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Solving a quartic equation

Special cases

Consider the quartic

$Q(x)=a_{4}x^{4}+a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}.\,$

Degenerate case

If a₀ = 0 then Q(0) = 0, and so x = 0 is a solution. It follows that Q(x) may be factorised as Q(x) = x·(a₄x³ + a₃x² + a₂x + a₁). The remaining three roots – see Fundamental Theorem of Algebra – can be found by solving the cubic equation a₄x³ + a₃x² + a₂x + a₁ = 0.

Evident roots: 1 and −1 and −k

If $a_{0}+a_{1}+a_{2}+a_{3}+a_{4}=0,\,$ then $Q(1)=0\,$, so $1\,$ is a root. Similarly, if $a_{4}-a_{3}+a_{2}-a_{1}+a_{0}=0,\,$ that is, $a_{0}+a_{2}+a_{4}=a_{1}+a_{3},\,$ then $-1\,$ is a root.

When $1\,$ is a root, we can divide $Q(x)\,$ by $x-1\,$ and get

$Q(x)=(x-1)p(x),\,$

where $p(x)\,$ is a cubic polynomial, which may be solved to find $Q\,$ 's other roots. Similarly, if $-1\,$ is a root,

$Q(x)=(x+1)p(x),\,$

where $p(x)\,$ is some cubic polynomial.

If $a_{2}=0,a_{3}=ka_{4},a_{0}=ka_{1},\,$ then −k is a root and we can factor out $x+k\,$,

${\begin{aligned}Q(x)&=a_{4}x^{4}+ka_{4}x^{3}+a_{1}x+ka_{1}\\&=(x+k)a_{4}x^{3}+(x+k)a_{1}\\&=(x+k)(a_{4}x^{3}+a_{1}).\end{aligned}}$

And if $a_{0}=0,a_{3}=ka_{4},a_{1}=ka_{2},\,$ then both $0\,$ and $-k\,$ are roots Now we can factor out $x(x+k)\,$ and get

${\begin{aligned}Q(x)&=x(a_{4}x^{3}+ka_{4}x^{2}+a_{2}x+ka_{2})\\&=x(x+k)(a_{4}x^{2}+a_{2}).\end{aligned}}$

To get Q 's other roots, we simply solve the quadratic factor.

Biquadratic equations

If $a_{3}=a_{1}=0,\,$ then

$Q(x)=a_{4}x^{4}+a_{2}x^{2}+a_{0}.\,\!$

We call such a polynomial a biquadratic, which is easy to solve.

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

$q(z)=a_{4}z^{2}+a_{2}z+a_{0}.\,\!$

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

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

Quasi-symmetric equations

$a_{0}x^{4}+a_{1}x^{3}+a_{2}x^{2}+a_{1}mx+a_{0}m^{2}=0\,$

Steps:

1) Divide by x ².

2) Use variable change z = x + m/x.

The general case, along Ferrari's lines

To begin, the quartic must first be converted to a depressed quartic.

Converting to a depressed quartic

Let

$Ax^{4}+Bx^{3}+Cx^{2}+Dx+E=0\qquad \qquad (1')$

be the general quartic equation we want to solve. Divide both sides by A to produce a monic polynomial,

$x^{4}+{B \over A}x^{3}+{C \over A}x^{2}+{D \over A}x+{E \over A}=0.$

The first step should be to eliminate the x³ term. To do this, change variables from x to u, such that

$x=u-{B \over 4A}$.

Then

$\left(u-{B \over 4A}\right)^{4}+{B \over A}\left(u-{B \over 4A}\right)^{3}+{C \over A}\left(u-{B \over 4A}\right)^{2}+{D \over A}\left(u-{B \over 4A}\right)+{E \over A}=0.$

Expanding the powers of the binomials produces

$\left(u^{4}-{B \over A}u^{3}+{6u^{2}B^{2} \over 16A^{2}}-{4uB^{3} \over 64A^{3}}+{B^{4} \over 256A^{4}}\right)+{B \over A}\left(u^{3}-{3u^{2}B \over 4A}+{3uB^{2} \over 16A^{2}}-{B^{3} \over 64A^{3}}\right)+{C \over A}\left(u^{2}-{uB \over 2A}+{B^{2} \over 16A^{2}}\right)+{D \over A}\left(u-{B \over 4A}\right)+{E \over A}=0.$

Collecting the same powers of u yields

$u^{4}+\left({-3B^{2} \over 8A^{2}}+{C \over A}\right)u^{2}+\left({B^{3} \over 8A^{3}}-{BC \over 2A^{2}}+{D \over A}\right)u+\left({-3B^{4} \over 256A^{4}}+{CB^{2} \over 16A^{3}}-{BD \over 4A^{2}}+{E \over A}\right)=0.$

Now rename the coefficients of u. Let

${\begin{aligned}\alpha &={-3B^{2} \over 8A^{2}}+{C \over A},\\\beta &={B^{3} \over 8A^{3}}-{BC \over 2A^{2}}+{D \over A},\\\gamma &={-3B^{4} \over 256A^{4}}+{CB^{2} \over 16A^{3}}-{BD \over 4A^{2}}+{E \over A}.\end{aligned}}$

The resulting equation is

$u^{4}+\alpha u^{2}+\beta u+\gamma =0\qquad \qquad (1)$

which is a depressed quartic equation.

If $\beta =0\$ then we have a biquadratic equation, which (as explained above) is easily solved; using reverse substitution we can find our values for $x$.

If $\gamma =0\$ then one of the roots is $u=0\$ and the other roots can be found by dividing by $u$, and solving the resulting depressed cubic equation,

$u^{3}+\alpha u+\beta =0\,.$

Using reverse substitution we can find our values for $x$.

Ferrari's solution

Otherwise, the depressed quartic can be solved by means of a method discovered by Lodovico Ferrari. Once the depressed quartic has been obtained, the next step is to add the valid identity

$(u^{2}+\alpha )^{2}-u^{4}-2\alpha u^{2}=\alpha ^{2}\,$

to equation (1), yielding

$(u^{2}+\alpha )^{2}+\beta u+\gamma =\alpha u^{2}+\alpha ^{2}.\qquad \qquad (2)$

The effect has been to fold up the u⁴ term into a perfect square: (u² + α)². The second term, αu² did not disappear, but its sign has changed and it has been moved to the right side.

The next step is to insert a variable y into the perfect square on the left side of equation (2), and a corresponding 2y into the coefficient of u² in the right side. To accomplish these insertions, the following valid formulas will be added to equation (2),

${\begin{matrix}(u^{2}+\alpha +y)^{2}-(u^{2}+\alpha )^{2}&=&2y(u^{2}+\alpha )+y^{2}\\\\&=&2yu^{2}+2y\alpha +y^{2},\end{matrix}}$

and

$0=(\alpha +2y)u^{2}-2yu^{2}-\alpha u^{2}\,$

These two formulas, added together, produce

$(u^{2}+\alpha +y)^{2}-(u^{2}+\alpha )^{2}=(\alpha +2y)u^{2}-\alpha u^{2}+2y\alpha +y^{2}\qquad \qquad (y-{\hbox{insertion}})\,$

which added to equation (2) produces

$(u^{2}+\alpha +y)^{2}+\beta u+\gamma =(\alpha +2y)u^{2}+(2y\alpha +y^{2}+\alpha ^{2}).\,$

This is equivalent to

$(u^{2}+\alpha +y)^{2}=(\alpha +2y)u^{2}-\beta u+(y^{2}+2y\alpha +\alpha ^{2}-\gamma ).\qquad \qquad (3)\,$

The objective now is to choose a value for y such that the right side of equation (3) becomes a perfect square. This can be done by letting the discriminant of the quadratic function become zero. To explain this, first expand a perfect square so that it equals a quadratic function:

$(su+t)^{2}=(s^{2})u^{2}+(2st)u+(t^{2}).\,$

The quadratic function on the right side has three coefficients. It can be verified that squaring the second coefficient and then subtracting four times the product of the first and third coefficients yields zero:

$(2st)^{2}-4(s^{2})(t^{2})=0.\,$

Therefore to make the right side of equation (3) into a perfect square, the following equation must be solved:

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

Multiply the binomial with the polynomial,

$\beta ^{2}-4(2y^{3}+5\alpha y^{2}+(4\alpha ^{2}-2\gamma )y+(\alpha ^{3}-\alpha \gamma ))=0\,$

Divide both sides by −4, and move the −β²/4 to the right,

$2y^{3}+5\alpha y^{2}+(4\alpha ^{2}-2\gamma )y+\left(\alpha ^{3}-\alpha \gamma -{\beta ^{2} \over 4}\right)=0\qquad \qquad$

This is a cubic equation for y. Divide both sides by 2,

$y^{3}+{5 \over 2}\alpha y^{2}+(2\alpha ^{2}-\gamma )y+\left({\alpha ^{3} \over 2}-{\alpha \gamma \over 2}-{\beta ^{2} \over 8}\right)=0.\qquad \qquad (4)$

Conversion of the nested cubic into a depressed cubic

Equation (4) is a cubic equation nested within the quartic equation. It must be solved to solve the quartic. To solve the cubic, first transform it into a depressed cubic by means of the substitution

$y=v-{5 \over 6}\alpha .$

Equation (4) becomes

$\left(v-{5 \over 6}\alpha \right)^{3}+{5 \over 2}\alpha \left(v-{5 \over 6}\alpha \right)^{2}+(2\alpha ^{2}-\gamma )\left(v-{5 \over 6}\alpha \right)+\left({\alpha ^{3} \over 2}-{\alpha \gamma \over 2}-{\beta ^{2} \over 8}\right)=0.$

Expand the powers of the binomials,

$\left(v^{3}-{5 \over 2}\alpha v^{2}+{25 \over 12}\alpha ^{2}v-{125 \over 216}\alpha ^{3}\right)+{5 \over 2}\alpha \left(v^{2}-{5 \over 3}\alpha v+{25 \over 36}\alpha ^{2}\right)+(2\alpha ^{2}-\gamma )\left(v-{5 \over 6}\alpha \right)+\left({\alpha ^{3} \over 2}-{\alpha \gamma \over 2}-{\beta ^{2} \over 8}\right)=0.$

Distribute, collect like powers of v, and cancel out the pair of v² terms,

$v^{3}+\left(-{\alpha ^{2} \over 12}-\gamma \right)v+\left(-{\alpha ^{3} \over 108}+{\alpha \gamma \over 3}-{\beta ^{2} \over 8}\right)=0.$

This is a depressed cubic equation.

Relabel its coefficients,

$P=-{\alpha ^{2} \over 12}-\gamma ,$

$Q=-{\alpha ^{3} \over 108}+{\alpha \gamma \over 3}-{\beta ^{2} \over 8}.$

The depressed cubic now is

$v^{3}+Pv+Q=0.\qquad \qquad (5)$

Solving the nested depressed cubic

The solutions (any solution will do, so pick any of the three complex roots) of equation (5) are computed as (see Cubic equation)

$y=-{5 \over 6}\alpha +U+V\qquad \qquad (6)$

where

$U={\sqrt[{3}]{-{Q \over 2}\pm {\sqrt {{Q^{2} \over 4}+{P^{3} \over 27}}}}}$

and V is computed according to the two defining equations $-U{\mbox{ }}^{3}-V{\mbox{ }}^{3}=Q$ and $-3UV=P$, so

$V={\begin{cases}-{\frac {P}{3U}}&{\text{ if }}U\neq 0{\text{ and }}\\-{\sqrt[{3}]{Q}}&{\text{ if }}U=0\ .\end{cases}}$

Folding the second perfect square

With the value for y given by equation (6), it is now known that the right side of equation (3) is a perfect square of the form

$(s^{2})u^{2}+(2st)u+(t^{2})=(s^{2})\left(u-{-(2st) \over 2(s^{2})}\right)^{2}=\left[{\sqrt {(s^{2})}}\left(u-{-(2st) \over 2(s^{2})}\right)\right]^{2},$

$(s^{2})u^{2}+(2st)u+(t^{2})=\left(\left({\sqrt {(s^{2})}}\right)u+{(2st) \over 2{\sqrt {(s^{2})}}}\right)^{2}$

(This is correct for both signs of square root, as long as the same sign is taken for both square roots. A ± is redundant, as it would be absorbed by another ± a few equations further down this page.)

so that it can be folded:

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

Note: If β ≠ 0 then α + 2y ≠ 0. If β = 0 then this would be a biquadratic equation, which we solved earlier.

Therefore equation (3) becomes

$(u^{2}+\alpha +y)^{2}=\left(\left({\sqrt {\alpha +2y}}\right)u-{\beta \over 2{\sqrt {\alpha +2y}}}\right)^{2}\qquad \qquad (7)$.

Equation (7) has a pair of folded perfect squares, one on each side of the equation. The two perfect squares balance each other.

If two squares are equal, then the sides of the two squares are also equal, as shown by:

$(u^{2}+\alpha +y)=\pm \left(\left({\sqrt {\alpha +2y}}\right)u-{\beta \over 2{\sqrt {\alpha +2y}}}\right)\qquad \qquad (7')$.

Collecting like powers of u produces

$u^{2}+\left(\mp _{s}{\sqrt {\alpha +2y}}\right)u+\left(\alpha +y\pm _{s}{\beta \over 2{\sqrt {\alpha +2y}}}\right)=0\qquad \qquad (8)$.

Note: The subscript s of $\pm _{s}$ and $\mp _{s}$ is to note that they are dependent.

Equation (8) is a quadratic equation for u. Its solution is

$u={\pm _{s}{\sqrt {\alpha +2y}}\pm _{t}{\sqrt {(\alpha +2y)-4(\alpha +y\pm _{s}{\beta \over 2{\sqrt {\alpha +2y}}})}} \over 2}.$

Simplifying, one gets

$u={\pm _{s}{\sqrt {\alpha +2y}}\pm _{t}{\sqrt {-\left(3\alpha +2y\pm _{s}{2\beta \over {\sqrt {\alpha +2y}}}\right)}} \over 2}.$

This is the solution of the depressed quartic, therefore the solutions of the original quartic equation are

$x=-{B \over 4A}+{\pm _{s}{\sqrt {\alpha +2y}}\pm _{t}{\sqrt {-\left(3\alpha +2y\pm _{s}{2\beta \over {\sqrt {\alpha +2y}}}\right)}} \over 2}.\qquad \qquad (8')$

Remember: The two $\pm _{s}$ come from the same place in equation (7'), and should both have the same sign, while the sign of $\pm _{t}$ is independent.

Summary of Ferrari's method

Given the quartic equation

$Ax^{4}+Bx^{3}+Cx^{2}+Dx+E=0,\,$

its solution can be found by means of the following calculations:

$\alpha =-{3B^{2} \over 8A^{2}}+{C \over A},$

$\beta ={B^{3} \over 8A^{3}}-{BC \over 2A^{2}}+{D \over A},$

$\gamma =-{3B^{4} \over 256A^{4}}+{CB^{2} \over 16A^{3}}-{BD \over 4A^{2}}+{E \over A}.$

If $\,\beta =0,$ then

$x=-{B \over 4A}\pm _{s}{\sqrt {-\alpha \pm _{t}{\sqrt {\alpha ^{2}-4\gamma }} \over 2}}\qquad {\mbox{(for }}\beta =0{\mbox{ only)}}.$

Otherwise, continue with

$P=-{\alpha ^{2} \over 12}-\gamma ,$

$Q=-{\alpha ^{3} \over 108}+{\alpha \gamma \over 3}-{\beta ^{2} \over 8},$

$R=-{Q \over 2}\pm {\sqrt {{Q^{2} \over 4}+{P^{3} \over 27}}},$

(either sign of the square root will do)

$U={\sqrt[{3}]{R}},$

(there are 3 complex roots, any one of them will do)

$y={\begin{cases}-{5 \over 6}\alpha +U-{\frac {P}{3U}}&{\text{if }}U\neq 0\\-{5 \over 6}\alpha +U-{\sqrt[{3}]{Q}}&{\text{if }}U=0\end{cases}}$

$W={\sqrt {\alpha +2y}}$

$x=-{B \over 4A}+{\pm _{s}W\mp _{t}{\sqrt {-\left(3\alpha +2y\pm _{s}{2\beta \over W}\right)}} \over 2}.$

Ferrari was the first to discover one of these labyrinthine solutions. The equation he solved was:

$\ x^{4}+6x^{2}-60x+36=0$

which was already in depressed form. It has a pair of solutions that can be found with the set of formulas shown above.

Ferrari's solution in the special case of real coefficients

If the coefficients of the quartic equation are real then the nested depressed cubic equation (5) also has real coefficients, thus it has at least one real root.

Furthermore the cubic function $\ C(v)=v^{3}+Pv+Q,$ where P and Q are given by (5) has the properties that

$C\left({\alpha \over 3}\right)={-\beta ^{2} \over 8}<0\$ and

$\lim _{v\to \infty }{C(v)}=\infty ,$ where α and β are given by (1).

This means that (5) has a real root greater than $\alpha \over 3$, and therefore that (4) has a real root greater than $-\alpha \over 2$.

Using this root the term ${\sqrt {\alpha +2y}}$ in (8) is always real, which ensures that the two quadratic equations (8) have real coefficients.

Obtaining alternative solutions by factoring out complex conjugate solutions

It could happen that one only obtained one solution through the seven formulae above, because not all four sign patterns are tried for four solutions, and the solution obtained is complex. It may also be the case that one is only looking for a real solution. Let x₁ denote the complex solution. If all the original coefficients A, B, C, D and E are real — which should be the case when one desires only real solutions — then there is another complex solution x₂, which is the complex conjugate of x₁. If the other two roots are denoted as x₃ and x₄ then the quartic equation can be expressed as

$(x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4})=0,\,$

but this quartic equation is equivalent to the product of two quadratic equations:

$(x-x_{1})(x-x_{2})=0\qquad \qquad (9)$

and

$(x-x_{3})(x-x_{4})=0.\qquad \qquad (10)$

Since

$x_{2}=x_{1}^{\star }$

then

${\begin{matrix}(x-x_{1})(x-x_{2})&=&x^{2}-(x_{1}+x_{1}^{\star })x+x_{1}x_{1}^{\star }\qquad \qquad \qquad \quad \\&=&x^{2}-2\,\mathrm {Re} (x_{1})x+[\mathrm {Re} (x_{1})]^{2}+[\mathrm {Im} (x_{1})]^{2}.\end{matrix}}$

Let

$a=-2\,\mathrm {Re} (x_{1}),$

$b=\left[\mathrm {Re} (x_{1})\right]^{2}+\left[\mathrm {Im} (x_{1})\right]^{2}$

so that equation (9) becomes

$x^{2}+ax+b=0.\qquad \qquad (11)$

Also let there be (unknown) variables w and v such that equation (10) becomes

$x^{2}+wx+v=0.\qquad \qquad (12)$

Multiplying equations (11) and (12) produces

$x^{4}+(a+w)x^{3}+(b+wa+v)x^{2}+(wb+va)x+vb=0.\qquad \qquad (13)$

Comparing equation (13) to the original quartic equation, it can be seen that

$a+w={B \over A},$

$b+wa+v={C \over A},$

$wb+va={D \over A},$

and

$vb={E \over A}.$

Therefore

$w={B \over A}-a={B \over A}+2\mathrm {Re} (x_{1}),$

$v={E \over Ab}={E \over A\left([\mathrm {Re} (x_{1})]^{2}+[\mathrm {Im} (x_{1})]^{2}\right)}.$

Equation (12) can be solved for x yielding

$x_{3}={-w+{\sqrt {w^{2}-4v}} \over 2},$

$x_{4}={-w-{\sqrt {w^{2}-4v}} \over 2}.$

These two solutions are the desired real solutions if real solutions exist.

Alternative methods

Quick and memorable solution from first principles

Most textbook solutions of the quartic equation require a substitution that is almost impossible to memorize. Here is a way to approach it that makes it easy to understand.

The job is done if we can factor the quartic equation into a product of two quadratics. Let

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

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

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

This is harder to solve than it looks, but if we start again with a depressed quartic where $b=0$, which can be obtained by substituting $(x-b/4)$ for $x$, then $r=-p$, and:

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

It's now easy to eliminate both $s$ and $q$ by doing the following:

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

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

$P^{3}+2cP^{2}+(c^{2}-4e)P-d^{2}=0\,$

which is solved elsewhere. Once you have $p$, then:

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

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

Galois theory and factorization

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. Suppose r_i for i from 0 to 3 are roots of

$x^{4}+bx^{3}+cx^{2}+dx+e=0\qquad (1)$

If we now set

${\begin{aligned}s_{0}&={\tfrac {1}{2}}(r_{0}+r_{1}+r_{2}+r_{3}),\\s_{1}&={\tfrac {1}{2}}(r_{0}-r_{1}+r_{2}-r_{3}),\\s_{2}&={\tfrac {1}{2}}(r_{0}+r_{1}-r_{2}-r_{3}),\\s_{3}&={\tfrac {1}{2}}(r_{0}-r_{1}-r_{2}+r_{3}),\end{aligned}}$

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 really only need the values for s₁, s₂ and s₃. These we may find by expanding the polynomial

$(z^{2}-s_{1}^{2})(z^{2}-s_{2}^{2})(z^{2}-s_{3}^{2})\qquad (2)$

which if we make the simplifying assumption that b=0, is equal to

${z}^{6}+2c\,{z}^{4}+({c}^{2}-4e)\,{z}^{2}-{d}^{2}\qquad (3)$

This polynomial is of degree six, but only of degree three in z², and so the corresponding equation is solvable. By trial we can determine which three roots are the correct ones, and hence find the solutions of the quartic.

We can remove any requirement for trial by using a root of the same resolvent polynomial for factoring; if w is any root of (3), and if

$F_{1}={x}^{2}+wx+1/2\,{w}^{2}+1/2\,c-1/2\,{\frac {{c}^{2}w}{d}}-1/2\,{\frac {{w}^{5}}{d}}-{\frac {c{w}^{3}}{d}}+2\,{\frac {ew}{d}}$

$F_{2}={x}^{2}-wx+1/2\,{w}^{2}+1/2\,c+1/2\,{\frac {{w}^{5}}{d}}+{\frac {c{w}^{3}}{d}}-2\,{\frac {ew}{d}}+1/2\,{\frac {{c}^{2}w}{d}}$

then

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

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

Algebraic geometry

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

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

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

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

the pencil is given by the forms $\lambda F_{1}+\mu F_{2}$ for any point $[\lambda ,\mu ]$ in the projective line – in other words, where $\lambda$ and $\mu$ are not both zero, and multiplying a quadratic form by a constant does not change its quadratic curve of zeros.

This pencil contains three reducible quadratics, each corresponding to a pair of lines, each passing through two of the four points, which can be done $\textstyle {{\binom {4}{2}}=3}$ different ways. Denote these $Q_{1}=L_{12}+L_{34},$ $Q_{2}=L_{13}+L_{24},$ $Q_{3}=L_{14}+L_{23}.$ Given any two of these, their intersection is exactly the four points.

The reducible quadratics, in turn, may be determined by expressing the quadratic form $\lambda F_{1}+\mu F_{2}$ as a 3×3 matrix: reducible quadratics correspond to this matrix being singular, which is a equivalent to its determinant being zero, and the determinant is a homogeneous degree three polynomial in $\lambda$ and $\mu ,$ and corresponds to the resolvent cubic.