quartic equation

A quartic or 4th degree polynomial equation, also traditionally called a biquadratic equation , has the form

${\displaystyle Ax^{4}+Bx^{3}+Cx^{2}+Dx+E=0}$

with coefficients and from a field with characteristic , where then comes from the -algebra . ${\displaystyle A,B,C,D,E}$ ${\displaystyle A\neq 0}$ ${\displaystyle K}$ ${\displaystyle 0}$ ${\displaystyle x}$ ${\displaystyle K}$

In the following, only the real or the complex numbers are considered as fields.

According to the fundamental theorem of algebra , the equation can be uniquely written in the form, except for the order

${\displaystyle A(x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4})=0}$

bring, where and are the not necessarily different four complex solutions of the equation. ${\displaystyle x_{1},x_{2},x_{3}}$ ${\displaystyle x_{4}}$

If and , then the equation can be reduced to a quadratic equation by substitution . Nowadays, especially in school mathematics, it is customary to call just this special form a biquadratic equation , although traditionally biquadratic has a more general meaning. ${\displaystyle B=0}$ ${\displaystyle D=0}$

story

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

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 carried out must be reversed at the end of the solutions in reverse order.

Requirement: Given is a quartic equation with and . ${\displaystyle Ax^{4}+Bx^{3}+Cx^{2}+Dx+E=0}$ ${\displaystyle A,B,C,D,E,x\in \mathbb {C} }$ ${\displaystyle A\neq 0}$

Statement: Then one can state their solutions in an algebraic way as follows:

Normalize and Reduce

First the equation with the substitution

${\displaystyle x=u-{\frac {B}{4A}}}$

simplified in such a way that the cubic coefficient vanishes ( Tschirnhaus transform ) and at the same time the leading coefficient is set by dividing the entire equation by to . ${\displaystyle B}$ ${\displaystyle A}$ ${\displaystyle 1}$

With the definitions

${\displaystyle {\begin{array}{rl}\alpha &=-{\dfrac {3B^{2}}{8A^{2}}}+{\dfrac {C}{A}}\\[1em]\beta &={\dfrac {B^{3}}{8A^{3}}}-{\dfrac {B\,C}{2A^{2}}}+{\dfrac {D}{A}}\\[1em]\gamma &=-{\dfrac {3B^{4}}{256A^{4}}}+{\dfrac {B^{2}\,C}{16A^{3}}}-{\dfrac {B\,D}{4A^{2}}}+{\dfrac {E}{A}}\end{array}}}$

the equation reduces to

${\displaystyle u^{4}+\alpha u^{2}+\beta u+\gamma =0}$.

At the end of the calculation, the roots of the output polynomial are recovered as . In the following it can be assumed that the third degree coefficient is zero. ${\displaystyle x_{1,2,3,4}=u_{1,2,3,4}-{\tfrac {1}{4}}\,{\tfrac {B}{A}}}$

case that only even exponents occur

If , then one obtains the special case of a (genuine) biquadratic equation ${\displaystyle \beta =0}$

${\displaystyle u^{4}+\alpha u^{2}+\gamma =0}$

and can use the zeros as square roots in both sign variants from the solutions of the quadratic equation obtained by the substitution ${\displaystyle u^{2}=m}$

${\displaystyle m^{2}+\alpha m+\gamma =0}$

determine.

If the coefficients are real and , then it makes more sense not to directly determine the then complex solutions of the quadratic equation in and to derive the square roots from them, but to first factor the equation in a real way in a different way, with the two quadratic factors again having real coefficients: ${\displaystyle \gamma >{\tfrac {\alpha ^{2}}{4}}\geq 0}$ ${\displaystyle m}$

${\displaystyle {\begin{array}{rl}u^{4}+\alpha u^{2}+\gamma &=\left[(u^{2}+{\sqrt {\gamma }})^{2}\right]-\left[(2{\sqrt {\gamma }}-\alpha )\,u^{2}\right]\\&=\left(\left[u^{2}+{\sqrt {\gamma }}\right]+\left[{\sqrt {2{\sqrt {\gamma }}-\alpha }}\,u\right]\right)\;\cdot \;\left(\left[u^{2}+{\sqrt {\gamma }}\right]-\left[{\sqrt {2{\sqrt {\gamma }}-\alpha }}\,u\right]\right)\\&=\left(u^{2}+{\sqrt {2{\sqrt {\gamma }}-\alpha }}\cdot u+{\sqrt {\gamma }}\right)\;\cdot \;\left(u^{2}-{\sqrt {2{\sqrt {\gamma }}-\alpha }}\cdot u+{\sqrt {\gamma }}\right)\end{array}}}$

The zeros can now be determined individually for each factor:

${\displaystyle u_{1}=-{\frac {\sqrt {2{\sqrt {\gamma }}-\alpha }}{2}}+{\sqrt {{\frac {2{\sqrt {\gamma }}-\alpha }{4}}-{\sqrt {\gamma }}}}}$	${\displaystyle u_{2}=-{\frac {\sqrt {2{\sqrt {\gamma }}-\alpha }}{2}}-{\sqrt {{\frac {2{\sqrt {\gamma }}-\alpha }{4}}-{\sqrt {\gamma }}}}}$
${\displaystyle u_{3}={\frac {\sqrt {2{\sqrt {\gamma }}-\alpha }}{2}}+{\sqrt {{\frac {2{\sqrt {\gamma }}-\alpha }{4}}-{\sqrt {\gamma }}}}}$	${\displaystyle u_{4}={\frac {\sqrt {2{\sqrt {\gamma }}-\alpha }}{2}}-{\sqrt {{\frac {2{\sqrt {\gamma }}-\alpha }{4}}-{\sqrt {\gamma }}}}}$

general case

If , one tries to write the equation as the difference of two complete squares. Complex parameters are introduced. The representation as a difference then leads directly to a factorization into quadratic factors with complex coefficients: ${\displaystyle \beta \neq 0}$ ${\displaystyle w,y,z}$

${\displaystyle {\begin{array}{rl}u^{4}+\alpha u^{2}+\beta u+\gamma &=(u^{2}+\alpha +y)^{2}-(w\,u-z)^{2}\\&=(u^{2}+w\,u-z+\alpha +y)\,(u^{2}-w\,u+z+\alpha +y)\end{array}}}$

By comparison with

${\displaystyle u^{4}+\alpha \,u^{2}+\beta \,u+\gamma =(u^{2}+\alpha +y)^{2}-[(\alpha +2y)\,u^{2}-\beta \,u+((\alpha +y)^{2}-\gamma )]}$

arise and as well as . ${\displaystyle w^{2}=\alpha +2y}$ ${\displaystyle z^{2}=(\alpha +y)^{2}-\gamma }$ ${\displaystyle \beta =2wz}$

In order for the second part of the difference to be a complete square in , the discriminant of this quadratic term must vanish: ${\displaystyle u}$

${\displaystyle 0=4w^{2}z^{2}-\beta ^{2}=4\,(\alpha +2y)\,((\alpha +y)^{2}-\gamma )-\beta ^{2}}$

${\displaystyle 0=y^{3}+{\tfrac {5}{2}}\alpha \,y^{2}+(2\alpha ^{2}-\gamma )\,y+{\tfrac {1}{2}}\alpha (\alpha ^{2}-\gamma )-{\tfrac {1}{8}}\beta ^{2}}$

This is a cubic equation in . ${\displaystyle y}$

One of the solutions for gives two quadratic equations in , which lead to a total of four solutions for and then , respectively . ${\displaystyle y}$ ${\displaystyle u}$ ${\displaystyle u}$ ${\displaystyle x}$

summary

Overall, the following calculation steps are carried out:

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

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

${\displaystyle y=-{\frac {5}{6}}\alpha +{\begin{cases}-{\sqrt[{3}]{Q}}&{\text{ falls }}P=0\\U-{\frac {P}{3U}}&{\text{ falls }}P\neq 0\end{cases}}}$With ${\displaystyle U={\sqrt[{3}]{-{\frac {Q}{2}}+{\sqrt {{\frac {Q^{2}}{4}}+{\frac {P^{3}}{27}}}}}}}$

${\displaystyle w={\sqrt {\alpha +2y}}}$

${\displaystyle z={\frac {\beta }{2w}}}$.

Now the zeros can be calculated as follows:

${\displaystyle u_{1,2,3,4}={\frac {1}{2}}\left[s\cdot w+r{\sqrt {w^{2}-4\left(\alpha +y+s\,z\right)}}\right]}$

and in the variable of the original equation

${\displaystyle x_{1,2,3,4}=-{\frac {B}{4A}}+{\frac {1}{2}}\left[s\cdot w+r{\sqrt {-(\alpha +2y)-2\left(\alpha +s\,{\tfrac {\beta }{w}}\right)}}\right]}$.

The parameters specify the sign to choose in the two square roots , all four combinations of and are needed to get the four solutions. ${\displaystyle r,s\in \{-1,1\}}$ ${\displaystyle r}$ ${\displaystyle s}$

Decomposition into quadratic factors

Here the decomposition into a product with two quadratic factors

${\displaystyle x^{4}+ax^{3}+bx^{2}+cx+d=\left(x^{2}+px+q\right)\cdot \left(x^{2}+sx+t\right)}$

traced back to the solution of the cubic equation ${\displaystyle u}$

${\displaystyle u^{3}-2bu^{2}+\left(ac+b^{2}-4d\right)u+c^{2}-abc+a^{2}d=0}$.

(For real coefficients and there is a real one with .) ${\displaystyle a,b,c}$ ${\displaystyle d}$ ${\displaystyle u}$ ${\displaystyle 4u\leq a^{2}}$

With a solution of this equation one calculates directly: ${\displaystyle u}$

${\displaystyle p={\frac {a+{\sqrt {a^{2}-4u}}}{2}}}$(special case see below) ${\displaystyle p={\tfrac {a}{2}}}$

${\displaystyle q={\frac {(b-u)\left({\sqrt {a^{2}-4u}}+a\right)-2c}{2{\sqrt {a^{2}-4u}}}}}$

${\displaystyle s={\frac {a-{\sqrt {a^{2}-4u}}}{2}}}$

${\displaystyle t={\frac {(b-u)\left({\sqrt {a^{2}-4u}}-a\right)+2c}{2{\sqrt {a^{2}-4u}}}}}$

In the special case is the solution ${\displaystyle 4ab-a^{3}=8c}$

${\displaystyle p={\frac {a}{2}}}$

${\displaystyle q={\frac {c}{a}}+{\sqrt {{\frac {c^{2}}{a^{2}}}-d}}}$(If , solve the original equation.) ${\displaystyle a=0}$ ${\displaystyle x^{4}+bx^{2}+d=0}$

${\displaystyle s={\frac {a}{2}}}$

${\displaystyle t={\frac {c}{a}}-{\sqrt {{\frac {c^{2}}{a^{2}}}-d}}}$

Example 1: For one arrives at the 3rd degree equation ${\displaystyle x^{4}-x^{3}-x^{2}+4x-2}$

${\displaystyle u^{3}+2u^{2}+5u+10=0}$.

A solution is . This results in the decomposition: ${\displaystyle u=-2}$

${\displaystyle x^{4}-x^{3}-x^{2}+4x-2=\left(x^{2}+x-1\right)\left(x^{2}-2x+2\right)}$.

Example 2: For one arrives at the 3rd degree equation ${\displaystyle x^{4}-3x^{3}-8x-6}$

${\displaystyle u^{3}+10=0}$. A solution is . This results in the decomposition: ${\displaystyle u=-{\sqrt[{3}]{10}}}$

${\displaystyle x^{4}-3x^{3}-8x-6=\left(x^{2}+px+q\right)\left(x^{2}+sx+t\right)}$With

${\displaystyle p={\frac {3+{\sqrt {9+4{\sqrt[{3}]{10}}}}}{2}}\approx 3{,}598674508}$

${\displaystyle q={\frac {{\sqrt {9+4{\sqrt[{3}]{10}}}}{\sqrt[{3}]{10}}+16+3{\sqrt[{3}]{10}}}{2{\sqrt {9+4{\sqrt[{3}]{10}}}}}}\approx 3{,}753109199}$

${\displaystyle s={\frac {3-{\sqrt {9+4{\sqrt[{3}]{10}}}}}{2}}\approx -0{,}598674508}$

${\displaystyle t={\frac {{\sqrt {9+4{\sqrt[{3}]{10}}}}{\sqrt[{3}]{10}}-16-3{\sqrt[{3}]{10}}}{2{\sqrt {9+4{\sqrt[{3}]{10}}}}}}\approx -1{,}598674508}$

Example 3: . ${\displaystyle x^{4}+4x^{3}-11x^{2}-30x+50=\left(x^{2}+2x-5\right)\left(x^{2}+2x-10\right)}$

Here is and . There is a special case . ${\displaystyle a=4,b=-11,c=-30}$ ${\displaystyle d=50}$ ${\displaystyle 4ab-a^{3}=8c}$

Example 4: ${\displaystyle x^{4}+x^{2}+4=\left(x^{2}-{\sqrt {3}}x+2\right)\left(x^{2}+{\sqrt {3}}x+2\right)}$

Here the values ​​are calculated and via the zeros: ${\displaystyle p=-(x_{1}+x_{2}),\,q=x_{1}x_{2},\,s=-(x_{3}+x_{4})}$ ${\displaystyle t=x_{3}x_{4}}$

${\displaystyle x_{1}=\ \ \ {\frac {1}{2}}{\sqrt {-2+2i{\sqrt {15}}}}=\ \ \ {\frac {1}{2}}{\sqrt {3}}+{\frac {1}{2}}{\sqrt {5}}i}$

${\displaystyle x_{2}=\ \ \ {\frac {1}{2}}{\sqrt {-2-2i{\sqrt {15}}}}=\ \ \ {\frac {1}{2}}{\sqrt {3}}-{\frac {1}{2}}{\sqrt {5}}i}$

${\displaystyle x_{3}=-{\frac {1}{2}}{\sqrt {-2-2i{\sqrt {15}}}}=-{\frac {1}{2}}{\sqrt {3}}+{\frac {1}{2}}{\sqrt {5}}i}$

${\displaystyle x_{4}=-{\frac {1}{2}}{\sqrt {-2+2i{\sqrt {15}}}}=-{\frac {1}{2}}{\sqrt {3}}-{\frac {1}{2}}{\sqrt {5}}i}$

Unusual decompositions of biquadratic equations

For purely biquadratic equations without odd exponents, it is better to use the above equations.

${\displaystyle {\begin{array}{rl}u^{4}+\alpha u^{2}+\gamma &=\left[\left(u^{2}+{\sqrt {\gamma }}\right)^{2}\right]-\left[\left(2{\sqrt {\gamma }}-\alpha \right)\,u^{2}\right]\\&=\left(\left[u^{2}+{\sqrt {\gamma }}\right]+\left[{\sqrt {2{\sqrt {\gamma }}-\alpha }}\,u\right]\right)\,\cdot \,\left(\left[u^{2}+{\sqrt {\gamma }}\right]-\left[{\sqrt {2{\sqrt {\gamma }}-\alpha }}\,u\right]\right)\end{array}}}$

Amazing decompositions result for when a square number is: ${\displaystyle u^{4}+u^{2}+\gamma }$ ${\displaystyle \gamma }$

${\displaystyle u^{4}+u^{2}+1=\left(u^{2}+u+1\right)\left(u^{2}-u+1\right)}$

${\displaystyle u^{4}+u^{2}+4=\left(u^{2}+{\sqrt {3}}u+2\right)\left(u^{2}-{\sqrt {3}}u+2\right)}$(see above)

${\displaystyle u^{4}+u^{2}+9=\left(u^{2}+{\sqrt {5}}u+3\right)\left(u^{2}-{\sqrt {5}}u+3\right)}$

${\displaystyle u^{4}+u^{2}+16=\left(u^{2}+{\sqrt {7}}u+4\right)\left(u^{2}-{\sqrt {7}}u+4\right)}$

and finally the not at all common decompositions with only integer coefficients

${\displaystyle u^{4}+u^{2}+25=\left(u^{2}+3u+5\right)\left(u^{2}-3u+5\right)}$

${\displaystyle u^{4}+u^{2}+169=\left(u^{2}+5u+13\right)\left(u^{2}-5u+13\right)}$

Here forms a Pythagorean triple , where the coefficient does not appear at all. Accordingly, the next such decompositions are also ${\displaystyle ({\sqrt {2{\sqrt {\gamma }}-1}},\,{\sqrt {\gamma }}-1,\,{\sqrt {\gamma }})}$ ${\displaystyle {\sqrt {\gamma }}-1}$

${\displaystyle u^{4}+u^{2}+625=\left(u^{2}+7u+25\right)\left(u^{2}-7u+25\right)}$

${\displaystyle u^{4}+u^{2}+1681=\left(u^{2}+9u+41\right)\left(u^{2}-9u+41\right)}$

${\displaystyle u^{4}+u^{2}+3721=\left(u^{2}+11u+61\right)\left(u^{2}-11u+61\right)}$etc.

Because of the decomposition of , a "Pythagorean triple" can even be defined as a special case, although it does not result in a right-angled triangle, but only two coincident sides of a triangle. ${\displaystyle u^{4}+u^{2}+1}$ ${\displaystyle (1,0,1)}$

Other special forms

B =0 and D =0

This type of quartic equation, which is the most common in school mathematics, can be reduced to a quadratic equation relatively easily by substitution. To do this, one substitutes with and obtains: . This can be solved using the quadratic solution formula. You get the solutions . From the back substitution follows: ${\displaystyle x^{2}=z}$ ${\displaystyle Az^{2}+Cz+E=0}$ ${\displaystyle z_{1},z_{2}}$

${\displaystyle x_{1,2}^{2}=z_{1}}$

${\displaystyle x_{3,4}^{2}=z_{2}}$

These purely quadratic equations each have two solutions:

${\displaystyle x_{1,2}=\pm {\sqrt {z_{1}}}}$

${\displaystyle x_{3,4}=\pm {\sqrt {z_{2}}}}$

E = 0

In this case is a solution to the equation. Then you can factor out the factor and get the equation ${\displaystyle x_{1}=0}$ ${\displaystyle x-x_{1}}$ ${\displaystyle x-0=x}$

${\displaystyle x\,\left(Ax^{3}+Bx^{2}+Cx+D\right)=0}$.

The solutions of the quartic equation are then and the three solutions of the cubic equation ${\displaystyle 0}$

${\displaystyle Ax^{3}+Bx^{2}+Cx+D=0}$.

real coefficients

If all coefficients are real, case distinctions can be made for the possible solutions. This is based on the following fact: If the non-real number with zero is any polynomial with real coefficients, then it is also the conjugate complex number ( proof ). Decomposing the associated polynomial gives the product of the two factors ${\displaystyle a+bi}$ ${\displaystyle b\neq 0}$ ${\displaystyle a-bi}$

${\displaystyle (x-(a+bi))(x-(a-bi))}$

is a quadratic polynomial with real coefficients, namely . So any polynomial with real coefficients, regardless of its degree, can be decomposed into linear and quadratic factors with real coefficients. There are three possibilities for the quartic equation: ${\displaystyle x^{2}-2ax+a^{2}+b^{2}}$

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

Four real solutions

Among the solutions there can be simple solutions or those with a multiplicity or . ( Explanation ). ${\displaystyle 2,3}$ ${\displaystyle 4}$

Specifically, there are these options:

• a multiplicity solution ${\displaystyle 4}$

Example: , disassembled ${\displaystyle 2x^{4}+8x^{3}+12x^{2}+8x+2=0}$ ${\displaystyle 2(x+1)^{4}=0}$

has the fourfold solution . ${\displaystyle x_{1,2,3,4}=-1}$

• a solution with multiplicity and a simple solution ${\displaystyle 3}$

Example: , disassembled ${\displaystyle {\tfrac {1}{2}}x^{4}-3x^{3}+6x^{2}-4x=0}$ ${\displaystyle {\tfrac {1}{2}}(x-2)^{3}\,x=0}$

has the triple solution and the simple solution . ${\displaystyle x_{1,2,3}=2}$ ${\displaystyle x_{4}=0}$

• two solutions, each with multiplicity ${\displaystyle 2}$

Example: , disassembled ${\displaystyle x^{4}+2x^{3}-11x^{2}-12x+36=0}$ ${\displaystyle (x-2)^{2}(x+3)^{2}=0}$

has the double solution and the double solution . ${\displaystyle x_{1,2}=2}$ ${\displaystyle x_{3,4}=-3}$

• one solution with multiplicity and two simple solutions ${\displaystyle 2}$

Example: , disassembled ${\displaystyle 5x^{4}-{\tfrac {15}{2}}x^{3}-{\tfrac {45}{2}}x^{2}-{\tfrac {5}{2}}x+{\tfrac {15}{2}}=0}$ ${\displaystyle 5(x+1)^{2}(x-3)\left(x-{\tfrac {1}{2}}\right)=0}$

has the double solution and the simple solutions . ${\displaystyle x_{1,2}=-1}$ ${\displaystyle x_{3}=3,\,x_{4}={\tfrac {1}{2}}}$

• four simple solutions

Example: , disassembled ${\displaystyle x^{4}+x^{3}-4x^{2}-4x=0}$ ${\displaystyle (x-2)(x+1)(x+2)x=0}$

has the simple solutions . ${\displaystyle x_{1}=2,x_{2}=-1,x_{3}=-2,x_{4}=0}$

Two real and two conjugate complex solutions

Here, too, the real solution can occur with multiples. So there are these two possibilities: ${\displaystyle 2}$

• a real solution with multiplicity and two conjugate complex solutions ${\displaystyle 2}$

Example: , disassembled ${\displaystyle x^{4}-4x^{3}+7x^{2}-6x+2=0}$ ${\displaystyle (x-1)^{2}(x-(1+i))(x-(1-i))=0}$

or with a real quadratic factor ${\displaystyle (x-1)^{2}\left(x^{2}-2x+2\right)=0}$

has the twofold solution and the conjugate complex solutions . ${\displaystyle x_{1,2}=1}$ ${\displaystyle x_{3}=1+i,x_{4}=1-i}$

• two simple real solutions and two conjugate complex solutions

Example: , disassembled ${\displaystyle -x^{4}+3x^{3}-2x^{2}-16x+16=0}$ ${\displaystyle -(x-1)(x+2)(x-(2+2i))(x-(2-2i))=0}$

or with a real quadratic factor ${\displaystyle -(x-1)(x+2)\left(x^{2}-4x+8\right)=0}$

has the simple solutions and the conjugate complex solutions . ${\displaystyle x_{1}=1,\,x_{2}=-2}$ ${\displaystyle x_{3}=2+2i,\,x_{4}=2-2i}$

Two pairs of conjugate complex solutions

There are two options here:

• two conjugate complex solutions with multiplicity ${\displaystyle 2}$

Example: , disassembled ${\displaystyle x^{4}-4x^{3}+14x^{2}-20x+25=0}$ ${\displaystyle (x-(1+2i))^{2}(x-(1-2i))^{2}=0}$

or with two real quadratic factors ${\displaystyle \left(x^{2}-2x+5\right)\left(x^{2}-2x+5\right)=0}$

has the twofold conjugate complex solutions . ${\displaystyle x_{1,2}=1+2i,\,x_{3,4}=1-2i}$

• two pairs of simple conjugate complex solutions

Example: , disassembled ${\displaystyle x^{4}-4x^{3}+{\tfrac {21}{4}}x^{2}-4x+{\tfrac {17}{4}}=0}$ ${\displaystyle (x-i)(x+i)\left(x-\left(2-{\tfrac {1}{2}}i\right)\right)\left(x-\left(2+{\tfrac {1}{2}}i\right)\right)=0}$

or with two real quadratic factors ${\displaystyle \left(x^{2}+1\right)\left(x^{2}-4x+{\tfrac {17}{4}}\right)=0}$

has the conjugate complex solutions and . ${\displaystyle x_{1}=i,\,x_{2}=-i}$ ${\displaystyle x_{3}=2-{\tfrac {1}{2}}i,\,x_{4}=2+{\tfrac {1}{2}}i}$

Compact formulation for real-valued coefficients

In the case of real coefficients, the equation can be solved as follows. An equation of the fourth degree is given

${\displaystyle ax^{4}+bx^{3}+cx^{2}+dx+e=0}$

with real coefficients and . Through the substitution ${\displaystyle a,b,c,d,e}$ ${\displaystyle a\neq 0}$

${\displaystyle y=x+{\frac {b}{4a}}}$

transfer this to the reduced equation

${\displaystyle y^{4}+py^{2}+qy+r=0}$

with real coefficients and . In case this equation is biquadratic and therefore easy to solve. In the general case , the solutions of the original equation are obtained from the solutions of the reduced equation by back substitution. The so-called cubic resolvent is formed using the coefficients of the reduced equation ${\displaystyle p,q}$ ${\displaystyle r}$ ${\displaystyle q=0}$ ${\displaystyle q\neq 0}$

${\displaystyle z^{3}+2pz^{2}+(p^{2}-4r)z-q^{2}=0}$.

The solutions of the fourth degree equation are related to the solutions of the cubic resolvents as follows:

cubic resolvents	Fourth degree equation
all solutions real and positive	four real solutions
all solutions real, one positive and two negative	two pairs of complex conjugate solutions
a positive real solution and two complex solutions conjugate to each other	two real and two conjugate complex solutions

Let the solutions of the cubic resolvents be . For each let be any one of the two complex roots from . Then one obtains the solutions of the reduced equation by ${\displaystyle z_{1},z_{2},z_{3}}$ ${\displaystyle j\in \{1,2,3\}}$ ${\displaystyle {\sqrt {z_{j}}}}$ ${\displaystyle z_{j}}$

${\displaystyle {\begin{array}{lcl}y_{1}&=&\sigma {\tfrac {1}{2}}\left({\sqrt {z_{1}}}+{\sqrt {z_{2}}}-{\sqrt {z_{3}}}\right)\\[5pt]y_{2}&=&\sigma {\tfrac {1}{2}}\left({\sqrt {z_{1}}}-{\sqrt {z_{2}}}+{\sqrt {z_{3}}}\right)\\[5pt]y_{3}&=&\sigma {\tfrac {1}{2}}\left(-{\sqrt {z_{1}}}+{\sqrt {z_{2}}}+{\sqrt {z_{3}}}\right)\\[5pt]y_{4}&=&\sigma {\tfrac {1}{2}}\left(-{\sqrt {z_{1}}}-{\sqrt {z_{2}}}-{\sqrt {z_{3}}}\right),\end{array}}}$

where it is to be chosen that ${\displaystyle \sigma \in \{-1,+1\}}$

${\displaystyle \sigma {\sqrt {z_{1}}}{\sqrt {z_{2}}}{\sqrt {z_{3}}}=q}$.

Through the back substitution

${\displaystyle x_{i}=y_{i}-{\frac {b}{4a}},\,i=1,2,3,4}$

one obtains the solutions of the original fourth-degree equation.

See also

• Linear equation
• Quadratic equation
• cubic equation
• solving equations
• Descartes' rule of signs

itemizations

1. ↑ ^{a b} Bronstein, Semendyaev: Pocket Book of Mathematics. 22nd edition, Verlag Harri Deutsch, Thun 1985, ISBN 3-87144-492-8 .
2. ^ Freely adapted from Ferrari.
3. ↑ Source: Solution formula by Joachim Mohr.
4. ↑ Implementable as
   w = sqrt(a^2 - 4 * u)
p = (a + w)/2
q = ((b - u) * (w + a) - 2 * c)/(2 * w)
s = (a - w)/2
t = ((b - u) * (w - a) + 2 * c)/(2 * w)
5. ↑ Source: kilchb.de .
6. ↑ In this case the diagram is the fourth degree parabola

   ${\displaystyle y=x^{4}+ax^{3}+bx^{2}+cx+d}$

   symmetric to the line with the equation

   ${\displaystyle x=-{\frac {a}{4}}}$.

   The solution is obtained by substitution

   ${\displaystyle x=u-{\frac {a}{4}}}$

   via the elementary solvable equation

   ${\displaystyle u^{4}+bu^{2}+d=0}$.

7. ↑ kilchb.de .

literature

• Jörg Bewersdorff : Algebra for beginners. From Equation Resolution to Galois Theory. 6th edition, Wiesbaden 2019, ISBN 978-3-658-26151-1 , DOI:10.1007/978-3-658-26152-8 , introduction (PDF; 319 kB).
• Johannes Beuriger: To solve the biquadratic equations. 1901 ( digital copy ).
• Heinrich Dörrie : Cubic and biquadratic equations. Munich 1948, doi:10.1515/9783486775990 .
• Ludwig Matthiessen : Fundamentals of ancient and modern algebra of literal equations. Leipzig 1896, doi:10.3931/e-rara-78944 (freely accessible).

web links

• Biquadratic Case, Symmetric Case, General Case (PDF; 65 kB).