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where a is nonzero; or in other words, a polynomial of degree four. Such a function is sometimes called a biquadratic function, but the latter term can occasionally also refer to a quadratic function of a square, having the form
or a product of two quadratic factors, having the form
Setting results in a quartic equation of the form:
where a ≠ 0.
Since a quartic function is a polynomial of even degree, it has the same limit when the argument goes to positive or negative infinity. If a is positive, then the function increases to positive infinity at both sides; and thus the function has a global minimum. Likewise, if a is negative, it decreases to negative infinity and has a global maximum.
The quartic is the highest order polynomial equation that can be solved by radicals in the general case (i.e., one where the coefficients can take any value).
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).
It is reported that even earlier, in 1486, Spanish mathematician Paolo Valmes was burned at the stake for claiming to have solved the quartic equation. Inquisitor General Tomás de Torquemada allegedly told him that it was the will of God that such a solution be inaccessible to human understanding. However, attempts to find corroborating evidence for this story, or for the existence of Paolo Valmes, have not succeeded.
The proof that four is the highest degree of a general polynomial for which such solutions can 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 prior to dying in a duel in 1832 later led to an elegant complete theory of the roots of polynomials, of which this theorem was one result.
Polynomials of high degrees often appear in problems involving optimization, and sometimes these polynomials happen to be quartics, but this is a coincidence.
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. 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 program demonstrating various analytic solutions to the quartic was provided in Graphics Gems Book V. However, none of the three algorithms implemented are unconditionally stable. In an updated version of the paper, which compares the 3 algorithms from the original paper and 2 others, it is demonstrated that computationally stable solutions exist only for 4 of the possible 16 sign combinations of the quartic coefficients.
The 4 roots () for any quartic equation;
where are equal to those of
where , , and .
The roots in terms of these 4 coefficients are given by the formula in the image below:
However, this formula is too unwieldy for general use, hence other methods or simpler formulas (given below) are generally used.
Consider the quartic
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.
If then , so is a root. Similarly, if that is, then is a root.
When is a root, we can divide by and get
where is a cubic polynomial, which may be solved to find 's other roots. Similarly, if is a root,
where is some cubic polynomial.
If then −k is a root and we can factor out ,
And if then both and are roots Now we can factor out and get
To get Q 's other roots, we simply solve the quadratic factor.
We call such a polynomial a biquadratic, which is easy to solve.
Let Then Q becomes a quadratic q in
Let and be the roots of q. Then the roots of our quartic Q are
To begin, the quartic must first be converted to a depressed quartic.
be the general quartic equation we want to solve. Divide both sides by A to produce a monic polynomial,
The first step should be to eliminate the x3 term. To do this, change variables from x to u, such that
Expanding the powers of the binomials produces
Collecting the same powers of u yields
Now rename the coefficients of u. Let
The resulting equation is
which is a depressed quartic equation.
If then we have a biquadratic equation, which (as explained above) is easily solved; using reverse substitution we can find our values for .
If then one of the roots is and the other roots can be found by dividing by , and solving the resulting depressed cubic equation,
Using reverse substitution we can find our values for .
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
to equation (1), yielding
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),
These two formulas, added together, produce
which added to equation (2) produces
This is equivalent to
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:
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:
Therefore to make the right side of equation (3) into a perfect square, the following equation must be solved:
Multiply the binomial with the polynomial,
Divide both sides by −4, and move the −β2/4 to the right,
This is a cubic equation for y. Divide both sides by 2,
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
Equation (4) becomes
Expand the powers of the binomials,
Distribute, collect like powers of v, and cancel out the pair of v2 terms,
This is a depressed cubic equation.
Relabel its coefficients,
The depressed cubic now is
The solutions (any solution will do, so pick any of the three complex roots) of equation (5) are computed as (see Cubic equation)
and V is computed according to the two defining equations and , so
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
so that it can be folded:
Therefore equation (3) becomes
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:
Collecting like powers of u produces
Equation (8) is a quadratic equation for u. Its solution is
Simplifying, one gets
This is the solution of the depressed quartic, therefore the solutions of the original quartic equation are
Given the quartic equation
its solution can be found by means of the following calculations:
Otherwise, continue with
(either sign of the square root will do)
(there are 3 complex roots, any one of them will do)
As stated above, Cardano credited Ferrari as the first to discover one of these labyrinthine solutions. The equation he solved was:
which was already in depressed form. It has a pair of solutions that can be found with the set of formulas shown above.
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 where P and Q are given by (5) has the properties that
where α and β are given by (1).
This means that (5) has a real root greater than , and therefore that (4) has a real root greater than .
Using this root the term in (8) is always real, which ensures that the two quadratic equations (8) have real coefficients.
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
but this quartic equation is equivalent to the product of two quadratic equations:
so that equation (9) becomes
Also let there be (unknown) variables w and v such that equation (10) becomes
Multiplying equations (11) and (12) produces
Comparing equation (13) to the original quartic equation, it can be seen that
Equation (12) can be solved for x yielding
These two solutions are the desired real solutions if real solutions exist.
By equating coefficients, this results in the following set of simultaneous equations:
This can be simplified by starting again with a depressed quartic where , which can be obtained by substituting for , then , and:
It's now easy to eliminate both and by doing the following:
If we set , then this equation turns into the resolvent cubic equation
which is solved elsewhere. Then:
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 for the square root of merely exchanges the two quadratics with one another.
The above solution shows that the quartic polynomial with a zero coefficient on the cubic term is factorable into quadratics with rational coefficients if and only if the resolvent cubic has a root which is the square of a rational; this can readily be checked using the rational root test.
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 ri for i from 0 to 3 are roots of
If we now set
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
which if we make the simplifying assumption that b=0, is equal to
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
We therefore can solve the quartic by solving for w and then solving for the roots of the two factors using the quadratic formula.
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 may also be expressed as the x coordinates of the intersections of the two quadratic equations i.e., using the substitution that two quadratics intersect in four points is an instance of Bézout's theorem. Explicitly, the four points are for the four roots of the quartic.
These four points are not collinear because they lie on the irreducible quadratic 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:
the pencil is given by the forms for any point in the projective line – in other words, where and 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 different ways. Denote these 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 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 and and corresponds to the resolvent cubic.
Alternative quartic solver Starting with a_0 x^4+a_1 x^3 〖+a〗_2 x^2+a_3 x+a_4 = 0 (1) where a0 is not zero, one proceeds as follows: 1. Simplify the equation by dividing throughout by a0 to get x^4+A_1 x^3 〖+A〗_2 x^2+A_3 x+A_4 = 0 (2) where A1 = a1/a0, etc. 2. Obtain the reduced form by the substitution: x=y- A_1/4 (3) to get y^4+py^2+qy+r = 0 (4) where p=A_2-(3A_1^2)/8 q= (A_1^3)/8-(A_1 A_2)/2+A_3 r= (A_1^2 A_2)/16-(A_1 A_3)/4-(3A_1^4)/256+A_4 3. Express equation (4) as a difference of two squares: (y^2+a)^2-(by+c)^2=0 (5) where a,b,c are related to p,q,r by 2a-b^2=p -2bc=q a^2-c^2=r
4. Eliminate b and c to get a cubic for a: k_0 a^3 〖+k〗_1 a^2+k_2 a+k_3 = 0 (6) where k_0=8 k_1=-4p k_2=-8r k_3= 4pr-q^2 Equation (6) has at least one real root that can be calculated using for example Cardano’s method (1545). Having found a, the other two can be deduced from: b=±√(2a-p) c=±√(a^2-r) Here, depending on the sign of q, q>0: b and c must have opposite signs q<0: b and c must have the same sign q=0: b or c must be zero 5. Solve the two quadratic equations from equation (5) y^2+by+a+c=0 y^2-by+a-c=0 to get the 4 roots for equation (4), and those for equation (2) and ultimately equation (1) can then be deduced from equation (3).
6. Numerical example The following is a typical computer printout: Coefficients of quartic are 1, 2, 3, 4, -45 p = 1.5 q = 2 r = -46.4375 Coefficients of cubic are 8, -6, 371.5, -282.625 Roots of cubic 1st root = 0.760635 2nd root is complex:
real part = -5.317331E-03 imag part = 6.815098
3rd root is conjugate of 2nd root a = 0.760635 abs(b) = 0.145840 abs(c) = 6.856826 Roots of quartic First pair of roots are 1.897203 and -3.043043 Root 3 is complex:
real part = -0.427080 imag part = 2.759011
Root 4 is conjugate of Root 3