symmetric quartic equation

0.1 Symmetric quartic

Besides the biquadratic equation, there are other of quartic equations

\displaystyle a_{0}z^{4}+a_{1}z^{3}+a_{2}z^{2}+a_{3}z+a_{4}\;=\;0\qquad(a_{0}% \;\neq\;0),

(1)

which can be reduced to quadratic equations. If the left hand side of (1) is $P(z)$ , one may write the identity

\displaystyle\frac{P(z)}{z^{2}}\;=\;\left(a_{0}z^{2}+\frac{a_{4}}{z^{2}}\right% )+\left(a_{1}z+\frac{a_{3}}{z}\right)+a_{2}.

(2)

If we assume first that $a_{4}=a_{0}$ and $a_{3}=a_{1}$ , the identity is

\frac{P(z)}{z^{2}}\;=\;a_{0}\left(z^{2}+\frac{1}{z^{2}}\right)+a_{1}\left(z+% \frac{1}{z}\right)+a_{2}.

We set $\displaystyle z+\frac{1}{z}:=x$ , whence $\displaystyle z^{2}+\frac{1}{z^{2}}=x^{2}-2$ ; hence the identity is simplified to

\frac{P(z)}{z^{2}}\;=\;a_{0}x^{2}+a_{1}x+a_{2}-2a_{0}.

Accordingly, we obtain the roots (http://planetmath.org/Equation) of the so-called symmetric quartic equation

\displaystyle a_{0}z^{4}+a_{1}z^{3}+a_{2}z^{2}+a_{1}z+a_{0}\;=\;0\qquad(a_{0}% \;\neq\;0),

(3)

if we first determine the roots $x_{1}$ and $x_{2}$ of the quadratic

a_{0}x^{2}+a_{1}x+a_{2}-2a_{0}\;=\;0

and then solve the equations $z+\frac{1}{z}=x_{1}$ and $z+\frac{1}{z}=x_{2}$ which can be written

\displaystyle z^{2}-x_{1}z+1\;=\;0,\quad z^{2}-x_{2}z+1\;=\;0.

(4)

Note, that the roots of either equations (4) are inverse numbers of each other (see properties of quadratic equations). Therefore, as well the inverse number of any root of the symmetric quartic (3) is a root of (3); this fact is, by the way, clear also because of the identity

z^{4}P\!\left(\frac{1}{z}\right)\;=\;P(z).

Example. The equation

2z^{4}-5z^{3}+4z^{2}-5z+2\;=\;0

is symmetric. Thus we solve first

2x^{2}-5x+4-2\cdot 2\;=\;0,

which yields $x_{1}=0$ , $x_{2}=\frac{5}{2}$ . Secondly we solve

z^{2}+1\;=\;0,\quad z^{2}-\frac{5}{2}z+1\;=\;0

which yield all the four roots $z=\pm i$ , $z=\frac{1}{2}$ , $z=2$ of the quartic.

0.2 Almost symmetric quartic

There is still the quartic equation

\displaystyle a_{0}z^{4}+a_{1}z^{3}+a_{2}z^{2}-a_{1}z+a_{0}\;=\;0\qquad(a_{0}% \neq 0),

(5)

which reduces to quadratics — the identity (2) for it reads

\frac{P(z)}{z^{2}}\;=\;a_{0}\left(z^{2}+\frac{1}{z^{2}}\right)+a_{1}\left(z-% \frac{1}{z}\right)+a_{2}.

The substitution $\displaystyle z-\frac{1}{z}:=x$ converts it to

\frac{P(z)}{z^{2}}\;=\;a_{0}x^{2}+a_{1}x+a_{2}+2a_{0}.

Thus, (5) can be solved by determining first the roots $x_{1}$ and $x_{2}$ of

a_{0}x^{2}+a_{1}x+a_{2}+2a_{0}\;=\;0,

then the roots $z$ of $z-\frac{1}{z}=x_{1}$ and $z-\frac{1}{z}=x_{2}$ which may written

z^{2}-x_{1}z-1\;=\;0,\quad z^{2}-x_{2}z-1\;=\;0.

Hence one infers, that if $z$ is a root of (5), so is also its opposite inverse $-\frac{1}{z}$ ; this is apparent also due to the identity

z^{4}P\!\left(-\frac{1}{z}\right)\;=\;P(z).

References

1 Ernst Lindelöf: Johdatus korkeampaan analyysiin. Fourth edition. Werner Söderström Osakeyhtiö, Porvoo ja Helsinki (1956).
2 E. Lindelöf: Einführung in die höhere Analysis. Nach der ersten schwedischen und zweiten finnischen Auflage auf deutsch herausgegeben von E. Ullrich. Teubner, Leipzig (1934).

Title	symmetric quartic equation
Canonical name	SymmetricQuarticEquation
Date of creation	2013-03-22 18:05:28
Last modified on	2013-03-22 18:05:28
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	16
Author	pahio (2872)
Entry type	Topic
Classification	msc 30-00
Classification	msc 12D99
Synonym	symmetric quartic
Related topic	AlgebraicEquation
Related topic	EulersDerivationOfTheQuarticFormula
Related topic	ErnstLindelof