Quartic function#Biquadratic equation

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this solution, like all algebraic solutions of the quartic, requires the solution of a cubic to be found, it could not be published immediately.{{MacTutor|id=Ferrari|title=Lodovico Ferrari}} The solution of the quartic was published together with that of the cubic by Ferrari's mentor Gerolamo Cardano in the book Ars Magna.{{Citation | last = Cardano | first = Gerolamo | author-link = Gerolamo Cardano | year = 1993 | orig-year = 1545 | title = Ars magna or The Rules of Algebra | publisher = Dover | isbn = 0-486-67811-3 | url-access = registration | url = https://archive.org/details/arsmagnaorruleso0000card }}

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.Stewart, Ian, Galois Theory, Third Edition (Chapman & Hall/CRC Mathematics, 2004)

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 examples of other geometric problems whose solution involves 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 {{math|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.{{Cite web|url=http://people.math.gatech.edu/~etnyre/class/4441Fall16/ShifrinDiffGeo.pdf|title=DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces, p. 36|website=math.gatech.edu}}

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.{{Cite web|last=Weisstein|first=Eric W.|title=Crossed Ladders Problem|url=https://mathworld.wolfram.com/CrossedLaddersProblem.html|access-date=2020-07-27|website=mathworld.wolfram.com|language=en}}

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.{{MacTutor|id=Al-Haytham|title=Abu Ali al-Hasan ibn al-Haytham}}{{citation|title=Scientific Method, Statistical Method and the Speed of Light|first1=R. J.|last1=MacKay|first2=R. W.|last2=Oldford|journal=Statistical Science|volume=15|issue=3|date=August 2000|pages=254–78|doi=10.1214/ss/1009212817|mr=1847825|doi-access=free}}{{Citation|last = Neumann|first = Peter M.|author-link = Peter M. Neumann|journal = American Mathematical Monthly|title = Reflections on Reflection in a Spherical Mirror|year = 1998|volume = 105|issue = 6|pages = 523–528|doi = 10.2307/2589403|jstor = 2589403}}

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.{{Cite book|last=Shabana|first=A. A.|url=https://books.google.com/books?id=G2UyBTji18oC&q=Timoshenko-Rayleigh+theory&pg=PA2|title=Theory of Vibration: An Introduction|date=1995-12-08|publisher=Springer Science & Business Media|isbn=978-0-387-94524-8|language=en}}

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

Letting {{mvar|F}} and {{mvar|G}} be the distinct inflection points of the graph of a quartic function, and letting {{mvar|H}} be the intersection of the inflection secant line {{mvar|FG}} and the quartic, nearer to {{mvar|G}} than to {{mvar|F}}, then {{mvar|G}} divides {{mvar|FH}} into the golden section:{{Citation|last = Aude|first = H. T. R.|journal = American Mathematical Monthly|year = 1949|issue = 3|volume = 56|title = Notes on Quartic Curves|jstor = 2305030|doi = 10.2307/2305030|pages=165–170}}

:$\backslash frac\{FG\}\{GH\}=\backslash frac\{1+\backslash sqrt\{5\}\}\{2\}=\; \backslash varphi\; \backslash ;\; (\backslash text\{the\; golden\; ratio\}).$

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

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

with real coefficients and {{math|a ≠ 0}} the nature of its roots is mainly determined by the sign of its discriminant

:$\backslash begin\{align\}$

\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 {{math|{{sfrac|P|8a²}}}} is the second degree coefficient of the associated depressed quartic (see below);

:$R=\; b^3+8da^2-4abc,$

such that {{math|{{sfrac|R|8a³}}}} is the first degree coefficient of the associated depressed quartic;

:$\backslash Delta\_0\; =\; c^2\; -\; 3bd\; +\; 12ae,$

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:{{cite journal|first= E. L.|last=Rees|title=Graphical Discussion of the Roots of a Quartic Equation|journal = The American Mathematical Monthly|volume=29|issue=2|year=1922|pages=51–55|doi=10.2307/2972804|jstor = 2972804}}

• If {{math|∆ < 0}} then the equation has two distinct real roots and two complex conjugate non-real roots.
• If {{math|∆ > 0}} then either the equation's four roots are all real or none is.
• If {{mvar|P}} < 0 and {{mvar|D}} < 0 then all four roots are real and distinct.
• If {{mvar|P}} > 0 or {{mvar|D}} > 0 then there are two pairs of non-real complex conjugate roots.{{Cite journal | last1 = Lazard | first1 = D. | doi = 10.1016/S0747-7171(88)80015-4 | title = Quantifier elimination: Optimal solution for two classical examples | journal = Journal of Symbolic Computation | volume = 5 | pages = 261–266 | year = 1988 | issue = 1–2 | doi-access = free }}
• If {{math|∆ {{=}} 0}} then (and only then) the polynomial has a multiple root. Here are the different cases that can occur:
• If {{mvar|P}} < 0 and {{mvar|D}} < 0 and {{math|∆₀ ≠ 0}}, there are a real double root and two real simple roots.
• If {{mvar|D}} > 0 or ({{mvar|P}} > 0 and ({{mvar|D}} ≠ 0 or {{mvar|R}} ≠ 0)), there are a real double root and two complex conjugate roots.
• If {{math|∆₀ {{=}} 0}} and {{mvar|D}} ≠ 0, there are a triple root and a simple root, all real.
• If {{mvar|D}} = 0, then:
• If {{mvar|P}} < 0, there are two real double roots.
• If {{mvar|P}} > 0 and {{mvar|R}} = 0, there are two complex conjugate double roots.
• If {{math|∆₀ {{=}} 0}}, all four roots are equal to {{math|−{{sfrac|b|4a}}}}

There are some cases that do not seem to be covered, but in fact they cannot occur. For example, {{math|∆₀ > 0}}, {{mvar|P}} = 0 and {{mvar|D}} ≤ 0 is not one of the cases. In fact, if {{math|∆₀ > 0}} and {{mvar|P}} = 0 then {{mvar|D}} > 0, since $16\; a^2\backslash Delta\_0\; =\; 3D\; +\; P^2;$ so this combination is not possible.

File:Quartic Formula.svg

The four roots {{math|x₁}}, {{math|x₂}}, {{math|x₃}}, and {{math|x₄}} for the general quartic equation

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

with {{mvar|a}} ≠ 0 are given in the following formula, which is deduced from the one in the section on Ferrari's method by back changing the variables (see {{slink||Converting to a depressed quartic}}) and using the formulas for the quadratic and cubic equations.

:$\backslash begin\{align\}$

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 {{mvar|p}} and {{mvar|q}} are the coefficients of the second and of the first degree respectively in the associated depressed quartic

:$\backslash begin\{align\}$

p &= \frac{8ac-3b^2}{8a^2}\\

q &= \frac{b^3 - 4abc + 8a^2d}{8a^3}

\end{align}

:

and where

:$\backslash begin\{align\}$

S &= \frac{1}{2}\sqrt{-\frac23\ p+\frac{1}{3a}\left(Q + \frac{\Delta_0}{Q}\right)}\\

Q &= \sqrt[3]{\frac{\Delta_1 + \sqrt{\Delta_1^2 - 4\Delta_0^3}}{2}}

\end{align}

(if {{math|S {{=}} 0}} or {{math|Q {{=}} 0}}, see {{slink||Special cases of the formula}}, below)

with

:$\backslash begin\{align\}$

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

\Delta_1 &= 2c^3 - 9bcd + 27b^2 e + 27ad^2 - 72ace

\end{align}

and

:$\backslash Delta\_1^2-4\backslash Delta\_0^3\; =\; -\; 27\; \backslash Delta\backslash \; ,$ where $\backslash Delta$ is the aforementioned discriminant. For the cube root expression for Q, 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 their cubic counterparts.

• If $\backslash Delta\; >\; 0,$ 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, despite being expressed in terms of $Q;$ this is casus 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 using trigonometric functions, as follows:

::$S\; =\; \backslash frac\{1\}\{2\}\; \backslash sqrt\{-\backslash frac23\backslash \; p+\backslash frac\{2\}\{3a\}\backslash sqrt\{\backslash Delta\_0\}\backslash cos\backslash frac\{\backslash varphi\}\{3\}\}$

:where

::$\backslash varphi\; =\; \backslash arccos\backslash left(\backslash frac\{\backslash Delta\_1\}\{2\backslash sqrt\{\backslash Delta\_0^3\}\}\backslash right).$

• If $\backslash Delta\; \backslash neq\; 0$ and $\backslash Delta\_0\; =\; 0,$ the sign of $\backslash sqrt\{\backslash Delta\_1^2\; -\; 4\; \backslash Delta\_0^3\}=\backslash sqrt\{\backslash Delta\_1^2\}$ has to be chosen to have $Q\; \backslash neq\; 0,$ that is one should define $\backslash sqrt\{\backslash Delta\_1^2\}$ as $\backslash Delta\_1,$ maintaining the sign of $\backslash Delta\_1.$
• If $S\; =\; 0,$ then one must change the choice of the cube root in $Q$ in order to have $S\; \backslash neq\; 0.$ This is always possible except if the quartic may be factored into $\backslash left(x+\backslash tfrac\{b\}\{4a\}\backslash right)^4.$ 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{{Clarify|reason=Both kinds or specific?|date=March 2024}} 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 $\backslash Delta\; =\; 0$ and $\backslash Delta\_0\; =\; 0,$ and thus also $\backslash Delta\_1\; =\; 0,$ at least three roots are equal to each other, and the roots are rational functions of the coefficients. The triple root $x\_0$ is a common root of the quartic and its second derivative $2(6ax^2+3bx+c);$ it is thus also the unique root of the remainder of the Euclidean division of the quartic by its second derivative, which is a linear polynomial. The simple root $x\_1$ can be deduced from $x\_1+3x\_0=-b/a.$
• If $\backslash Delta=0$ and $\backslash Delta\_0\; \backslash neq\; 0,$ 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.

Consider the general quartic

:$Q(x)\; =\; a\_4x^4+a\_3x^3+a\_2x^2+a\_1x+a\_0.$

It is reducible if {{math|Q(x) {{=}} R(x)×S(x)}}, where {{math|R(x)}} and {{math|S(x)}} are non-constant polynomials with rational coefficients (or more generally with coefficients in the same field as the coefficients of {{math|Q(x)}}). Such a factorization will take one of two forms:

:$Q(x)\; =\; (x-x\_1)(b\_3x^3+b\_2x^2+b\_1x+b\_0)$

or

:$Q(x)\; =\; (c\_2x^2+c\_1x+c\_0)(d\_2x^2+d\_1x+d\_0).$

In either case, the roots of {{math|Q(x)}} are the roots of the factors, which may be computed using the formulas for the roots of a quadratic function or cubic function.

Detecting the existence of such factorizations can be done Resolvent cubic#Factoring quartic polynomials. It turns out that:

• if we are working over {{math|R}} (that is, if coefficients are restricted to be real numbers) (or, more generally, over some real closed field) then there is always such a factorization;
• if we are working over {{math|Q}} (that is, if coefficients are restricted to be rational numbers) then there is an algorithm to determine whether or not {{math|Q(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.

If {{math|a₃ {{=}} a₁ {{=}} 0}} then the function

:$$

Q(x) = a_4x^4+a_2x^2+a_0

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

Let the auxiliary variable {{math|z {{=}} x²}}.

Then {{math|Q(x)}} becomes a quadratic {{math|q}} in {{math|z}}: {{math|q(z) {{=}} a₄z² + a₂z + a₀}}. Let {{math|z₊}} and {{math|z₋}} be the roots of {{math|q(z)}}. Then the roots of the quartic {{math|Q(x)}} are

:$$

\begin{align}

x_1&=+\sqrt{z_+},

\\

x_2&=-\sqrt{z_+},

\\

x_3&=+\sqrt{z_-},

\\

x_4&=-\sqrt{z_-}.

\end{align}

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 {{math|P(mx) {{=}} {{sfrac|x⁴|m²}}P({{sfrac|m|x}})}} (it is palindromic if {{math|m {{=}} 1}}). The change of variables {{math|z {{=}} x + {{sfrac|m|x}}}} in {{math|{{sfrac|P(x)|x²}} {{=}} 0}} produces the quadratic equation {{math|a₀z² + a₁z + a₂ − 2ma₀ {{=}} 0}}. Since {{math|x² − xz + m {{=}} 0}}, the quartic equation {{math|P(x) {{=}} 0}} may be solved by applying the quadratic formula twice.

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 {{math|a₄}}, provides the equivalent equation {{math|x⁴ + bx³ + cx² + dx + e {{=}} 0}}, with {{math|b {{=}} {{sfrac|a₃|a₄}}}}, {{math|c {{=}} {{sfrac|a₂|a₄}}}}, {{math|d {{=}} {{sfrac|a₁|a₄}}}}, and {{math|e {{=}} {{sfrac|a₀|a₄}}}}.

Substituting {{math|y − {{sfrac|b|4}}}} for {{mvar|x}} gives, after regrouping the terms, the equation {{math|y⁴ + py² + qy + r {{=}} 0}},

where

:$\backslash begin\{align\}$

p&=\frac{8c-3b^2}{8} =\frac{8a_2a_4-3{a_3}^2}{8{a_4}^2}\\

q&=\frac{b^3-4bc+8d}{8} =\frac{{a_3}^3-4a_2a_3a_4+8a_1{a_4}^2}{8{a_4}^3}\\

r&=\frac{-3b^4+256e-64bd+16b^2c}{256}=\frac{-3{a_3}^4+256a_0{a_4}^3-64a_1a_3{a_4}^2+16a_2{a_3}^2a_4}{256{a_4}^4}.

\end{align}

If {{math|y₀}} is a root of this depressed quartic, then {{math|y₀ − {{sfrac|b|4}}}} (that is {{math|y₀ − {{sfrac|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 the depressed quartic equation

:$y^4\; +\; p\; y^2\; +\; q\; y\; +\; r\; =\; 0.$

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) as

:$\backslash left(y^2\; +\; \backslash frac\; p2\backslash right)^2\; =\; -q\; y\; -\; r\; +\; \backslash frac\{p^2\}4.$

Then, we introduce a variable {{mvar|m}} into the factor on the left-hand side by adding {{math|2y²m + pm + m²}} to both sides. After regrouping the coefficients of the power of {{mvar|y}} on the right-hand side, this gives the equation

{{NumBlk|:|$\backslash left(y^2\; +\; \backslash frac\; p2\; +\; m\backslash right)^2\; =\; 2\; m\; y^2\; -\; q\; y\; +\; m^2\; +\; m\; p\; +\; \backslash frac\{p^2\}4\; -\; r,$|{{EquationRef|1}}}}

which is equivalent to the original equation, whichever value is given to {{mvar|m}}.

As the value of {{mvar|m}} may be arbitrarily chosen, we will choose it in order to complete the square on the right-hand side. This implies that the discriminant in {{mvar|y}} of this quadratic equation is zero, that is {{mvar|m}} is a root of the equation

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

which may be rewritten as

{{NumBlk|:|$8m^3+\; 8pm^2\; +\; (2p^2\; -8r)m-\; q^2\; =0.$|{{EquationRef|1a}}}}

This is the resolvent cubic of the quartic equation. The value of {{mvar|m}} may thus be obtained from Cardano's formula. When {{mvar|m}} is a root of this equation, the right-hand side of equation ({{EquationNote|1}}) is the square

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

However, this induces a division by zero if {{math|m {{=}} 0}}. This implies {{math|q {{=}} 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 that {{math|m ≠ 0}}. This is always possible except for the depressed equation {{math|y⁴ {{=}} 0}}.

Now, if {{mvar|m}} is a root of the cubic equation such that {{math|m ≠ 0}}, equation ({{EquationNote|1}}) becomes

:$\backslash left(y^2\; +\; \backslash frac\; p2\; +\; m\backslash right)^2\; =\; \backslash left(y\backslash sqrt\{2\; m\}-\backslash frac\{q\}\{2\backslash sqrt\{2\; m\}\}\backslash right)^2.$

This equation is of the form {{math|M² {{=}} N²}}, which can be rearranged as {{math|M² − N² {{=}} 0}} or {{math|(M + N)(M − N) {{=}} 0}}. Therefore, equation ({{EquationNote|1}}) may be rewritten as

:$\backslash left(y^2\; +\; \backslash frac\; p2\; +\; m\; +\; \backslash sqrt\{2\; m\}y-\backslash frac\; q\{2\backslash sqrt\{2\; m\}\}\backslash right)\; \backslash left(y^2\; +\; \backslash frac\; p2\; +\; m\; -\; \backslash sqrt\{2\; m\}y+\backslash frac\; q\{2\backslash sqrt\{2\; m\}\}\backslash right)=0.$

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

:$y=\{\backslash pm\_1\backslash sqrt\{2\; m\}\; \backslash pm\_2\; \backslash sqrt\{-\backslash left(2p\; +\; 2m\; \backslash pm\_1\; \{\backslash sqrt\; 2q\; \backslash over\; \backslash sqrt\{m\}\}\; \backslash right)\}\; \backslash over\; 2\},$

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

Therefore, the solutions of the original quartic equation are

:$x=-\{a\_3\; \backslash over\; 4a\_4\}\; +\; \{\backslash pm\_1\backslash sqrt\{2\; m\}\; \backslash pm\_2\; \backslash sqrt\{-\backslash left(2p\; +\; 2m\; \backslash pm\_1\; \{\backslash sqrt2q\; \backslash over\; \backslash sqrt\{m\}\}\; \backslash right)\}\; \backslash over\; 2\}.$

A comparison with the general formula above shows that {{math|{{sqrt|2m}} {{=}} 2S}}.

Descartes{{Citation|last = Descartes|first = René|author-link = René Descartes|title = The Geometry of Rene Descartes with a facsimile of the first edition|isbn = 0-486-60068-8|publisher = Dover|year = 1954|jfm = 51.0020.07|chapter = Book III: On the construction of solid and supersolid problems|orig-year = 1637}} introduced in 1637 the method of finding the roots of a quartic polynomial by factoring it into two quadratic ones. Let

:$$

\begin{align}

x^4 + bx^3 + cx^2 + dx + e & = (x^2 + sx + t)(x^2 + ux + v) \\

& = x^4 + (s + u)x^3 + (t + v + su)x^2 + (sv + tu)x + tv

\end{align}

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

:$$

\left\{\begin{array}{l}

b = s + u \\

c = t + v + su \\

d = sv + tu \\

e = tv

\end{array}\right.

This can be simplified by starting again with the depressed quartic {{math|y⁴ + py² + qy + r}}, which can be obtained by substituting {{math|y − b/4}} for {{math|x}}. Since the coefficient of {{math|y³}} is {{math|0}}, we get {{math|s {{=}} −u}}, and:

:$$

\left\{\begin{array}{l}

p + u^2 = t + v \\

q = u (t - v) \\

r = tv

\end{array}\right.

One can now eliminate both {{mvar|t}} and {{mvar|v}} by doing the following:

:$$

\begin{align}

u^2(p + u^2)^2 - q^2 & = u^2(t + v)^2 - u^2(t - v)^2 \\

& = u^2 [(t + v + (t - v))(t + v - (t - v))]\\

& = u^2(2t)(2v) \\

& = 4u^2tv \\

& = 4u^2r

\end{align}

If we set {{math|U {{=}} u²}}, then solving this equation becomes finding the roots of the resolvent cubic

{{NumBlk|:|$U^3\; +\; 2pU^2\; +\; (p^2-4r)U\; -\; q^2,$|{{EquationRef|2}}}}

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

If {{math|u}} is a square root of a non-zero root of this resolvent (such a non-zero root exists except for the quartic {{math|x⁴}}, which is trivially factored),

:$$

\left\{\begin{array}{l}

s = -u \\

2t = p + u^2 + q/u \\

2v = p + u^2 - q/u

\end{array}\right.

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 {{mvar|u}} for the square root of {{mvar|U}} 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 ({{EquationNote|2}}) has a non-zero root which is the square of a rational, or {{math|p² − 4r}} is the square of rational and {{math|q {{=}} 0}}; this can readily be checked using the rational root test.{{cite journal |author=Brookfield, G. |title=Factoring quartic polynomials: A lost art |journal=Mathematics Magazine |volume=80 |issue=1 |year=2007 |pages=67–70|doi=10.1080/0025570X.2007.11953453 |s2cid=53375377 |url = https://www.maa.org/sites/default/files/Brookfield2007-103574.pdf}}

A variant of the previous method is due to Euler.{{Citation|last = van der Waerden|first=Bartel Leendert|author-link = Bartel Leendert van der Waerden|title = Algebra|volume = 1|publisher=Springer-Verlag|edition = 7th|isbn = 0-387-97424-5|year = 1991|section = The Galois theory: Equations of the second, third, and fourth degrees|zbl = 0724.12001}}{{Citation|last = Euler|first = Leonhard|author-link = Leonhard Euler|title = Elements of Algebra|chapter= Of a new method of resolving equations of the fourth degree|publisher=Springer-Verlag|orig-year = 1765|year = 1984|zbl = 0557.01014|isbn = 978-1-4613-8511-0}} Unlike the previous methods, both of which use some root of the resolvent cubic, Euler's method uses all of them. Consider a depressed quartic {{math|x⁴ + px² + qx + r}}. Observe that, if

• {{math|x⁴ + px² + qx + r {{=}} (x² + sx + t)(x² − sx + v)}},
• {{math|r₁}} and {{math|r₂}} are the roots of {{math|x² + sx + t}},
• {{math|r₃}} and {{math|r₄}} are the roots of {{math|x² − sx + v}},

then

• the roots of {{math|x⁴ + px² + qx + r}} are {{math|r₁}}, {{math|r₂}}, {{math|r₃}}, and {{math|r₄}},
• {{math|r₁ + r₂ {{=}} −s}},
• {{math|r₃ + r₄ {{=}} s}}.

Therefore, {{math|(r₁ + r₂)(r₃ + r₄) {{=}} −s²}}. In other words, {{math|−(r₁ + r₂)(r₃ + r₄)}} is one of the roots of the resolvent cubic ({{EquationNote|2}}) and this suggests that the roots of that cubic are equal to {{math|−(r₁ + r₂)(r₃ + r₄)}}, {{math|−(r₁ + r₃)(r₂ + r₄)}}, and {{math|−(r₁ + r₄)(r₂ + r₃)}}. This is indeed true and it follows from Vieta's formulas. It also follows from Vieta's formulas, together with the fact that we are working with a depressed quartic, that {{math|r₁ + r₂ + r₃ + r₄ {{=}} 0}}. (Of course, this also follows from the fact that {{math|r₁ + r₂ + r₃ + r₄ {{=}} −s + s}}.) Therefore, if {{math|α}}, {{math|β}}, and {{math|γ}} are the roots of the resolvent cubic, then the numbers {{math|r₁}}, {{math|r₂}}, {{math|r₃}}, and {{math|r₄}} are such that

:$\backslash left\backslash \{\backslash begin\{array\}\{l\}r\_1+r\_2+r\_3+r\_4=0\backslash \backslash (r\_1+r\_2)(r\_3+r\_4)=-\backslash alpha\backslash \backslash (r\_1+r\_3)(r\_2+r\_4)=-\backslash beta\backslash \backslash (r\_1+r\_4)(r\_2+r\_3)=-\backslash gamma\backslash text\{.\}\backslash end\{array\}\backslash right.$

It is a consequence of the first two equations that {{math|r₁ + r₂}} is a square root of {{math|α}} and that {{math|r₃ + r₄}} is the other square root of {{math|α}}. For the same reason,

• {{math|r₁ + r₃}} is a square root of {{math|β}},
• {{math|r₂ + r₄}} is the other square root of {{math|β}},
• {{math|r₁ + r₄}} is a square root of {{math|γ}},
• {{math|r₂ + r₃}} is the other square root of {{math|γ}}.

Therefore, the numbers {{math|r₁}}, {{math|r₂}}, {{math|r₃}}, and {{math|r₄}} are such that

:$\backslash left\backslash \{\backslash begin\{array\}\{l\}r\_1+r\_2+r\_3+r\_4=0\backslash \backslash r\_1+r\_2=\backslash sqrt\{\backslash alpha\}\backslash \backslash r\_1+r\_3=\backslash sqrt\{\backslash beta\}\backslash \backslash r\_1+r\_4=\backslash sqrt\{\backslash gamma\}\backslash text\{;\}\backslash end\{array\}\backslash right.$

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

:$\backslash left\backslash \{\backslash begin\{array\}\{l\}r\_1=\backslash frac\{\backslash sqrt\{\backslash alpha\}+\backslash sqrt\{\backslash beta\}+\backslash sqrt\{\backslash gamma\}\}2\backslash \backslash [2mm]r\_2=\backslash frac\{\backslash sqrt\{\backslash alpha\}-\backslash sqrt\{\backslash beta\}-\backslash sqrt\{\backslash gamma\}\}2\backslash \backslash [2mm]r\_3=\backslash frac\{-\backslash sqrt\{\backslash alpha\}+\backslash sqrt\{\backslash beta\}-\backslash sqrt\{\backslash gamma\}\}2\backslash \backslash [2mm]r\_4=\backslash frac\{-\backslash sqrt\{\backslash alpha\}-\backslash sqrt\{\backslash beta\}+\backslash sqrt\{\backslash gamma\}\}2\backslash text\{.\}\backslash end\{array\}\backslash right.$

Since, in general, there are two choices for each square root, it might look as if this provides {{math|8 ({{=}} 2³)}} choices for the set {{math|{r₁, r₂, r₃, r₄}}}, but, in fact, it provides no more than {{math|2}} such choices, because the consequence of replacing one of the square roots by the symmetric one is that the set {{math|{r₁, r₂, r₃, r₄}}} becomes the set {{math|{−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 numbers {{math|α}}, {{math|β}}, and {{math|γ}} and uses them to compute the numbers {{math|r₁}}, {{math|r₂}}, {{math|r₃}}, and {{math|r₄}} from the previous equalities. Then, one computes the number {{math|{{sqrt|α}}{{sqrt|β}}{{sqrt|γ}}}}. Since {{math|α}}, {{math|β}}, and {{math|γ}} are the roots of ({{EquationNote|2}}), it is a consequence of Vieta's formulas that their product is equal to {{math|q²}} and therefore that {{math|{{sqrt|α}}{{sqrt|β}}{{sqrt|γ}} {{=}} ±q}}. But a straightforward computation shows that

:{{math|{{sqrt|α}}{{sqrt|β}}{{sqrt|γ}} {{=}} r₁r₂r₃ + r₁r₂r₄ + r₁r₃r₄ + r₂r₃r₄.}}

If this number is {{math|−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 {{math|−r₁}}, {{math|−r₂}}, {{math|−r₃}}, and {{math|−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:

• pick any square root {{math|{{sqrt|α}}}} of {{math|α}} and any square root {{math|{{sqrt|β}}}} of {{math|β}};
• define {{math|{{sqrt|γ}}}} as $-\backslash frac\; q\{\backslash sqrt\{\backslash alpha\}\backslash sqrt\{\backslash beta\}\}$.

Of course, this will make no sense if {{math|α}} or {{math|β}} is equal to {{math|0}}, but {{math|0}} is a root of ({{EquationNote|2}}) only when {{math|q {{=}} 0}}, that is, only when we are dealing with a biquadratic equation, in which case there is a much simpler approach.

The symmetric group {{math|S₄}} on four elements has the Klein four-group as a normal subgroup. This suggests using a {{visible anchor|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 {{math|x_i}}, for {{math|i}} from {{math|0}} to {{math|3}}, the four roots of {{math|x⁴ + bx³ + cx² + dx + e}}. If we set

: $\backslash begin\{align\}$

s_0 &= \tfrac12(x_0 + x_1 + x_2 + x_3), \\[4pt]

s_1 &= \tfrac12(x_0 - x_1 + x_2 - x_3), \\[4pt]

s_2 &= \tfrac12(x_0 + x_1 - x_2 - x_3), \\[4pt]

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 {{math|s_i}} in exactly the same way. Since we know the value {{math|s₀ {{=}} −{{sfrac|b|2}}}}, we only need the values for {{math|s₁}}, {{math|s₂}} and {{math|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 {{math|s_i}} by their values in term of the {{math|x_i}}, this polynomial may be expanded in a polynomial in {{math|s}} whose coefficients are symmetric polynomials in the {{math|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 {{math|b {{=}} 0}}, this results in the polynomial

{{NumBlk|:|$s^6+2cs^4+(c^2-4e)s^2-d^2$|{{EquationRef|3}}}}

This polynomial is of degree six, but only of degree three in {{math|s²}}, and so the corresponding equation is solvable by the method described in the article about cubic function. By substituting the roots in the expression of the {{math|x_i}} in terms of the {{math|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 {{math|x_i}}.

These expressions are unnecessarily complicated, involving the cubic roots of unity, which can be avoided as follows. If {{math|s}} is any non-zero root of ({{EquationNote|3}}), and if we set

:$\backslash begin\{align\}$

F_1(x) & = x^2 + sx + \frac{c}{2} + \frac{s^2}{2} - \frac{d}{2s} \\

F_2(x) & = x^2 - sx + \frac{c}{2} + \frac{s^2}{2} + \frac{d}{2s}

\end{align}

then

:$F\_1(x)\backslash times\; F\_2(x)\; =\; x^4\; +\; cx^2\; +\; dx\; +\; e.$

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

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

There is an alternative solution using algebraic geometry{{Citation|last = Faucette|first = William M.|journal = American Mathematical Monthly|pages = 51–57|title = A Geometric Interpretation of the Solution of the General Quartic Polynomial|volume = 103|year = 1996|issue = 1|doi = 10.2307/2975214|jstor = 2975214|mr = 1369151}} 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 {{math|x⁴ + px² + qx + r {{=}} 0}} may also be expressed as the {{mvar|x}} coordinates of the intersections of the two quadratic equations {{math|y² + py + qx + r {{=}} 0}} and {{math|y − x² {{=}} 0}} i.e., using the substitution {{math|y {{=}} x²}} that two quadratics intersect in four points is an instance of Bézout's theorem. Explicitly, the four points are {{math|P_i ≔ (x_i, x_i²)}} for the four roots {{math|x_i}} of the quartic.

These four points are not collinear because they lie on the irreducible quadratic {{math|y {{=}} x²}} 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:

:$\backslash begin\{align\}$

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 {{math|λF₁ + μF₂}} for any point {{math|[λ, μ]}} in the projective line — in other words, where {{math|λ}} and {{math|μ}} 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 $\backslash textstyle\{\backslash binom\{4\}\{2\}\}$ = {{math|6}} different ways. Denote these {{math|Q₁ {{=}} L₁₂ + L₃₄}}, {{math|Q₂ {{=}} L₁₃ + L₂₄}}, and {{math|Q₃ {{=}} 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 form {{math|λF₁ + μF₂}} as a {{math|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 {{math|λ}} and {{math|μ}} and corresponds to the resolvent cubic.

• {{annotated link|Linear function}}
• {{annotated link|Quadratic function}}
• {{annotated link|Cubic function}}
• {{annotated link|Quintic function}}

:{{note|Alpha|α}} For the purposes of this article, e is used as a variable as opposed to its conventional use as Euler's number (except when otherwise specified).

{{reflist}}

• {{cite journal |author=Carpenter, W. |title=On the solution of the real quartic |journal=Mathematics Magazine |volume=39 |year=1966 |issue=1 |pages=28–30 |doi=10.2307/2688990|jstor=2688990 }}
• {{cite journal |author1=Yacoub, M.D.|author2=Fraidenraich, G. |title=A solution to the quartic equation |journal=Mathematical Gazette |volume=96|date=July 2012|pages=271–275 |doi=10.1017/s002555720000454x|s2cid=124512391 }}

• {{PlanetMath | urlname = QuarticFormula | title = Quartic formula as four single equations }}
• [http://members.tripod.com/l_ferrari/quartic_equation.htm Ferrari's achievement]

{{Polynomials}}

Category:Elementary algebra

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