Quartic equation

From Wikipedia, the free encyclopedia

In mathematics, a quartic equation is the result of setting a quartic function equal to zero. The general form of a quartic equation is

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

where $a\ne 0$ .

The quartic is the highest order polynomial equation that can be solved by radicals in the general case (ie one where the coefficients can take any value).

[edit] History

Quartic equations were first discovered by Jaina Mathematicians in ancient India between 400 BC and 200 AD. See History of Cubic equation for more details.^{[citation needed]}

Lodovico Ferrari is attributed 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 couldn't be published immediately.[1] The solution of the quartic was published together with that of the cubic by Ferrari's mentor Gerolamo Cardano in the book Ars Magna (1545).

The proof that this was the highest order general polynomial for which such solutions could 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 in 1832 later lead to the complete theory of the roots of polynomials, of which this theorem was one result. ^[1]

[edit] Applications

Polynomials of high degrees often appears in problems involving Optimization (mathematics), and sometimes these polynomials happen to be quartics, but this is a coincidence.

Quartics often arise in computer graphics and during ray-tracing against surfaces such as quadrics or tori surfaces, which are the next level beyond the sphere and developable surfaces.[2]

Another frequent generator of quartics is the intersection of two ellipses.

In Computer-aided manufacturing, the torus is a common shape associated with the endmill cutter. In order to calculate its location relative to a triangulated surface, the position of a horizontal torus on the 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. Over 10% of the computational time in a CAM system can be consumed simply calculating the solution to millions of quartic equations.

A programmed version of a stable solution to the quartic was provided in Graphics Gems Book V.[3]

[edit] Solving a quartic equation

[edit] Special cases

[edit] Quartics in name only

If a₄ = 0, then one of the roots is x = 0, and the other roots can be found by dividing by x, and solving the resulting cubic equation,

a 0 x 3 + a 1 x 2 + a 2 x + a 3 = 0.

[edit] Evident roots: 1 and -1

The equation $a 0 x 4 + a 1 x 3 + a 2 x 2 + a 3 x + a 4 = 0$ has 1 as a root if $a 0 + a 1 + a 2 + a 3 + a 4 = 0.$

In this case, $a 0 x 4 + a 1 x 3 + a 2 x 2 + a 3 x + a 4$ can be divided by $x - 1$ and then the other roots can be found in the quotient.

The equation $a 0 x 4 + a 1 x 3 + a 2 x 2 + a 3 x + a 4 = 0$ has -1 as a root if $a 0 + a 2 + a 4 = a 1 + a 3$

In this case, $a 0 x 4 + a 1 x 3 + a 2 x 2 + a 3 x + a 4$ can be divided by $x + 1$ and then the other roots can be found in the quotient.

[edit] Biquadratic equations

A quartic equation where a₃ and a₁ are equal to 0 takes the form

$a_0x^4+a_2x^2+a_4=0\,\!$

and thus is a biquadratic equation, very easy to solve. Let $z = x 2$ , so our equation turns to

$a_0z^2+a_2z+a_4=0\,\!$

which is a simple quadratic equation, whose solutions are easily found using the quadratic formula:

$z={{-a_2\pm\sqrt{a_2^2-4a_0a_4}} \over {2a_0}}\,\!$

When we've solved it (i.e. found these two z values), we can extract x from them

$x_1=+\sqrt{z_1}\,\!$

$x_2=-\sqrt{z_1}\,\!$

$x_3=+\sqrt{z_2}\,\!$

$x_4=-\sqrt{z_2}\,\!$

If either of the z solutions were negative or complex numbers, some of the x solutions are complex numbers.

[edit] Other particular case: Quasi-symmetric equations

This kind of equation

$x 4 + a 1 x 3 + a 2 x 2 + a 3 x + m 2 = 0$ , where $m = a 3 / a 1$ , can be solved using Ana Flores` method:

Dividing the equation by $x 2$ , we obtain

$x 2 + a 1 x + a 2 + a 3 / x + m 2 / x 2 = 0$

$x 2 + m 2 / x 2 + a 1 x + a 3 / x + a 2 = 0$

$(x 2 + m 2 / x 2) + a 1 (x + m / x) + a 2 = 0$

and then using this variable change:

$z = x + m / x .$

And

$z 2 - 2 m = x 2 + m 2 / x 2$

So:

$(z 2 - 2 m) + a 1 z + a 2 = 0.$

This equation gives up to 2 different real roots

$z 1$ and $z 2$

The root of the original equation can be found solving the equations

$x 2 - z 1 x + m = 0.$

and

$x 2 - z 2 x + m = 0.$

If $a 0$ is different from 1 in

$a 0 x 4 + a 1 x 3 + a 2 x 2 + a 3 x + a 0 m 2 = 0$

this method can still be applied. The whole equation has then to be divided by $a 0$ as a first step.

The quasi symmetric equation has the following property: lets $x 1$ , $x 2$ , and $x 3$ , $x 4$ be the roots of the equation, therefore $x 1 x 2 = m$ .

Since que product of the 4 roots is $m 2$ , then $x 3 x 4 = m$ too.

[edit] The general case, along Ferrari's lines

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

[edit] Converting to a depressed quartic

Let

$A x^4 + B x^3 + C x^2 + D x + E = 0 \qquad\qquad(1')$

be the general quartic equation which it is desired to solve. Divide both sides by A,

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

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

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

Then

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

Expanding the powers of the binomials produces

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

Collecting the same powers of u yields

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

Now rename the coefficients of u. Let

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

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

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

The resulting equation is

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

which is a depressed quartic equation.

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

[edit] Ferrari's solution

Otherwise, the depressed quartic can be solved by means of a method discovered by 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

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

These two formulas, added together, produce

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

which added to equation (2) produces

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

This is equivalent to

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

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:

$(s u + t)^2 = (s^2) u^2 + (2 s t) u + (t^2).\,$

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:

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

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

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

Multiply the binomial with the polynomial,

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

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

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

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

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

[edit] Conversion of the nested cubic into a depressed cubic

Equation (4) is a cubic equation nested within the quartic equation. It must be solved in order 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,

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

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

$v^3 + P v + Q = 0. \qquad \qquad (5)$

[edit] Solving the nested depressed cubic

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

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

(taken from Cubic equation)

$v = {P\over 3U} - U$

therefore the solution of the original nested cubic is

$y = - {5 \over 6} \alpha + {P\over 3U} - U \qquad \qquad (6)$

Remember 1: $P=0 \Longleftarrow {Q\over 2} + \sqrt{{Q^{2}\over 4}+{P^{3}\over 27}}=0$

Remember 2: $\lim_{P\to 0}{P \over \sqrt[3]{{Q\over 2} + \sqrt{{Q^{2}\over 4}+{P^{3}\over 27}}}}=0$

[edit] 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

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

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:

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

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

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

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:

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

Collecting like powers of u produces

$u^2 + \left(\mp_s \sqrt{\alpha + 2 y}\right)u + \left( \alpha + y \pm_s {\beta \over 2\sqrt{\alpha + 2 y}} \right) = 0 \qquad\qquad (8)$ .

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

$u={\pm_s\sqrt{\alpha + 2 y} \pm_t \sqrt{(\alpha + 2y) - 4(\alpha + y \pm_s {\beta \over 2\sqrt{\alpha + 2 y}})} \over 2}.$

Simplifying, one gets

$u={\pm_s\sqrt{\alpha + 2 y} \pm_t \sqrt{-\left(3\alpha + 2y \pm_s {2\beta \over \sqrt{\alpha + 2 y}} \right)} \over 2}.$

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

$x=-{B \over 4A} + {\pm_s\sqrt{\alpha + 2 y} \pm_t \sqrt{-\left(3\alpha + 2y \pm_s {2\beta \over \sqrt{\alpha + 2 y}} \right)} \over 2}. \qquad\qquad (8')$

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.

[edit] Summary of Ferrari's method

Given the quartic equation

A x 4 + B x 3 + C x 2 + D x + E = 0,

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

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

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

$\gamma = {-3 B^4 \over 256 A^4} + {C B^2 \over 16 A^3} - {B D \over 4 A^2} + {E \over A},$

β = 0

solve

u 4 + α u 2 + γ = 0

and substitute $x=u-{B\over 4A}$ finding the roots

$x=-{B\over 4A}\pm_s\sqrt{-\alpha\pm_t\sqrt{\alpha^2-4\gamma}\over 2},\qquad\beta=0$ .

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

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

$R = {Q\over 2} \pm \sqrt{{Q^{2}\over 4}+{P^{3}\over 27}}$ , (either sign of the square root will do)

$U = \sqrt[3]{R}$ , (there are 3 complex roots, any one of them will do)

$y = - {5 \over 6} \alpha -U + \begin{cases}U=0 &\to 0\\U\ne 0 &\to {P\over 3U}\end{cases},$

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

$x = - {B \over 4 A} + { \pm_s W \pm_t \sqrt{-\left(3\alpha + 2 y \pm_s {2\beta\over W} \right) }\over 2 }.$

The two ±_s must have the same sign, the ±_t is independent. To get all roots, find x for ±_s,±_t = +,+ and for +,− and for −,+ and for −,−. Double roots will be given twice, triple roots 3 times and quadruple roots would be given 4 times (although then β = 0, which is a special case). The order of the roots depends on which cubic root U one chose. (see note for (8) vis-à-vis (8'))

Quod Erat Faciendum. There are other methods of solving the quartic equations, perhaps more optimal. Ferrari was the first to discover one of these labyrinthine solutions. The equation which he solved was

x 4 + 6 x 2 - 60 x + 36 = 0

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

[edit] Obtaining alternative solutions the hard way

It could happen that one only obtained one solution through the seven formulae above, because one doesn't like trying all four sign patterns to get all four solutions, and the solution one 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

$\begin{matrix} (x-x_1)(x-x_2)&=&x^2-(x_1+x_1^\star)x+x_1x_1^\star\qquad\qquad\qquad\quad \\ &=&x^2-2\,\mathrm{Re}(x_1)x+[\mathrm{Re}(x_1)]^2+[\mathrm{Im}(x_1)]^2. \end{matrix}$

Let

$a = - 2 \, \mathrm{Re}(x_1),$

b = [Re(x 1)] 2 + [Im(x 1)] 2

so that equation (9) becomes

$x^2 + a x + b = 0. \qquad \qquad (11)$

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

$x^2 + w x + v = 0. \qquad \qquad (12)$

Multiplying equations (11) and (12) produces

$x^4 + (a + w) x^3 + (b + w a + v) x^2 + (w b + v a) x + v b = 0. \qquad \qquad (13)$

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

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

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

$w b + v a = {D \over A},$

and

$v b = {E \over A}.$

Therefore

$w = {B \over A} - a = {B \over A} + 2 \mathrm{Re}(x_1),$

$v = {E \over A b} = {E \over A \left( [ \mathrm{Re}(x_1) ]^2 + [ \mathrm{Im}(x_1) ]^2 \right) }.$

Equation (12) can be solved for x yielding

$x_3 = {-w + \sqrt{w^2 - 4 v} \over 2},$

$x_4 = {-w - \sqrt{w^2 - 4 v} \over 2}.$

One of these two solutions should be the desired real solution.

[edit] Alternative methods

[edit] Quick and memorable solution from first principles

Most textbook solutions of the quartic equation require a magic 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 factorize 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.

[edit] 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 whose roots may be variously described as a discrete Fourier transform or a Hadamard matrix transform of the roots. 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

s 0 = (r 0 + r 1 + r 2 + r 3) / 2

s 1 = (r 0 - r 1 + r 2 - r 3) / 2

s 2 = (r 0 + r 1 - r 2 - r 3) / 2

s 3 = (r 0 - r 1 - r 2 + r 3) / 2

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.

[edit] See also

[edit] References

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

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