symmetric quartic equation

symmetric quartic equation

0.1

Symmetric quartic

Besides the

biquadratic equation

, there are other of quartic equations

a

0

z

4

+

a

1

z

3

+

a

2

z

2

+

a

3

z

+

a

4

=

0

(

a

0

≠

0

)

,

(1)

which can be reduced to

quadratic equations

.  If the

left hand side

of (1) is

P

⁢

(

z

)

, one may write the

identity

P

⁢

(

z

)

z

2

=

(

a

0

⁢

z

2

+

a

4

z

2

)

+

(

a

1

⁢

z

+

a

3

z

)

+

a

2

.

(2)

If we assume first that

a

4

=

a

0

and

a

3

=

a

1

,  the identity is

P

⁢

(

z

)

z

2

=

a

0

⁢

(

z

2

+

1

z

2

)

+

a

1

⁢

(

z

+

1

z

)

+

a

2

.

We set

z

+

1

z

:=

x

,  whence

z

2

+

1

z

2

=

x

2

-

2

;  hence the identity is simplified to

P

⁢

(

z

)

z

2

=

a

0

⁢

x

2

+

a

1

⁢

x

+

a

2

-

2

⁢

a

0

.

Accordingly, we obtain the roots (

http://planetmath.org/Equation

) of the so-called

symmetric quartic equation

a

0

z

4

+

a

1

z

3

+

a

2

z

2

+

a

1

z

+

a

0

=

0

(

a

0

≠

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

-

2

⁢

a

0

=

0

and then solve the equations

z

+

1

z

=

x

1

and

z

+

1

z

=

x

2

which can be written

z

2

-

x

1

⁢

z

+

1

=

0

,

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

⁢

(

1

z

)

=

P

⁢

(

z

)

.

Example.

The equation

2

⁢

z

4

-

5

⁢

z

3

+

4

⁢

z

2

-

5

⁢

z

+

2

=

0

is symmetric.  Thus we solve first

2

⁢

x

2

-

5

⁢

x

+

4

-

2

⋅

2

=

0

,

which yields

x

1

=

0

,

x

2

=

5

2

.  Secondly we solve

z

2

+

1

=

0

,

z

2

-

5

2

⁢

z

+

1

=

0

which yield all the four roots

z

=

±

i

,

z

=

1

2

,

z

=

2

of the quartic.

0.2

Almost symmetric quartic

There is still the quartic equation

a

0

z

4

+

a

1

z

3

+

a

2

z

2

-

a

1

z

+

a

0

=

0

(

a

0

≠

0

)

,

(5)

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

P

⁢

(

z

)

z

2

=

a

0

⁢

(

z

2

+

1

z

2

)

+

a

1

⁢

(

z

-

1

z

)

+

a

2

.

The substitution

z

-

1

z

:=

x

converts it to

P

⁢

(

z

)

z

2

=

a

0

⁢

x

2

+

a

1

⁢

x

+

a

2

+

2

⁢

a

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

+

2

⁢

a

0

=

0

,

then the roots

z

of

z

-

1

z

=

x

1

and

z

-

1

z

=

x

2

which may written

z

2

-

x

1

⁢

z

-

1

=

0

,

z

2

-

x

2

⁢

z

-

1

=

0

.

Hence one infers, that if

z

is a root of (5), so is also its opposite

inverse

-

1

z

; this is apparent also due to the identity

z

4

⁢

P

⁢

(

-

1

z

)

=

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

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