Quartic equation
The quartic is the highest order polynomial equation that can be solved by radicals in the general case (i.e., one in which the coefficients can take any value).
[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 before his death in a duel in 1832 later led to an elegant complete theory of the roots of polynomials, of which this theorem was one result.
: There exists a general formula for finding the roots to quartic equations, provided the coefficient of the leading term is non-zero.
However, since the general method is quite complex and susceptible to errors in execution, it is better to apply one of the special cases listed below if possible.
If the constant term 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, Call our quartic polynomial Q(x).
, so our equation becomes which is a simple quadratic equation, whose solutions are easily found using the quadratic formula: When we've solved it (i.e. found these two z values), we can extract x from them If either of the z solutions were negative or complex numbers, then some of the x solutions are complex numbers.
Steps: This leads to: If the quartic has a double root, it can be found by taking the polynomial greatest common divisor with its derivative.
In general, there exist only four possible cases of quartic equations with multiple roots, which are listed below:[3] So, if the three non-monic coefficients of the depressed quartic equation,
, then the criteria to identify a priori each case of quartic equations with multiple roots and their respective solutions are shown below.
Let be the general quartic equation which it is desired to solve.
then we have the special case of a biquadratic equation, which is easily solved, as explained above.
Note that the general solution, given below, will not work for the special case
After converting to a depressed quartic equation and excluding the special case b = 0, which is solved as a biquadratic, we assume from here on that b ≠ 0 .
Let y be any solution of this cubic equation: Then (since b ≠ 0) so we may divide by it, giving Then Subtracting, we get the difference of two squares which is the product of the sum and difference of their roots which can be solved by applying the quadratic formula to each of the two factors.
A wise strategy is to choose the sign of the square-root that makes the absolute value of w as large as possible.
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 + a)2.
The second term, au2 did not disappear, but its sign has changed and it has been moved to the right side.
To accomplish these insertions, the following valid formulas will be added to equation (2), and 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.
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 −b2/4 to the right, Divide both sides by 2, This is a cubic equation in y.
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 (6) is a quadratic equation for u.
It has a pair of solutions which can be found with the set of formulas shown above.
Most textbook solutions of the quartic equation require a substitution that is hard to memorize.
The job is done if we can factor the quartic equation into a product of two quadratics.
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
The symmetric group S4 on four elements has the Klein four-group as a normal subgroup.
By trial we can determine which three roots are the correct ones, and hence find the solutions of the quartic.
Iterative methods are the only ones available for quintic and higher-order equations, beyond trivial or special cases.
