Editing Quartic function (section)

==Solution==
===Nature of the roots===
Given the general quartic equation
:<math>ax^4 + bx^3 + cx^2 + dx + e = 0</math>
with real coefficients and {{math|''a'' ≠ 0}} the nature of its roots is mainly determined by the sign of its [[discriminant]] 
:<math>\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} </math> 
This may be refined by considering the signs of four other polynomials:
:<math>P = 8ac - 3b^2</math>
such that {{math|{{sfrac|''P''|8''a''<sup>2</sup>}}}} is the second degree coefficient of the associated depressed quartic (see [[#Converting_to_a_depressed_quartic|below]]);
:<math>R= b^3+8da^2-4abc,</math>
such that {{math|{{sfrac|''R''|8''a''<sup>3</sup>}}}} is the first degree coefficient of the associated depressed quartic; 
:<math>\Delta_0 = c^2 - 3bd + 12ae,</math>
which is 0 if the quartic has a triple root; and
:<math>D = 64 a^3 e - 16 a^2 c^2 + 16 a b^2 c - 16 a^2 bd - 3 b^4</math>
which is 0 if the quartic has two double roots.

The possible cases for the nature of the roots are as follows:<ref>{{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}}</ref>

* 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.<ref>{{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 }}</ref>
* If {{math|∆ {{=}} 0}} then (and only then) the polynomial has a [[multiplicity (mathematics)|multiple]] root. Here are the different cases that can occur:
** If {{mvar|P}} < 0 and {{mvar|D}} < 0 and {{math|∆<sub>0</sub> ≠ 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|∆<sub>0</sub> {{=}} 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|∆<sub>0</sub> {{=}} 0}}, all four roots are equal to {{math|−{{sfrac|''b''|4''a''}}}}

There are some cases that do not seem to be covered, but in fact they cannot occur.  For example, {{math|∆<sub>0</sub> > 0}}, {{mvar|P}} = 0 and {{mvar|D}} ≤ 0 is one of the cases.  In fact, if {{math|∆<sub>0</sub> > 0}} and {{mvar|P}} = 0 then  {{mvar|D}} > 0, since <math>16 a^2\Delta_0 = 3D + P^2; </math> so this combination is not possible.

===General formula for roots===
[[File:Quartic Formula.svg|thumb|600px|right|Solution of <math>x^4+ax^3+bx^2+cx+d=0</math> written out in full. This formula is too unwieldy for general use; hence other methods, or simpler formulas for special cases, are generally used.]]

The four roots {{math|''x''<sub>1</sub>}}, {{math|''x''<sub>2</sub>}}, {{math|''x''<sub>3</sub>}}, and {{math|''x''<sub>4</sub>}} for the general quartic equation
:<math>ax^4+bx^3+cx^2+dx+e=0 \,</math>
with {{mvar|a}} ≠ 0 are given in the following formula, which is deduced from the one in the section on [[#Ferrari's solution|Ferrari's method]] by back changing the variables (see {{slink||Converting to a depressed quartic}}) and using the formulas for the [[Quadratic function|quadratic]] and [[Cubic function#General formula for roots|cubic equation]]s.
:<math>\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}</math>

where {{mvar|p}} and {{mvar|q}} are the coefficients of the second and of the first degree respectively in the [[#Converting to a depressed quartic|associated depressed quartic]]
:<math>\begin{align}
p &= \frac{8ac-3b^2}{8a^2}\\
q &= \frac{b^3 - 4abc + 8a^2d}{8a^3} 
\end{align}</math>
:
and where
:<math>\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}</math>
(if {{math|''S'' {{=}} 0}} or {{math|''Q'' {{=}} 0}}, see {{slink||Special cases of the formula}}, below)

with
:<math>\begin{align}
\Delta_0 &= c^2 - 3bd + 12ae\\
\Delta_1 &= 2c^3 - 9bcd + 27b^2 e + 27ad^2 - 72ace
\end{align}</math>
and
:<math>\Delta_1^2-4\Delta_0^3 = - 27 \Delta\ ,</math> where <math>\Delta</math> 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 function#Algebraic solution|cubic counterparts]].

====Special cases of the formula====
*If <math>\Delta > 0,</math> the value of <math>Q</math> 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 <math>S</math> is also real, despite being expressed in terms of <math>Q;</math> 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:
::<math>S = \frac{1}{2} \sqrt{-\frac23\ p+\frac{2}{3a}\sqrt{\Delta_0}\cos\frac{\varphi}{3}}</math>
:where
::<math>\varphi = \arccos\left(\frac{\Delta_1}{2\sqrt{\Delta_0^3}}\right).</math>

*If <math>\Delta \neq 0</math> and <math>\Delta_0 = 0,</math> the sign of <math>\sqrt{\Delta_1^2 - 4 \Delta_0^3}=\sqrt{\Delta_1^2} </math>  has to be chosen to have <math>Q \neq 0,</math> that is one should define <math>\sqrt{\Delta_1^2}</math> as <math>\Delta_1,</math> maintaining the sign of <math>\Delta_1.</math>
*If <math>S = 0,</math> then one must change the choice of the cube root in <math>Q</math> in order to have <math>S \neq 0.</math> This is always possible except if the quartic may be factored into <math>\left(x+\tfrac{b}{4a}\right)^4.</math> 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 <math>q</math> is zero, in which case the associated [[#Converting to a depressed quartic|depressed quartic]] is biquadratic; it may thus be solved by the method described [[#Biquadratic equation|below]].
*If <math>\Delta = 0</math> and <math>\Delta_0 = 0,</math> and thus also <math>\Delta_1 = 0,</math> at least three roots are equal to each other, and the roots are [[rational function]]s of the coefficients. The triple root <math>x_0</math> is a common root of the quartic and its second derivative <math>2(6ax^2+3bx+c);</math> 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 <math>x_1</math> can be deduced from <math>x_1+3x_0=-b/a.</math>
*If <math>\Delta=0</math> and <math> \Delta_0 \neq 0,</math> the above expression for the roots is correct but misleading, hiding the fact that the polynomial is [[irreducible polynomial|reducible]] and no cube root is needed to represent the roots.

===Simpler cases===
====Reducible quartics====
Consider the general quartic
:<math>Q(x) = a_4x^4+a_3x^3+a_2x^2+a_1x+a_0.</math>
It is [[irreducible polynomial|reducible]] if {{math|''Q''(''x'') {{=}} ''R''(''x'')×''S''(''x'')}}, where {{math|''R''(''x'')}} and {{math|''S''(''x'')}} are non-constant polynomials with [[rational number|rational]] coefficients (or more generally with coefficients in the same [[field (mathematics)|field]] as the coefficients of {{math|''Q''(''x'')}}). Such a factorization will take one of two forms:

:<math>Q(x) = (x-x_1)(b_3x^3+b_2x^2+b_1x+b_0)</math>
or
:<math>Q(x) = (c_2x^2+c_1x+c_0)(d_2x^2+d_1x+d_0).</math>
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|using the resolvent cubic of {{math|''Q''(''x'')}}]]. 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 ([[Quartic function#Ferrari's solution|Ferrari's method]], [[Quartic function#Descartes' solution|Descartes' method]], and, to a lesser extent, [[Quartic function#Euler's solution|Euler's method]]) are based upon finding such factorizations.

====Biquadratic equation====
If {{math|''a''<sub>3</sub> {{=}} ''a''<sub>1</sub> {{=}} 0}} then the function 
:<math>
Q(x) = a_4x^4+a_2x^2+a_0
</math> 
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'' {{=}} ''x''<sup>2</sup>}}.
Then {{math|''Q''(''x'')}} becomes a [[Quadratic function|quadratic]] {{math|''q''}} in {{math|''z''}}: {{math|''q''(''z'') {{=}} ''a''<sub>4</sub>''z''<sup>2</sup> + ''a''<sub>2</sub>''z'' + ''a''<sub>0</sub>}}. Let {{math|''z''<sub>+</sub>}} and {{math|''z''<sub>−</sub>}} be the roots of {{math|''q''(''z'')}}. Then the roots of the quartic {{math|''Q''(''x'')}} are
:<math>
\begin{align}
x_1&=+\sqrt{z_+},
\\
x_2&=-\sqrt{z_+},
\\
x_3&=+\sqrt{z_-},
\\
x_4&=-\sqrt{z_-}.
\end{align}
</math>

==== Quasi-palindromic equation ====
The polynomial
: <math>P(x)=a_0x^4+a_1x^3+a_2x^2+a_1 m x+a_0 m^2</math>
is almost [[Reciprocal polynomial#Palindromic polynomial|palindromic]], as {{math|''P''(''mx'') {{=}} {{sfrac|''x''<sup>4</sup>|''m''<sup>2</sup>}}''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'')|''x''<sup>2</sup>}} {{=}} 0}} produces the [[quadratic equation]]  {{math|''a''<sub>0</sub>''z''<sup>2</sup> + ''a''<sub>1</sub>''z'' + ''a''<sub>2</sub> − 2''ma''<sub>0</sub> {{=}} 0}}. Since {{math|''x''<sup>2</sup> − ''xz'' + ''m'' {{=}} 0}}, the quartic equation {{math|''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
:<math> a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0 = 0 </math>
be the general quartic equation we want to solve.

Dividing by {{math|''a''<sub>4</sub>}}, provides the equivalent equation {{math|''x''<sup>4</sup> + ''bx''<sup>3</sup> + ''cx''<sup>2</sup> + ''dx'' + ''e'' {{=}} 0}}, with {{math|''b'' {{=}} {{sfrac|''a''<sub>3</sub>|''a''<sub>4</sub>}}}}, {{math|''c'' {{=}} {{sfrac|''a''<sub>2</sub>|''a''<sub>4</sub>}}}}, {{math|''d'' {{=}} {{sfrac|''a''<sub>1</sub>|''a''<sub>4</sub>}}}}, and {{math|''e'' {{=}} {{sfrac|''a''<sub>0</sub>|''a''<sub>4</sub>}}}}.
Substituting {{math|''y'' − {{sfrac|''b''|4}}}} for {{mvar|x}} gives, after regrouping the terms, the equation {{math|''y''<sup>4</sup> + ''py''<sup>2</sup> + ''qy'' + ''r'' {{=}} 0}},
where
:<math>\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}
</math>

If {{math|''y''<sub>0</sub>}} is a root of this depressed quartic, then {{math|''y''<sub>0</sub> − {{sfrac|''b''|4}}}} (that is {{math|''y''<sub>0</sub> − {{sfrac|''a''<sub>3</sub>|4''a''<sub>4</sub>}})}} is a root of the original quartic and every root of the original quartic can be obtained by this process.

====Ferrari's solution====
As explained in the preceding section, we may start with the ''depressed quartic equation''
:<math> y^4 + p y^2 + q y + r = 0. </math>
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
:<math> \left(y^2 + \frac p2\right)^2 = -q y - r + \frac{p^2}4. </math>
Then, we introduce a variable {{mvar|m}} into the factor on the left-hand side by adding {{math|2''y''<sup>2</sup>''m'' + ''pm'' + ''m''<sup>2</sup>}} to both sides. After regrouping the coefficients of the power of {{mvar|y}} on the right-hand side, this gives the equation 
{{NumBlk|:|<math> \left(y^2 + \frac p2 + m\right)^2 = 2 m y^2 - q y + m^2 + m p + \frac{p^2}4 - r, </math>|{{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
:<math> (-q)^2 - 4 (2m)\left(m^2 + p m + \frac{p^2}4 - r\right) = 0,\,</math>
which may be rewritten as

{{NumBlk|:|<math>8m^3+ 8pm^2 + (2p^2 -8r)m- q^2 =0.</math>|{{EquationRef|1a}}}}

This is the [[resolvent cubic]] of the quartic equation. The value of {{mvar|m}} may thus be obtained from [[Cubic equation#Cardano's method|Cardano's formula]]. When {{mvar|m}} is a root of this equation, the right-hand side of equation (''{{EquationNote|1}}'') is the square
:<math>\left(\sqrt{2m}y-\frac q{2\sqrt{2m}}\right)^2.</math>
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|''y''<sup>4</sup> {{=}} 0}}.

Now, if {{mvar|m}} is a root of the cubic equation such that {{math|''m'' ≠ 0}}, equation (''{{EquationNote|1}}'') becomes
:<math> \left(y^2 + \frac p2 + m\right)^2 = \left(y\sqrt{2 m}-\frac{q}{2\sqrt{2 m}}\right)^2. </math>

This equation is of the form {{math|''M''<sup>2</sup> {{=}} ''N''<sup>2</sup>}}, which can be rearranged as {{math|''M''<sup>2</sup> − ''N''<sup>2</sup> {{=}} 0}} or {{math|(''M'' + ''N'')(''M'' − ''N'') {{=}} 0}}. Therefore, equation (''{{EquationNote|1}}'')  may be rewritten as
:<math> \left(y^2 + \frac p2 + m + \sqrt{2 m}y-\frac q{2\sqrt{2 m}}\right) \left(y^2 + \frac p2 + m - \sqrt{2 m}y+\frac q{2\sqrt{2 m}}\right)=0.</math>

This equation is easily solved by applying to each factor the [[quadratic formula]]. Solving them we may write the four roots as
:<math>y={\pm_1\sqrt{2 m} \pm_2 \sqrt{-\left(2p + 2m \pm_1 {\sqrt 2q \over \sqrt{m}} \right)} \over 2},</math>
where {{math|±<sub>1</sub>}} and {{math|±<sub>2</sub>}} denote either {{math|+}} or {{math|−}}. As the two occurrences of {{math|±<sub>1</sub>}} must denote the same sign, this leaves four possibilities, one for each root.

Therefore, the solutions of the original quartic equation are
:<math>x=-{a_3 \over 4a_4} + {\pm_1\sqrt{2 m} \pm_2 \sqrt{-\left(2p + 2m \pm_1 {\sqrt2q \over \sqrt{m}} \right)} \over 2}.</math> 
A comparison with the [[#General_formula_for_roots|general formula]] above shows that {{math|{{sqrt|2''m''}} {{=}} 2''S''}}.

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

:<math>
  \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}
 </math>

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

:<math>
  \left\{\begin{array}{l}
   b = s + u \\
   c = t + v + su \\
   d = sv + tu \\
   e = tv
  \end{array}\right.
 </math>

This can be simplified by starting again with the [[#Converting to a depressed quartic|depressed quartic]] {{math|''y''<sup>4</sup> + ''py''<sup>2</sup> + ''qy'' + ''r''}}, which can be obtained by substituting {{math|''y'' − ''b''/4}} for {{math|''x''}}. Since the coefficient of {{math|''y''<sup>3</sup>}} is&nbsp;{{math|0}}, we get {{math|''s'' {{=}} −''u''}}, and:

:<math>
  \left\{\begin{array}{l}
   p + u^2 = t + v \\
   q = u (t - v) \\
   r = tv
  \end{array}\right.
 </math>

One can now eliminate both {{mvar|t}} and {{mvar|v}} by doing the following:
:<math>
  \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}
 </math>

If we set {{math|''U'' {{=}} ''u''<sup>2</sup>}}, then solving this equation becomes finding the roots of the [[resolvent cubic]]

{{NumBlk|:|<math> U^3 + 2pU^2 + (p^2-4r)U - q^2,</math>|{{EquationRef|2}}}}

which is [[Cubic function#General solution to the cubic equation with real coefficients|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|''x''<sup>4</sup>}}, which is trivially factored),

:<math>
  \left\{\begin{array}{l}
  s = -u \\
 2t = p + u^2 + q/u \\
 2v = p + u^2 - q/u
  \end{array}\right.
 </math>

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|''p''<sup>2</sup> − 4''r''}} is the square of rational and {{math|''q'' {{=}} 0}}; this can readily be checked using the [[rational root test]].<ref name=Brookfield>{{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}}</ref>

====Euler's solution====
A variant of the previous method is due to [[Leonhard Euler|Euler]].<ref>{{Citation|last = van der Waerden|first=Bartel Leendert|author-link = Bartel Leendert van der Waerden|title = [[Moderne Algebra|Algebra]]|volume = 1|publisher=[[Springer Science+Business Media|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}}</ref><ref>{{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 Science+Business Media|Springer-Verlag]]|orig-year = 1765|year = 1984|zbl = 0557.01014|isbn = 978-1-4613-8511-0}}</ref> 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|''x''<sup>4</sup> + ''px''<sup>2</sup> + ''qx'' + ''r''}}. Observe that, if
* {{math|''x''<sup>4</sup> + ''px''<sup>2</sup> + ''qx'' + ''r'' {{=}} (''x''<sup>2</sup> + ''sx'' + ''t'')(''x''<sup>2</sup> − ''sx'' + ''v'')}},
* {{math|''r''<sub>1</sub>}} and {{math|''r''<sub>2</sub>}} are the roots of {{math|''x''<sup>2</sup> + ''sx'' + ''t''}},
* {{math|''r''<sub>3</sub>}} and {{math|''r''<sub>4</sub>}} are the roots of {{math|''x''<sup>2</sup> − ''sx'' + ''v''}},
then
* the roots of {{math|''x''<sup>4</sup> + ''px''<sup>2</sup> + ''qx'' + ''r''}} are {{math|''r''<sub>1</sub>}}, {{math|''r''<sub>2</sub>}}, {{math|''r''<sub>3</sub>}}, and {{math|''r''<sub>4</sub>}},
* {{math|''r''<sub>1</sub> + ''r''<sub>2</sub> {{=}} −''s''}},
* {{math|''r''<sub>3</sub> + ''r''<sub>4</sub> {{=}} ''s''}}.
Therefore, {{math|(''r''<sub>1</sub> + ''r''<sub>2</sub>)(''r''<sub>3</sub> + ''r''<sub>4</sub>) {{=}} −''s''<sup>2</sup>}}. In other words, {{math|−(''r''<sub>1</sub> + ''r''<sub>2</sub>)(''r''<sub>3</sub> + ''r''<sub>4</sub>)}} is one of the roots of the resolvent cubic (''{{EquationNote|2}}'') and this suggests that the roots of that cubic are equal to {{math|−(''r''<sub>1</sub> + ''r''<sub>2</sub>)(''r''<sub>3</sub> + ''r''<sub>4</sub>)}}, {{math|−(''r''<sub>1</sub> + ''r''<sub>3</sub>)(''r''<sub>2</sub> + ''r''<sub>4</sub>)}}, and {{math|−(''r''<sub>1</sub> + ''r''<sub>4</sub>)(''r''<sub>2</sub> + ''r''<sub>3</sub>)}}. 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|''r''<sub>1</sub> + ''r''<sub>2</sub> + ''r''<sub>3</sub> + ''r''<sub>4</sub> {{=}} 0}}. (Of course, this also follows from the fact that {{math|''r''<sub>1</sub> + ''r''<sub>2</sub> + ''r''<sub>3</sub> + ''r''<sub>4</sub> {{=}} −''s'' + ''s''}}.) Therefore, if {{math|''α''}}, {{math|''β''}}, and {{math|''γ''}} are the roots of the resolvent cubic, then the numbers {{math|''r''<sub>1</sub>}}, {{math|''r''<sub>2</sub>}}, {{math|''r''<sub>3</sub>}}, and {{math|''r''<sub>4</sub>}} are such that
:<math>\left\{\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}\right.</math>
It is a consequence of the first two equations that {{math|''r''<sub>1</sub> + ''r''<sub>2</sub>}} is a square root of {{math|''α''}} and that {{math|''r''<sub>3</sub> + ''r''<sub>4</sub>}} is the other square root of {{math|''α''}}. For the same reason,
* {{math|''r''<sub>1</sub> + ''r''<sub>3</sub>}} is a square root of {{math|''β''}},
* {{math|''r''<sub>2</sub> + ''r''<sub>4</sub>}} is the other square root of {{math|''β''}},
* {{math|''r''<sub>1</sub> + ''r''<sub>4</sub>}} is a square root of {{math|''γ''}},
* {{math|''r''<sub>2</sub> + ''r''<sub>3</sub>}} is the other square root of {{math|''γ''}}.
Therefore, the numbers {{math|''r''<sub>1</sub>}}, {{math|''r''<sub>2</sub>}}, {{math|''r''<sub>3</sub>}}, and {{math|''r''<sub>4</sub>}} are such that
:<math>\left\{\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}\right.</math>
the sign of the square roots will be dealt with below. The only solution of this system is:
:<math>\left\{\begin{array}{l}r_1=\frac{\sqrt{\alpha}+\sqrt{\beta}+\sqrt{\gamma}}2\\[2mm]r_2=\frac{\sqrt{\alpha}-\sqrt{\beta}-\sqrt{\gamma}}2\\[2mm]r_3=\frac{-\sqrt{\alpha}+\sqrt{\beta}-\sqrt{\gamma}}2\\[2mm]r_4=\frac{-\sqrt{\alpha}-\sqrt{\beta}+\sqrt{\gamma}}2\text{.}\end{array}\right.</math>
Since, in general, there are two choices for each square root, it might look as if this provides {{math|8 ({{=}} 2<sup>3</sup>)}} choices for the set {{math|{''r''<sub>1</sub>, ''r''<sub>2</sub>, ''r''<sub>3</sub>, ''r''<sub>4</sub>}}}, but, in fact, it provides no more than {{math|2}}&nbsp;such choices, because the consequence of replacing one of the square roots by the symmetric one is that the set {{math|{''r''<sub>1</sub>, ''r''<sub>2</sub>, ''r''<sub>3</sub>, ''r''<sub>4</sub>}}} becomes the set {{math|{−''r''<sub>1</sub>, −''r''<sub>2</sub>, −''r''<sub>3</sub>, −''r''<sub>4</sub>}}}.

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|''r''<sub>1</sub>}}, {{math|''r''<sub>2</sub>}}, {{math|''r''<sub>3</sub>}}, and {{math|''r''<sub>4</sub>}} 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|''q''<sup>2</sup>}} and therefore that {{math|{{sqrt|''α''}}{{sqrt|''β''}}{{sqrt|''γ''}} {{=}} ±''q''}}. But a straightforward computation shows that
:{{math|{{sqrt|''α''}}{{sqrt|''β''}}{{sqrt|''γ''}} {{=}} ''r''<sub>1</sub>''r''<sub>2</sub>''r''<sub>3</sub> + ''r''<sub>1</sub>''r''<sub>2</sub>''r''<sub>4</sub> + ''r''<sub>1</sub>''r''<sub>3</sub>''r''<sub>4</sub> + ''r''<sub>2</sub>''r''<sub>3</sub>''r''<sub>4</sub>.}}
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|−''r''<sub>1</sub>}}, {{math|−''r''<sub>2</sub>}}, {{math|−''r''<sub>3</sub>}}, and {{math|−''r''<sub>4</sub>}}, 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 <math>-\frac q{\sqrt{\alpha}\sqrt{\beta}}</math>.
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 [[Quartic function#Biquadratic equation|biquadratic equation]], in which case there is a much simpler approach.

====Solving by Lagrange resolvent====
The [[symmetric group]] {{math|''S''<sub>4</sub>}} 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<sub>i</sub>''}}, for {{math|''i''}} from&nbsp;{{math|0}} to&nbsp;{{math|3}}, the four roots of {{math|''x''<sup>4</sup> + ''bx''<sup>3</sup> + ''cx''<sup>2</sup> + ''dx'' + ''e''}}. If we set

: <math> \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}</math>

then since the transformation is an [[Involution (mathematics)|involution]] we may express the roots in terms of the four {{math|''s<sub>i</sub>''}} in exactly the same way. Since we know the value {{math|''s''<sub>0</sub> {{=}} −{{sfrac|''b''|2}}}}, we only need the values for {{math|''s''<sub>1</sub>}}, {{math|''s''<sub>2</sub>}} and {{math|''s''<sub>3</sub>}}. These are the roots of the polynomial

:<math>(s^2 - {s_1}^2)(s^2-{s_2}^2)(s^2-{s_3}^2).</math>

Substituting the {{math|''s<sub>i</sub>''}} by their values in term of the {{math|''x<sub>i</sub>''}}, this polynomial may be expanded in a polynomial in {{math|''s''}} whose coefficients are [[symmetric polynomial]]s in the {{math|''x<sub>i</sub>''}}. 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|:|<math> s^6+2cs^4+(c^2-4e)s^2-d^2 </math>|{{EquationRef|3}}}}
This polynomial is of degree six, but only of degree three in {{math|''s''<sup>2</sup>}}, 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<sub>i</sub>''}} in terms of the {{math|''s<sub>i</sub>''}}, 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<sub>i</sub>''}}.

These expressions are unnecessarily complicated, involving the [[root of unity|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

:<math>\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}</math>

then

:<math>F_1(x)\times F_2(x) = x^4 + cx^2 + dx + e.</math>

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 [[Quartic function#Descartes' solution|Descartes' method]].

====Solving with algebraic geometry====
There is an alternative solution using algebraic geometry<ref>{{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}}</ref> In brief, one interprets the roots as the intersection of two quadratic curves, then finds the three [[degenerate conic|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|''x''<sup>4</sup> + ''px''<sup>2</sup> + ''qx'' + ''r'' {{=}} 0}} may also be expressed as the {{mvar|x}} coordinates of the intersections of the two quadratic equations {{math|''y''<sup>2</sup> + ''py'' + ''qx'' + ''r'' {{=}} 0}} and {{math|''y'' − ''x''<sup>2</sup> {{=}} 0}} i.e., using the substitution {{math|''y'' {{=}} ''x''<sup>2</sup>}} that two quadratics intersect in four points is an instance of [[Bézout's theorem]]. Explicitly, the four points are {{math|''P<sub>i</sub>'' ≔ (''x<sub>i</sub>'', ''x<sub>i</sub>''<sup>2</sup>)}} for the four roots {{math|''x<sub>i</sub>''}} of the quartic.

These four points are not collinear because they lie on the irreducible quadratic {{math|''y'' {{=}} ''x''<sup>2</sup>}} 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 form]]s in three variables:
:<math>\begin{align}
F_1(X,Y,Z) &:= Y^2 + pYZ + qXZ + rZ^2,\\
F_2(X,Y,Z) &:= YZ - X^2
\end{align}</math>
the pencil is given by the forms {{math|''λF''<sub>1</sub> + ''μF''<sub>2</sub>}} 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 <math>\textstyle{\binom{4}{2}}</math>&nbsp;=&nbsp;{{math|6}} different ways. Denote these {{math|''Q''<sub>1</sub> {{=}} ''L''<sub>12</sub> + ''L''<sub>34</sub>}}, {{math|''Q''<sub>2</sub> {{=}} ''L''<sub>13</sub> + ''L''<sub>24</sub>}}, and  {{math|''Q''<sub>3</sub> {{=}} ''L''<sub>14</sub> + ''L''<sub>23</sub>}}. 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|''λF''<sub>1</sub> + ''μF''<sub>2</sub>}} as a {{math|3×3}}&nbsp;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.