)\left(m^2+pm+\frac{p^2}{4}-r\right)=0,}$

which may be rewritten as

${\displaystyle 8m^3+8p{m}^2+(2p^2-8r)m-q^2=0.}$

1a

This is theresolvent cubic of the quartic equation. The value ofm may thus be obtained fromCardano's formula. Whenm is a root of this equation, the right-hand side of equation (1) is the square

${\displaystyle {\left(\sqrt{2m}y-\frac{q}{2\sqrt{2m}}\right)}^2.}$

However, this induces a division by zero ifm = 0. This impliesq = 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 needs to choose a root of the cubic equation such thatm ≠ 0. This is always possible except for the depressed equationy⁴ = 0.

Now, ifm is a root of the cubic equation such thatm ≠ 0, equation (1) becomes

${\displaystyle {\left(y^2+\frac{p}{2}+m\right)}^2={\left(y\sqrt{2m}-\frac{q}{2\sqrt{2m}}\right)}^2.}$

This equation is of the formM² =N², which can be rearranged asM² −N² = 0 or(M +N)(M −N) = 0. Therefore, equation (1) may be rewritten as

${\displaystyle \left(y^2+\frac{p}{2}+m+\sqrt{2m}y-\frac{q}{2\sqrt{2m}}\right)\left(y^2+\frac{p}{2}+m-\sqrt{2m}y+\frac{q}{2\sqrt{2m}}\right)=0.}$

This equation is easily solved by applying to each factor thequadratic formula. Solving them we may write the four roots as

${\displaystyle y=\frac{{\pm }_1\sqrt{2m}{\pm }_2\sqrt{-\left(2p+2m{\pm }_1\frac{\sqrt{2}q}{\sqrt{m}}\right)}}{2},}$

where±₁ and±₂ denote either+ or−. As the two occurrences of±₁ must denote the same sign, this leaves four possibilities, one for each root.

Therefore, the solutions of the original quartic equation are

${\displaystyle x=-\frac{a_3}{4a_4}+\frac{{\pm }_1\sqrt{2m}{\pm }_2\sqrt{-\left(2p+2m{\pm }_1\frac{\sqrt{2}q}{\sqrt{m}}\right)}}{2}.}$

A comparison with thegeneral formula above shows that√2m = 2S.

Descartes^[14] introduced in 1637 the method of finding the roots of a quartic polynomial by factoring it into two quadratic ones. Let

Byequating coefficients, this results in the following system of equations:

${\displaystyle \{\begin{array}{l}b=s+u\\ c=t+v+su\\ d=sv+tu\\ e=tv\end{array}}$

This can be simplified by starting again with thedepressed quarticy⁴ +py² +qy +r, which can be obtained by substitutingy −b/4 forx. Since the coefficient ofy³ is 0, we gets = −u, and:

${\displaystyle \{\begin{array}{l}p+u^2=t+v\\ q=u(t-v)\\ r=tv\end{array}}$

One can now eliminate botht andv by doing the following:

If we setU =u², then solving this equation becomes finding the roots of theresolvent cubic

${\displaystyle U^3+2p{U}^2+(p^2-4r)U-q^2,}$

2

which isdone elsewhere. This resolvent cubic is equivalent to the resolvent cubic given above (equation (1a)), as can be seen by substituting U = 2m.

Ifu is a square root of a non-zero root of this resolvent (such a non-zero root exists except for the quarticx⁴, which is trivially factored),

${\displaystyle \{\begin{array}{l}s=-u\\ 2t=p+u^2+q/u\\ 2v=p+u^2-q/u\end{array}}$

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 ofu for the square root ofU merely exchanges the two quadratics with one another.

The above solution shows that a quartic polynomial with rational coefficients and a zero coefficient on the cubic term is factorable into quadratics with rational coefficients if and only if either the resolvent cubic (2) has a non-zero root which is the square of a rational, orp² − 4r is the square of rational andq = 0; this can readily be checked using therational root test.^[15]

A variant of the previous method is due toEuler.^[16][17] Unlike the previous methods, both of which usesome root of the resolvent cubic, Euler's method uses all of them. Consider a depressed quarticx⁴ +px² +qx +r. Observe that, if

• x⁴ +px² +qx +r = (x² +sx +t)(x² −sx +v),
• r₁ andr₂ are the roots ofx² +sx +t,
• r₃ andr₄ are the roots ofx² −sx +v,

then

• the roots ofx⁴ +px² +qx +r arer₁,r₂,r₃, andr₄,
• r₁ +r₂ = −s,
• r₃ +r₄ =s.

Therefore,(r₁ +r₂)(r₃ +r₄) = −s². In other words,−(r₁ +r₂)(r₃ +r₄) is one of the roots of the resolvent cubic (2) and this suggests that the roots of that cubic are equal to−(r₁ +r₂)(r₃ +r₄),−(r₁ +r₃)(r₂ +r₄), and−(r₁ +r₄)(r₂ +r₃). This is indeed true and it follows fromVieta's formulas. It also follows from Vieta's formulas, together with the fact that we are working with a depressed quartic, thatr₁ +r₂ +r₃ +r₄ = 0. (Of course, this also follows from the fact thatr₁ +r₂ +r₃ +r₄ = −s +s.) Therefore, ifα,β, andγ are the roots of the resolvent cubic, then the numbersr₁,r₂,r₃, andr₄ are such that

${\displaystyle \{\begin{array}{l}{r}_1+r_2+r_3+r_4=0\\ (r_1+r_2)(r_3+r_4)=-\alpha \\ (r_1+r_3)(r_2+r_4)=-\beta \\ (r_1+r_4)(r_2+r_3)=-\gamma \text{.}\end{array}}$

It is a consequence of the first two equations thatr₁ +r₂ is a square root ofα and thatr₃ +r₄ is the other square root ofα. For the same reason,

• r₁ +r₃ is a square root ofβ,
• r₂ +r₄ is the other square root ofβ,
• r₁ +r₄ is a square root ofγ,
• r₂ +r₃ is the other square root ofγ.

Therefore, the numbersr₁,r₂,r₃, andr₄ are such that

${\displaystyle \{\begin{array}{l}{r}_1+r_2+r_3+r_4=0\\ r_1+r_2=\sqrt{\alpha }\\ r_1+r_3=\sqrt{\beta }\\ r_1+r_4=\sqrt{\gamma }\text{;}\end{array}}$

the sign of the square roots will be dealt with below. The only solution of this system is:

${\displaystyle \{\begin{array}{l}{r}_1=\frac{\sqrt{\alpha }+\sqrt{\beta }+\sqrt{\gamma }}{2}\\ r_2=\frac{\sqrt{\alpha }-\sqrt{\beta }-\sqrt{\gamma }}{2}\\ r_3=\frac{-\sqrt{\alpha }+\sqrt{\beta }-\sqrt{\gamma }}{2}\\ r_4=\frac{-\sqrt{\alpha }-\sqrt{\beta }+\sqrt{\gamma }}{2}\text{.}\end{array}}$

Since, in general, there are two choices for each square root, it might look as if this provides8 (= 2³) choices for the set{r₁,r₂,r₃,r₄}, but, in fact, it provides no more than2 such choices, because the consequence of replacing one of the square roots by the symmetric one is that the set{r₁,r₂,r₃,r₄} becomes the set{−r₁, −r₂, −r₃, −r₄}.

In order to determine the right sign of the square roots, one simply chooses some square root for each of the numbersα,β, andγ and uses them to compute the numbersr₁,r₂,r₃, andr₄ from the previous equalities. Then, one computes the number√α√β√γ. Sinceα,β, andγ are the roots of (2), it is a consequence of Vieta's formulas that their product is equal toq² and therefore that√α√β√γ = ±q. But a straightforward computation shows that

√α√β√γ =r₁r₂r₃ +r₁r₂r₄ +r₁r₃r₄ +r₂r₃r₄.

If this number is−q, then the choice of the square roots was a good one (again, by Vieta's formulas); otherwise, the roots of the polynomial will be−r₁,−r₂,−r₃, and−r₄, which are the numbers obtained if one of the square roots is replaced by the symmetric one (or, what amounts to the same thing, if each of the three square roots is replaced by the symmetric one).

This argument suggests another way of choosing the square roots:

• pickany square root√α ofα andany square root√β ofβ;
• define√γ as${\displaystyle -\frac{q}{\sqrt{\alpha }\sqrt{\beta }}}$.

Of course, this will make no sense ifα orβ is equal to0, but0 is a root of (2) only whenq = 0, that is, only when we are dealing with abiquadratic equation, in which case there is a much simpler approach.

Thesymmetric groupS₄ on four elements has theKlein four-group as anormal subgroup. This suggests using aresolvent cubic whose roots may be variously described as a discrete Fourier transform or aHadamard matrix transform of the roots; seeLagrange resolvents for the general method. Denote byx_i, fori from 0 to 3, the four roots ofx⁴ +bx³ +cx² +dx +e. If we set

then since the transformation is aninvolution we may express the roots in terms of the fours_i in exactly the same way. Since we know the values₀ = −⁠b/2⁠, we only need the values fors₁,s₂ ands₃. These are the roots of the polynomial

${\displaystyle (s^2-{s_1}^2)(s^2-{s_2}^2)(s^2-{s_3}^2).}$

Substituting thes_i by their values in term of thex_i, this polynomial may be expanded in a polynomial ins whose coefficients aresymmetric polynomials in thex_i. By thefundamental 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 isb = 0, this results in the polynomial

${\displaystyle s^6+2c{s}^4+(c^2-4e)s^2-d^2}$

3

This polynomial is of degree six, but only of degree three ins², and so the corresponding equation is solvable by the method described in the article aboutcubic function. By substituting the roots in the expression of thex_i in terms of thes_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 thex_i.

These expressions are unnecessarily complicated, involving thecubic roots of unity, which can be avoided as follows. Ifs is any non-zero root of (3), and if we set

then

${\displaystyle F_1(x)\times F_2(x)=x^4+c{x}^2+dx+e.}$

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

This gives exactly the same formula for the roots as the one provided byDescartes' method.

There is an alternative solution using algebraic geometry^[18] In brief, one interprets the roots as the intersection of two quadratic curves, then finds the threereducible 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 quarticx⁴ +px² +qx +r = 0 may also be expressed as thex coordinates of the intersections of the two quadratic equationsy² +py +qx +r = 0 andy −x² = 0 i.e., using the substitutiony =x² that two quadratics intersect in four points is an instance ofBézout's theorem. Explicitly, the four points areP_i ≔ (x_i,x_i²) for the four rootsx_i of the quartic.

These four points are not collinear because they lie on the irreducible quadraticy =x² and thus there is a 1-parameter family of quadratics (apencil of curves) passing through these points. Writing the projectivization of the two quadratics asquadratic forms in three variables:

the pencil is given by the formsλF₁ +μF₂ for any point[λ,μ] in the projective line — in other words, whereλ andμ 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${\displaystyle (\genfrac{}{}{0ex}{}{4}{2})}$ = 6 different ways. Denote theseQ₁ =L₁₂ +L₃₄,Q₂ =L₁₃ +L₂₄, andQ₃ =L₁₄ +L₂₃. Given any two of these, their intersection has exactly the four points.

The reducible quadratics, in turn, may be determined by expressing the quadratic formλF₁ +μF₂ as a3×3 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 inλ andμ and corresponds to the resolvent cubic.

• Linear function – Linear map or polynomial function of degree one
• Quadratic function – Polynomial function of degree two
• Cubic function – Polynomial function of degree 3
• Quintic function – Polynomial function of degree 5

^α For the purposes of this article,e is used as avariable as opposed to its conventional use asEuler's number (except when otherwise specified).

• Carpenter, W. (1966). "On the solution of the real quartic".Mathematics Magazine.39 (1):28–30.doi:10.2307/2688990.JSTOR 2688990.
• Yacoub, M.D.; Fraidenraich, G. (July 2012). "A solution to the quartic equation".Mathematical Gazette.96:271–275.doi:10.1017/s002555720000454x.S2CID 124512391.

• Quartic formula as four single equations atPlanetMath.
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