Quartic function

A quartic equation or polynomial equation of degree 4, traditionally known as biquadratic equation has the form

With coefficients and from a body with characteristic 0, then being derived from the K- algebra. In the following, but we consider as a body only the quantities and with the usual addition and multiplication.

By the fundamental theorem of algebra, the equation up to the order can be uniquely in the form

Bring, where and are the - not necessarily different - four solutions of the equation are.

If and, then can the equation by substitution of a quadratic equation traced. Today - especially in school mathematics - it's common, only this special form to name a biquadratic equation, although biquad traditionally has a more general meaning.

  • 4.1 B = 0 and D = 0
  • 4.2 E = 0
  • 4.3 Real coefficients 4.3.1 Four real solutions
  • 4.3.2 Two real and two complex conjugate solutions
  • 4.3.3 Two pairs of complex conjugate solutions

History

The first closed-form solution of the quartic equation was the Italian mathematician Lodovico Ferrari ( 1522-1565 ). This solution released his teacher Gerolamo Cardano in 1545 in the book Ars magna de Regulis Algebraicis. Another solution method with a different approach was published by Leonhard Euler in 1738 in Saint Petersburg, in an effort to find a general solution formula for equations of higher degrees. That this is impossible, it was proved by Niels Henrik Abel in 1824 ( set of Abel - Ruffini ).

Solution formula and proof

Since the general solution formula is confusing, the general equation is gradually converted into more specific, equivalent forms. The transformations of the variables it performed must be reversed at the end of the solutions in the reverse order.

Prerequisite: Consider using a quartic equation.

Statement: Then one can specify their solutions to algebraic way as follows:

Normalizing and reducing

First, the equation with substitution of the

Simplified so that the cubic coefficient B vanishes ( Tschirnhaus transformation), while the leading coefficient by dividing the entire equation by A is set to 1.

With the definitions

The equation reduces to

At the end of the statement, the zeros of the Ausgangspolynoms be as recovered. Hereinafter may thus be assumed that the third order coefficient is zero.

Case that only just exponents occur

If, then we obtain the special case of a (real) quartic equation

And can determine the zeros as square roots in both sign variants from the solutions of the quadratic equation.

If the coefficients are real and so it makes more sense, does not directly determine the then complex solutions of the quadratic equation in z, and from that the square. But the equation is factored only real in a different way, the two quadratic factors again have real coefficients.

The zeros can be determined for each factor individually again.

General case

If so you're trying to write the equation as a difference of two complete squares. This complex parameters are introduced. The representation as a difference then leads directly to a factorization into quadratic factors with complex coefficients,

By comparison with

Surrender and well.

So that the second part of the difference is a complete square in u, the discriminant must this quadratic term disappear

This is a cubic equation in y.

From one of the solutions for y are two quadratic equations in u, leading to a total of four solutions for u and then x arise.

Summary

Altogether, the following calculation steps are performed:

Now the zeros can be calculated as follows:

And the variables of the original equation

The parameters passed to the to be selected in the two square roots of sign, all four combinations of and are needed to get all four solutions.

Decomposition into quadratic factors

Here, the decomposition into a product of two quadratic factors

Attributed to the solution of the cubic equation u

( In real coefficients a, b, c and d there is a real u 4u ≤ a2)

With a solution u of this equation is calculated directly:

In the special case is the solution

Example 1: For one arrives at the equation of 3rd degree

One solution is u = -2. This results in the decomposition:

Example 2: For one arrives at the equation of 3rd degree

EXAMPLE 3. ( Here a = 4, b = -11, c = -30 and d = 50 Here is the special case 4ab -a3 = 8c ago. )

Example 4:

Here, calculate the values ​​p = - ( x1 x2), q = x1x2, s = - ( x3 x4 ) and t = x3x4 on the zeros:

Other special forms

B = 0 and D = 0

This most common in school mathematics kind of quartic equations is relatively easy to perform back substitution to a quadratic equation. To do this, substituted with and receives. This can be solved by the quadratic formula solution. This gives the solutions. From the back substitution follows: and by the square root is obtained amounts has to be resolved, and is replaced by: and.

E = 0

In this case, a solution of the equation. Then you can factor, so factor out and get the equation

The solutions of the quartic equation are then 0 and the three solutions of the cubic equation

Real coefficients

If all the coefficients are real, to case distinctions for the possible solutions can specify. This is based on the following fact: If the non-real number with zeros of any polynomial with real coefficients, so it is the complex conjugate of ( proof). In the decomposition of the corresponding polynomial, the product of the two factors results in

So every polynomial with real coefficients can be regardless of his degree in linear and quadratic factors with real coefficients decompose. There are for quartic equation thus three possibilities:

  • The equation has four real solutions. It is divided into four linear factors with real coefficients.
  • The equation has two real and two complex conjugate solutions. It is divided into two linear factors and a quadratic factor with real coefficients.
  • The equation has two pairs of complex conjugate solutions. It is divided into two quadratic factors with real coefficients.

Four real solutions

Among the solutions simple solutions or those having a multiplicity of 2, 3 or 4 can be. (Note ).

Specifically, there are these options:

  • A solution with multiplicity 4
  • A solution with multiplicity 3 and a simple solution
  • Two solutions, each with multiplicity 2
  • A solution with multiplicity 2 and two simple solutions
  • Four simple solutions

Two real and two complex conjugate solutions

Here, too, may occur with multiplicity 2, the real solution. So there are these two options:

  • A real solution with multiplicity 2 and two complex conjugate solutions
  • Two simple real roots and two complex conjugate solutions

Two pairs of complex conjugate solutions

There are these two options:

  • Two complex conjugate solutions with multiplicity 2
  • Two pairs of simple complex conjugate solutions

Compact formulation for real-valued coefficients

For the case of real coefficients of the equation can be solved as follows. Given a quartic equation

We convert it into the reduced equation

With real coefficients, and. In the case of this equation is biquadratic and thus easy to solve. We are interested in the general case. From the solutions of the reduced equation we obtained by back- substitution, the solution of the original equation. By means of the coefficients of the reduced equation, we form the so-called cubic resolvent

The solutions of the quartic equation depend follows along with the solutions of the cubic resolvent:

The solutions of the cubic resolvent are,, . Then we get the solutions of the reduced equation by

Which is to be chosen so that

By back substitution

We obtain the solutions of the original equation of the fourth degree.

Characteristic (algebra) Change of variables Abel–Ruffini theorem Descartes' rule of signs
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