Special cases of the formula

• If ${\displaystyle \Delta >0,}$ 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 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:

${\displaystyle S={\frac {1}{2}}{\sqrt {-{\frac {2}{3}}\ p+{\frac {2}{3a}}{\sqrt {\Delta _{0}}}\cos {\frac {\varphi }{3}}}}}$

where

${\displaystyle \varphi =\arccos \left({\frac {\Delta _{1}}{2{\sqrt {\Delta _{0}^{3}}}}}\right).}$

• If ${\displaystyle \Delta \neq 0}$ and ${\displaystyle \Delta _{0}=0,}$ the sign of ${\displaystyle {\sqrt {\Delta _{1}^{2}-4\Delta _{0}^{3}}}={\sqrt {\Delta _{1}^{2}}}}$ has to be chosen to have ${\displaystyle Q\neq 0,}$ that is one should define ${\displaystyle {\sqrt {\Delta _{1}^{2}}}}$ as ${\displaystyle \Delta _{1},}$ maintaining the sign of ${\displaystyle \Delta _{1}.}$
• If ${\displaystyle S=0,}$ then one must change the choice of the cube root in ${\displaystyle Q}$ in order to have ${\displaystyle S\neq 0.}$ This is always possible except if the quartic may be factored into ${\displaystyle \left(x+{\tfrac {b}{4a}}\right)^{4}.}$ The result is then correct, but misleading because it hides the fact that no cube root is needed in this case. In fact this case may occur only if the numerator of ${\displaystyle q}$ is zero, in which case the associated depressed quartic is biquadratic; it may thus be solved by the method described below.
• If ${\displaystyle \Delta =0}$ and ${\displaystyle \Delta _{0}=0,}$ and thus also ${\displaystyle \Delta _{1}=0,}$ at least three roots are equal to each other, and the roots are rational functions of the coefficients. The triple root ${\displaystyle x_{0}}$ is a common root of the quartic and its second derivative ${\displaystyle 2(6ax^{2}+3bx+c);}$ 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 ${\displaystyle x_{1}}$ can be deduced from ${\displaystyle x_{1}+3x_{0}=-b/a.}$
• If ${\displaystyle \Delta =0}$ and ${\displaystyle \Delta _{0}\neq 0,}$ the above expression for the roots is correct but misleading, hiding the fact that the polynomial is reducible and no cube root is needed to represent the roots.

Reducible quartics

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 is reducible if Q(x) = R(x)×S(x), where R(x) and S(x) are non-constant polynomials with rational coefficients (or more generally with coefficients in the same field as the coefficients of Q(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 of 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 using the resolvent cubic of Q(x). It turns out that:

• if we are working over 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 Q (that is, if coefficients are restricted to be rational numbers) then there is an algorithm to determine whether or not 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.

Biquadratic equation

If a₃ = a₁ = 0 then the function

${\displaystyle Q(x)=a_{4}x^{4}+a_{2}x^{2}+a_{0}}$

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

Let the auxiliary variable z = x². Then Q(x) becomes a quadratic q in z: q(z) = a₄z² + a₂z + a₀. Let z₊ and z₋ be the roots of q(z). Then the roots of the quartic Q(x) are

${\displaystyle {\begin{aligned}x_{1}&=+{\sqrt {z_{+}}},\\x_{2}&=-{\sqrt {z_{+}}},\\x_{3}&=+{\sqrt {z_{-}}},\\x_{4}&=-{\sqrt {z_{-}}}.\end{aligned}}}$

Quasi-palindromic equation

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 almost palindromic, as P(mx) = x⁴/m²P(m/x) (it is palindromic if m = 1). The change of variables z = x + m/x in P(x)/x² = 0 produces the quadratic equation a₀z² + a₁z + a₂ − 2ma₀ = 0. Since x² − xz + m = 0, the quartic equation P(x) = 0 may be solved by applying the quadratic formula twice.

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

${\displaystyle a_{4}x^{4}+a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}=0}$

be the general quartic equation we want to solve.

Dividing by a₄, provides the equivalent equation x⁴ + bx³ + cx² + dx + e = 0, with b = a₃/a₄, c = a₂/a₄, d = a₁/a₄, and e = a₀/a₄. Substituting y − b/4 for x gives, after regrouping the terms, the equation y⁴ + py² + qy + r = 0, where

${\displaystyle {\begin{aligned}p&={\frac {8c-3b^{2}}{8}}={\frac {8a_{2}a_{4}-3{a_{3}}^{2}}{8{a_{4}}^{2}}}\\q&={\frac {b^{3}-4bc+8d}{8}}={\frac {{a_{3}}^{3}-4a_{2}a_{3}a_{4}+8a_{1}{a_{4}}^{2}}{8{a_{4}}^{3}}}\\r&={\frac {-3b^{4}+256e-64bd+16b^{2}c}{256}}={\frac {-3{a_{3}}^{4}+256a_{0}{a_{4}}^{3}-64a_{1}a_{3}{a_{4}}^{2}+16a_{2}{a_{3}}^{2}a_{4}}{256{a_{4}}^{4}}}.\end{aligned}}}$

If y₀ is a root of this depressed quartic, then y₀ − b/4 (that is y₀ − a₃/4a₄) 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

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

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

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

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

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

(1)

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

As the value of 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 y of this quadratic equation is zero, that is m 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

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

(1a)

This is the resolvent cubic of the quartic equation. The value of m may thus be obtained from Cardano's formula. When m 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 if m = 0. This implies 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 m ≠ 0. This is always possible except for the depressed equation y⁴ = 0.

Now, if m is a root of the cubic equation such that m ≠ 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 form M² = N², which can be rearranged as M² − 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 the quadratic formula. Solving them we may write the four roots as

${\displaystyle y={\pm _{1}{\sqrt {2m}}\pm _{2}{\sqrt {-\left(2p+2m\pm _{1}{{\sqrt {2}}q \over {\sqrt {m}}}\right)}} \over 2},}$

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

${\displaystyle x=-{a_{3} \over 4a_{4}}+{\pm _{1}{\sqrt {2m}}\pm _{2}{\sqrt {-\left(2p+2m\pm _{1}{{\sqrt {2}}q \over {\sqrt {m}}}\right)}} \over 2}.}$

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

Descartes' solution

Descartes^[18] introduced in 1637 the method of finding the roots of a quartic polynomial by factoring it into two quadratic ones. Let

${\displaystyle {\begin{aligned}x^{4}+bx^{3}+cx^{2}+dx+e&=(x^{2}+sx+t)(x^{2}+ux+v)\\&=x^{4}+(s+u)x^{3}+(t+v+su)x^{2}+(sv+tu)x+tv\end{aligned}}}$

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

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

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

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

One can now eliminate both t and v by doing the following:

${\displaystyle {\begin{aligned}u^{2}(p+u^{2})^{2}-q^{2}&=u^{2}(t+v)^{2}-u^{2}(t-v)^{2}\\&=u^{2}[(t+v+(t-v))(t+v-(t-v))]\\&=u^{2}(2t)(2v)\\&=4u^{2}tv\\&=4u^{2}r\end{aligned}}}$

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

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

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

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

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 u for the square root of 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 (2) has a non-zero root which is the square of a rational, or p² − 4r is the square of rational and q = 0; this can readily be checked using the rational root test.^[19]

Euler's solution

A variant of the previous method is due to Euler.^[20][21] 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 x⁴ + px² + qx + r. Observe that, if

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

then

• the roots of x⁴ + px² + qx + r are r₁, r₂, r₃, and r₄,
• 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 from Vieta's formulas. It also follows from Vieta's formulas, together with the fact that we are working with a depressed quartic, that r₁ + r₂ + r₃ + r₄ = 0. (Of course, this also follows from the fact that r₁ + r₂ + r₃ + r₄ = −s + s.) Therefore, if α, β, and γ are the roots of the resolvent cubic, then the numbers r₁, r₂, r₃, and r₄ are such that

${\displaystyle \left\{{\begin{array}{l}r_{1}+r_{2}+r_{3}+r_{4}=0\\(r_{1}+r_{2})(r_{3}+r_{4})=-\alpha \\(r_{1}+r_{3})(r_{2}+r_{4})=-\beta \\(r_{1}+r_{4})(r_{2}+r_{3})=-\gamma {\text{.}}\end{array}}\right.}$

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

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

Therefore, the numbers r₁, r₂, r₃, and r₄ are such that

${\displaystyle \left\{{\begin{array}{l}r_{1}+r_{2}+r_{3}+r_{4}=0\\r_{1}+r_{2}={\sqrt {\alpha }}\\r_{1}+r_{3}={\sqrt {\beta }}\\r_{1}+r_{4}={\sqrt {\gamma }}{\text{;}}\end{array}}\right.}$

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

${\displaystyle \left\{{\begin{array}{l}r_{1}={\frac {{\sqrt {\alpha }}+{\sqrt {\beta }}+{\sqrt {\gamma }}}{2}}\\[2mm]r_{2}={\frac {{\sqrt {\alpha }}-{\sqrt {\beta }}-{\sqrt {\gamma }}}{2}}\\[2mm]r_{3}={\frac {-{\sqrt {\alpha }}+{\sqrt {\beta }}-{\sqrt {\gamma }}}{2}}\\[2mm]r_{4}={\frac {-{\sqrt {\alpha }}-{\sqrt {\beta }}+{\sqrt {\gamma }}}{2}}{\text{.}}\end{array}}\right.}$

Since, in general, there are two choices for each square root, it might look as if this provides 8 (= 2³) choices for the set {r₁, r₂, r₃, r₄}, but, in fact, it provides no more than 2 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 numbers α, β, and γ and uses them to compute the numbers r₁, r₂, r₃, and r₄ from the previous equalities. Then, one computes the number √α√β√γ. Since α, β, and γ are the roots of (2), it is a consequence of Vieta's formulas that their product is equal to q² 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:

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

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

Solving by Lagrange resolvent

The symmetric group S₄ on four elements has the Klein four-group as a normal subgroup. This suggests using a resolvent cubic whose roots may be variously described as a discrete Fourier transform or a Hadamard matrix transform of the roots; see Lagrange resolvents for the general method. Denote by x_i, for i from 0 to 3, the four roots of x⁴ + bx³ + cx² + dx + e. If we set

${\displaystyle {\begin{aligned}s_{0}&={\tfrac {1}{2}}(x_{0}+x_{1}+x_{2}+x_{3}),\\[4pt]s_{1}&={\tfrac {1}{2}}(x_{0}-x_{1}+x_{2}-x_{3}),\\[4pt]s_{2}&={\tfrac {1}{2}}(x_{0}+x_{1}-x_{2}-x_{3}),\\[4pt]s_{3}&={\tfrac {1}{2}}(x_{0}-x_{1}-x_{2}+x_{3}),\end{aligned}}}$

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

${\displaystyle (s^{2}-{s_{1}}^{2})(s^{2}-{s_{2}}^{2})(s^{2}-{s_{3}}^{2}).}$

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

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

(3)

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

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

${\displaystyle {\begin{aligned}F_{1}(x)&=x^{2}+sx+{\frac {c}{2}}+{\frac {s^{2}}{2}}-{\frac {d}{2s}}\\F_{2}(x)&=x^{2}-sx+{\frac {c}{2}}+{\frac {s^{2}}{2}}+{\frac {d}{2s}}\end{aligned}}}$

then

${\displaystyle F_{1}(x)\times F_{2}(x)=x^{4}+cx^{2}+dx+e.}$

We therefore can solve the quartic by solving for 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.

Solving with algebraic geometry

There is an alternative solution using algebraic geometry^[22] 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 x⁴ + px² + qx + r = 0 may also be expressed as the x coordinates of the intersections of the two quadratic equations y² + py + qx + r = 0 and y − x² = 0 i.e., using the substitution y = x² that two quadratics intersect in four points is an instance of Bézout's theorem. Explicitly, the four points are P_i ≔ (x_i, x_i²) for the four roots x_i of the quartic.

These four points are not collinear because they lie on the irreducible quadratic 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:

${\displaystyle {\begin{aligned}F_{1}(X,Y,Z)&:=Y^{2}+pYZ+qXZ+rZ^{2},\\F_{2}(X,Y,Z)&:=YZ-X^{2}\end{aligned}}}$

the pencil is given by the forms λF₁ + μF₂ for any point [λ, μ] in the projective line — in other words, where λ and 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 ${\displaystyle \textstyle {\binom {4}{2}}}$ = 6 different ways. Denote these Q₁ = L₁₂ + L₃₄, Q₂ = L₁₃ + L₂₄, and 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 λF₁ + μF₂ as a 3×3 matrix: reducible quadratics correspond to this matrix being singular, which is equivalent to its determinant being zero, and the determinant is a homogeneous degree three polynomial in λ and μ and corresponds to the resolvent cubic.