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Encyclopedia > Quartic equation

In mathematics, a quartic equation is the result of setting a quartic function equal to zero. The general form of a quartic equation is Euclid, Greek mathematician, 3rd century BC, as imagined by by Raphael in this detail from The School of Athens. ... Polynomial of degree 4: f(x) = (x+4)(x+1)(x-1)(x-3)/14+0. ... $ax^4+bx^3+cx^2+dx+e=0 ,$

where $ane 0$.

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). In mathematics, an nth root of a number a is a number b, such that bn=a. ...

Quartic equations were first discovered by Jaina Mathematicians in ancient India between 400 BC and 200 CE. See History of Cubic equation for more details. The chronology of Indian mathematics spans from the Indus Valley civilization (3300-1500 BCE) and Vedic civilization (1500-500 BCE) to modern India (21st century CE). ... Satellite image of the Indian subcontinent Map of South Asia (see note) This article deals with the geophysical region in Asia. ... The Celtics claim Vienna, Austria. ... For other uses, see number 200. ... Graph of a cubic polynomial: y = x3/4 + 3x2/4 âˆ’ 3x/2 âˆ’ 2 = (1/4)(x + 4)(x + 1)(x âˆ’ 2) In mathematics, a cubic equation is a polynomial equation in which the highest occurring power of the unknown is the third power. ...

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. 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). Lodovico Ferrari (February 2, 1522 - October 5, 1565) was an Italian mathematician. ... Graph of a cubic polynomial: y = x3/4 + 3x2/4 âˆ’ 3x/2 âˆ’ 2 = (1/4)(x + 4)(x + 1)(x âˆ’ 2) In mathematics, a cubic equation is a polynomial equation in which the highest occurring power of the unknown is the third power. ... Gerolamo Cardano or Jerome Cardan or Girolamo Cardan (September 24, 1501 - September 21, 1576) was a celebrated Italian Renaissance mathematician, physician, astrologer, and gambler. ... The Ars Magna (Latin: Great Work) is an important book on mathematics written by Gerolamo Cardano. ...

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. The Abelâ€“Ruffini theorem (also known as Abels Impossibility Theorem) states that there is no general solution in radicals to polynomial equations of degree five or higher. ... Galois at the age of fifteen from the pencil of a classmate. ...

## 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. In mathematics, the term optimization refers to the study of problems that have the form Given: a function f : A R from some set A to the real numbers Sought: an element x0 in A such that f(x0) â‰¤ f(x) for all x in A (minimization) or such that...

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. A ray-traced scene Ray tracing is a general technique from geometrical optics of modelling the path taken by light by following rays of light as they interact with optical surfaces. ... Ellipsoid Elliptic Paraboloid Hyperbolic Paraboloid Hyperboloid of One Sheet Hyperboloid of Two Sheets Cone Elliptic Cylinder Hyperbolic Cylinder Parabolic Cylinder In mathematics a quadric, or quadric surface, is any D-dimensional (hyper-)surface represented by a second-order equation in spatial variables (coordinates). ... In geometry, a torus (pl. ... A sphere is a perfectly symmetrical geometrical object. ... A developable surface is a surface that can be flattened onto a plane without distortion (i. ...

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. Computer-aided manufacturing (CAM) refers to the software used to generate the instruction codes for a CNC machine in order for it to cut out a shape designed in a computer-aided design (CAD) system. ... In geometry, a torus (pl. ... Several types of endmills An endmill is a type of Milling cutter, a cutting tool used in industrial milling applications. ...

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

## Solving a quartic equation

### Special cases

#### 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, Graph of a cubic polynomial: y = x3/4 + 3x2/4 âˆ’ 3x/2 âˆ’ 2 = (1/4)(x + 4)(x + 1)(x âˆ’ 2) In mathematics, a cubic equation is a polynomial equation in which the highest occurring power of the unknown is the third power. ...

a0x3 + a1x2 + a2x + a3 = 0.

#### Evident roots: 1 and -1

The equation a0x4 + a1x3 + a2x2 + a3x + a4 = 0 has 1 as a root if a0 + a1 + a2 + a3 + a4 = 0.

In this case, a0x4 + a1x3 + a2x2 + a3x + a4 can be divided by x − 1 and then the other roots can be found in the quotient.

The equation a0x4 + a1x3 + a2x2 + a3x + a4 = 0 has -1 as a root if a0 + a2 + a4 = a1 + a3

In this case, a0x4 + a1x3 + a2x2 + a3x + a4 can be divided by x + 1 and then the other roots can be found in the quotient.

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 = x2, 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: $z={{-a_2pmsqrt{a_2^2-4a_0a_4}} over {2a_0}},!$

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.

#### Other particular case: Quasi-symmetric equations

This kind of equation

x4 + a1x3 + a2x2 + a3x + m2 = 0, where m = a3 / a1, can be solved using Ana Flores` method:

Dividing the equation by x2, we obtain

x2 + a1x + a2 + a3 / x + m2 / x2 = 0

x2 + m2 / x2 + a1x + a3 / x + a2 = 0

(x2 + m2 / x2) + a1(x + m / x) + a2 = 0

and then using this variable change:

z = x + m / x.

And

z2 − 2m = x2 + m2 / x2

So:

(z2 − 2m) + a1z + a2 = 0.

This equation gives up to 2 different real roots

z1 and z2

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

x2z1x + m = 0.

and

x2z2x + m = 0.

If a0 is different from 1 in

a0x4 + a1x3 + a2x2 + a3x + a0m2 = 0

this method can still be applied. The whole equation has then to be divided by a0 as a first step.

The quasi symmetric equation has the following property: lets x1, x2, and x3,x4 be the roots of the equation, therefore x1x2 = m.

Since que product of the 4 roots is m2, then x3x4 = m too.

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

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

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

#### 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 Lodovico Ferrari (February 2, 1522 - October 5, 1565) was an Italian mathematician. ...

(u2 + α)2u4 − 2αu2 = α2

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, α u2 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), $begin{matrix} (u^2+alpha+y)^2-(u^2+alpha)^2 & = & 2y(u^2+alpha)+ y^2 & = & 2yu^2+2yalpha+y^2, end{matrix}$

and

0 = (α + 2y)u2 − 2yu2 − αu2

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

(u2 + α + y)2 + βu + γ = (α + 2y)u2 + (2yα + y2 + α2).

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:

(su + t)2 = (s2)u2 + (2st)u + (t2).

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(s2)(t2) = 0.

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

( − β)2 − 4(2y + α)(y2 + 2yα + α2 − γ) = 0.

Multiply the binomial with the polynomial,

β2 − 4(2y3 + 5αy2 + (4α2 − 2γ)y + (α3 − αγ)) = 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, Graph of a cubic polynomial: y = x3/4 + 3x2/4 âˆ’ 3x/2 âˆ’ 2 = (1/4)(x + 4)(x + 1)(x âˆ’ 2) In mathematics, a cubic equation is a polynomial equation in which the highest occurring power of the unknown is the third power. ... $y^3 + {5 over 2} alpha y^2 + (2 alpha^2 - gamma) y + left( {alpha^3 over 2} - {alpha gamma over 2} - {beta^2 over 8} right) = 0. qquad qquad (4)$

##### Conversion of the nested cubic into a depressed cubic

Equation (4) is a cubic equation nested within the quartic equation. It must be solved 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 $v^3 + P v + Q = 0. qquad qquad (5)$

##### Solving the nested depressed cubic

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

let $U=sqrt{{Qover 2}pm sqrt{{Q^{2}over 4}+{P^{3}over 27}}}$
(taken from Cubic equation) $v = {Pover 3U} - U$

therefore the solution of the original nested cubic is Graph of a cubic polynomial: y = x3/4 + 3x2/4 âˆ’ 3x/2 âˆ’ 2 = (1/4)(x + 4)(x + 1)(x âˆ’ 2) In mathematics, a cubic equation is a polynomial equation in which the highest occurring power of the unknown is the third power. ... $y = - {5 over 6} alpha + {Pover 3U} - U qquad qquad (6)$
Remember 1: $P=0 Longleftarrow {Qover 2} + sqrt{{Q^{2}over 4}+{P^{3}over 27}}=0$
Remember 2: $lim_{Pto 0}{P over sqrt{{Qover 2} + sqrt{{Q^{2}over 4}+{P^{3}over 27}}}}=0$

##### 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 $(s^2)u^2+(2st)u+(t^2) = left(left(sqrt{(s^2)}right)u + {(2st) over 2sqrt{(s^2)}}right)^2$
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: $(alpha + 2 y) u^2 + (- beta) u + (y^2 + 2 y alpha + alpha^2 - gamma ) = left( left(sqrt{(alpha + 2y)}right)u + {(-beta) over 2sqrt{(alpha + 2 y)}} right)^2$.
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 $(u^2 + alpha + y)^2 = left( left(sqrt{alpha + 2 y}right)u - {beta over 2sqrt{alpha + 2 y}} right)^2 qquadqquad (7)$.

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: $(u^2 + alpha + y) = pmleft( left(sqrt{alpha + 2 y}right)u - {beta over 2sqrt{alpha + 2 y}} right) qquadqquad (7')$.

Collecting like powers of u produces $u^2 + left(mp_s sqrt{alpha + 2 y}right)u + left( alpha + y pm_s {beta over 2sqrt{alpha + 2 y}} right) = 0 qquadqquad (8)$.
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 In mathematics, a quadratic equation is a polynomial equation of the second degree. ... $u={pm_ssqrt{alpha + 2 y} pm_t sqrt{(alpha + 2y) - 4(alpha + y pm_s {beta over 2sqrt{alpha + 2 y}})} over 2}.$

Simplifying, one gets $u={pm_ssqrt{alpha + 2 y} pm_t sqrt{-left(3alpha + 2y pm_s {2beta over sqrt{alpha + 2 y}} right)} over 2}.$

This is the solution of the depressed quartic, therefore the solutions of the original quartic equation are $x=-{B over 4A} + {pm_ssqrt{alpha + 2 y} pm_t sqrt{-left(3alpha + 2y pm_s {2beta over sqrt{alpha + 2 y}} right)} over 2}. qquadqquad (8')$
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

Ax4 + Bx3 + Cx2 + Dx + E = 0,

its solution can be found by means of the following calculations: $alpha = - {3 B^2 over 8 A^2} + {C over A},$ $beta = {B^3 over 8 A^3} - {B C over 2 A^2} + {D over A},$ $gamma = {-3 B^4 over 256 A^4} + {C B^2 over 16 A^3} - {B D over 4 A^2} + {E over A},$
if β = 0 solve u4 + αu2 + γ = 0 and substitute $x=u-{Bover 4A}$ finding the roots $x=-{Bover 4A}pm_ssqrt{-alphapm_tsqrt{alpha^2-4gamma}over 2},qquadbeta=0$. $P = - {alpha^2 over 12} - gamma,$ $Q = - {alpha^3 over 108} + {alpha gamma over 3} - {beta^2 over 8},$ $R = {Qover 2} pm sqrt{{Q^{2}over 4}+{P^{3}over 27}}$, (either sign of the square root will do) $U = sqrt{R}$, (there are 3 complex roots, any one of them will do) $y = - {5 over 6} alpha -U + begin{cases}U=0 &to 0Une 0 &to {Pover 3U}end{cases},$ $W=sqrt{ alpha + 2 y}$ $x = - {B over 4 A} + { pm_s W pm_t sqrt{-left(3alpha + 2 y pm_s {2betaover W} right) }over 2 }.$
The two ±s must have the same sign, the ±t is independent. To get all roots, find x for ±st = +,+ 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 A Roman mosaic picturing Theseus and the Minotaur. ...

x4 + 6x2 − 60x + 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.

#### 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 In mathematics, a complex number is a number of the form where a and b are real numbers, and i is the imaginary unit, with the property i 2 = âˆ’1. ... In mathematics, the complex conjugate of a complex number is given by changing the sign of the imaginary part. ...

(xx1)(xx2)(xx3)(xx4) = 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 $begin{matrix} (x-x_1)(x-x_2)&=&x^2-(x_1+x_1^star)x+x_1x_1^starqquadqquadqquadquad &=&x^2-2,mathrm{Re}(x_1)x+[mathrm{Re}(x_1)]^2+[mathrm{Im}(x_1)]^2. end{matrix}$

Let $a = - 2 , mathrm{Re}(x_1),$
b = [Re(x1)]2 + [Im(x1)]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.

### Alternative methods

#### 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 In mathematics, a quadratic equation is a polynomial equation of the second degree. ... $begin{array}{lcl} 0 = x^4 + bx^3 + cx^2 + dx + e & = & (x^2 + px + q)(x^2 + rx + s) & = & x^4 + (p + r)x^3 + (q + s + pr)x^2 + (ps + qr)x + qs end{array}$

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 (xb / 4) for x, then r = − p, and: In mathematics, a quartic equation is the result of setting a quartic function equal to zero. ... $begin{array}{lcl} c + p^2 & = & s + q d/p & = & s - q e & = & sq end{array}$

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 = p2, then this equation turns into the cubic equation: Graph of a cubic polynomial: y = x3/4 + 3x2/4 âˆ’ 3x/2 âˆ’ 2 = (1/4)(x + 4)(x + 1)(x âˆ’ 2) In mathematics, a cubic equation is a polynomial equation in which the highest occurring power of the unknown is the third power. ... $P^3 + 2cP^2 + (c^2 - 4e)P - d^2 = 0,$

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 whose roots may be variously described as a discrete Fourier transform or a Hadamard matrix transform of the roots. Suppose ri for i from 0 to 3 are roots of In mathematics, the symmetric group on a set X, denoted by SX or Sym(X), is the group whose underlying set is the set of all bijective functions from X to X, in which the group operation is that of composition of functions, i. ... This article is about the mathematical group. ... In mathematics, a normal subgroup N of a group G is a subgroup invariant under conjugation; that is, for each element n in N and each g in G, the element gâˆ’1ng is still in N. The statement N is a normal subgroup of G is written: . There are... In mathematics, the resolvent formalism is a technique for applying concepts from complex analysis to the study of the spectrum of operators on Hilbert spaces and more general spaces. ... In mathematics, a Hadamard matrix is a square matrix whose entries are either +1 or âˆ’1 and whose rows are mutually orthogonal. ... $x^4 + bx^3 + cx^2 + dx + e = 0qquad (1)$

If we now set

s0 = (r0 + r1 + r2 + r3) / 2
s1 = (r0r1 + r2r3) / 2
s2 = (r0 + r1r2r3) / 2
s3 = (r0r1r2 + r3) / 2

then since the transformation is an involution we may express the roots in terms of the four si 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 In mathematics, an involution is a function that is its own inverse, so that f(f(x)) = x for all x in the domain of f. ... $(z^2 - s_1^2)(z^2-s_2^2)(z^2-s_3^2)qquad (2)$

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.

A linear equation is an equation in which each term is either a constant or the product of a constant times the first power of a variable. ... In mathematics, a quadratic equation is a polynomial equation of the second degree. ... Graph of a cubic polynomial: y = x3/4 + 3x2/4 âˆ’ 3x/2 âˆ’ 2 = (1/4)(x + 4)(x + 1)(x âˆ’ 2) In mathematics, a cubic equation is a polynomial equation in which the highest occurring power of the unknown is the third power. ... Polynomial of degree 5: f(x) = (x+4)(x+2)(x+1)(x-1)(x-3)/20+2 In mathematics, a quintic equation is a polynomial equation in which the greatest exponent on the independent variable is five. ... In mathematics, a polynomial is an expression in which constants and variables are combined using only addition, subtraction, multiplication, and positive whole number exponents (raising to a power). ... Lodovico Ferrari (February 2, 1522 - October 5, 1565) was an Italian mathematician. ... Gerolamo Cardano or Jerome Cardan (September 24, 1501 - September 21, 1576) was a celebrated Renaissance mathematician, physician, astrologer, and gambler. ... Results from FactBites:

 Klein's Quartic Curve (0 words) Klein's quartic curve is a surface of genus 3, which is to say that it is like a 3-holed torus. The 56 triangular faces of the dual tiling of Klein's quartic curve can be divided into 7 sets of 8, which have the same symmetries as the 8 corners of a truncated cube. Klein's quartic curve also has symmetries of order 7, which are not very clear in any of the pictures shown so far.
 PlanetMath: quartic formula (236 words) The formulas for the roots are much too unwieldy to be used for solving quartic equations by radicals, even with the help of a computer. A practical algorithm for solving quartic equations by radicals is given in the concluding paragraph of the Galois-theoretic derivation of the quartic formula. This is version 3 of quartic formula, born on 2002-01-22, modified 2005-07-07.
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