Third Fourth Fifth S Antoniou

Solving Third, Fourth and Fifth Degree

Polynomial Equations

Solomon M. Antoniou

SKEMSYS Scientific Knowledge Engineering

and Management Systems

37 Κoliatsou Street, Corinthos 20100, Greece [email protected]

Abstract

This is a report on solution methods of third, fourth and fifth degree polynomial

equations with both real and complex coefficients. The third degree equation is

first converted into the so-called reduced form (second order term absent) and then

the reduced form equation is solved using a number of different methods. The

methods used include among others (apart of the classical Cardano approach) the

use of trigonometric and hyperbolic functions, the Vieta and Tschirnhaus

transformation methods as well. A necessary ingredient of the above methods is

the algebraic way of evaluation of the cubic roots of a complex number. The

fourth degree equation is solved using the Ferrari, Descartes, Euler and Simpson

methods. Finally we comment on the solution methods of the quintic.

S. Antoniou: Cubic, Quartic and Quintic 2

Contents

1. Solutions of the equation az3

2. Solving the equation az3 using the algebraic method

3. Solution of an equation of the third degree

4. Solving the reduced third degree equation using Vieta’s method

5. Solving the reduced third degree equation using trigonometric functions

6. Solving the reduced third degree equation using hyperbolic functions

7. The Tschirnhaus method of solving the reduced third degree equation

8. The Glasser method

9. The method of differential resolvents

10. Fourth degree equations - The Ferrari, Descartes, Euler and Simpson methods

11. The quintic.


Section 1

Solution of the equation az3

1. Solution of the equation az3

1.I. Case I. The three roots (solutions) of the equation az3 , where a is any

complex number, are expressed by the formula

3

k2sini

3

k2cosrz 3

k (1.1)

2,1,0k

where r is the modulus, |a|r and the argument, )aarg(φ , πφπ

of the complex number a.

Example 1. Solve the equation 8z3

Solution. Since )sini(cos88 , ( )8(arg and 8|8|r ), the roots

are given by the formula

3

k2sini

3

k2cos8z 3

k , 2,1,0k

For 0k , 3i12

3i

2

12

3sini

3cos8z 3

0

For 1k , 2)sini(cos23

2sini

3

2cos8z 3

1

For 2k ,

3

sini3

cos23

4sini

3

4cos8z 3

2

3i12

3i

2

12


Example 2. Solve the equation 3i1z3

Solution. Since

3sini

3cos23i1 , the three roots are given by

3

3k2

sini3

3k2

cos2z 3k , 2,1,0k

For 0k ,

9

sini9

cos23

3sini3

3cos2z 330

)20sini20(cos29

sini9

cos2 0033

For 1k ,

9

5sini

9

5cos2

3

32

sini3

32

cos2z 331

)80sini80cos(29

4sini

9

4cos2 0033

For 2k ,

9

11sini

9

11cos2

3

34

sini3

34

cos2z 332

)40sini40(cos29

2sini

9

2cos2 0033

1.II. Case II. Solutions of the equation 1w3 .

Since

0sini0cos1 ( 0φ )

the cubic roots of unit are given by the formula

3

πk2sini

3

πk2coswk , 2,1,0k (1.2)


For 0k we find

10sini0cosw0 (1.3)

For 1k we find

2

3i

2

1

3

π2sini

3

π2cosw1 (1.4)

For 2k we find

2

3i

2

1

3

π4sini

3

π4cosw2 (1.5)

We see immediately that 1w and 2w are complex conjugate each other:

21 ww and 12 ww (1.6)

Since

0)1ww)(1w(01w1w 233

the 1w and 2w are the roots of the equation

01ww2

Therefore we have the known Vieta’s relations

1ww 21 and 1ww 21 (1.7)

On the other hand, since

01ww 12

1 and 01ww 22

2

we find, since 21 w1w and 12 w1w (from Vieta), that

0ww 22

1 and 0ww 12

2

or

22

1 ww and 12

2 ww (1.8)

Therefore denoting by ω any of the numbers 2

3i

2

1 or

2

3i

2

1

we see that the roots of the equation 1w3 are given by

1, ω and 2ω


1.III. Case III. We now consider the formula of the cubic roots of any complex

number a

3

φπk2sini

3

φπk2cosrz 3

k

2,1,0k

written as

3

k2sini

3

k2cos

3sini

3cosrz 3

k (1.9)

by exploiting the formula

)φsiniφ(cos)φsiniφ(cos)φφsin(i)φφcos( 22112121

The formula (1.2) gives us

For 0k

3

φsini

3

φcosrz 3

0 (1.10)

For 1k

103

1 wz3

π2sini

3

π2cos

3

φsini

3

φcosrz

(1.11)

For 2k

203

2 wz3

π4sini

3

π4cos

3

φsini

3

φcosrz

(1.12)

where 1w and 2w are given by (1.4) and (1.5) respectively.

Therefore the roots of the equation az3 are given by

0z , ωz0 and 2

0 ωz (1.13)

where 0z is the primitive root of the equation and

2

3i

2

1,

2

3i

2

1ω


Section 2

Solving the equation az3

using the algebraic method

2. Solving the equation az3 using the algebraic method

According to the previous section, the roots of the equation az3 are given by

0z , ωz0 and 2

0 ωz

where 0z is the primitive root of the equation and

2

3i

2

1,

2

3i

2

1ω

The equation az3 can be solved explicitly as long as we are able to determine

explicitly the argument of the complex number a.

However there are cases where the argument of a cannot be found explicitly and

the number a appears to have cubic roots determined explicitly. Some examples

are listed below.

Example 1. The number i469 has three cubic roots given by

i23 , )i23( and 2)i23(

despite the fact that the argument of the number i469 cannot be determined

explicitly, since equation 9

46xtan ,

x

2 cannot be solved analytically.

Example 2. The number 66i28 has three cubic roots given by

6i2 , )6i2( and 2)6i2(


despite the fact that the argument of the number 66i28 cannot be

determined explicitly, since equation 28

66xtan ,

x

2 cannot be solved

analytically.


i25 , )i25( and 2)i25(

despite the fact that the argument of the number i2257 cannot be


22xtan , 0x

2

cannot be solved

analytically.


i322 , )i322( and 2)i322(

despite the fact that the argument of the number i312234 cannot be


312xtan , 0x

2

cannot be solved

analytically.

The previous Examples imply that we have to find another way of determining the

cubic root of a complex number. That means that we have to determine the real

numbers x and y by solving the equation

yixbia3 (2.1)

for known real numbers a and b.

In this respect, we remind the reader that the square roots of the complex number

bia , are given by the formula

)b(sign2

abai

2

ababia

2222

(2.2)

where )x(sign is the function defined by


0x

0x

0x

if

if

if

1

0

1

)x(sign

(2.3)

The formula (2.2) can easily be established by considering the equation

yixbia

and solving with respect to x and y, after squaring both members and equating

real and imaginary parts. In fact we have

2)yix(bia

which can be written as

ixy2)yx(bia 22

From the above identity, equating real and imaginary parts, we obtain the system

}byx2,ayx{ 22

Solving the above system, we determine x and y in terms of a and b and we

arrive at the formula (2.2).

We shall use essentially the same method for determining the cubic roots of the

complex number bia . However we have to consider a number of cases, since

we cannot find a unique formula like (2.2) in this case (see in this respect

Appendix I).

2.I. Case I. The cubic roots 3 bia when a and b are integers.

Considering the identity

yixbia3 (x and y integers)

and taking the cube of both members, we obtain

3)yix(bia (2.4)

which can be written, after expanding and rearranging

)yyx3(i)yx3x(bia 3223 (2.5)

Equating real and imaginary parts, we obtain from the previous identity the system


}byyx3,ayx3x{ 3223 (2.6)

The above is a third degree homogeneous system, which can be solved in principle

with respect to x and y. However we cannot find a general solution of this system

(Appendix I). There is however a way out of this situation.

Considering the equivalence

yixbiayixbia 33 (2.7)

and multiplying the relations

yixbia3 and yixbia3

we obtain

)yix()yix()bia()bia(3

which is equivalent to

3 2222 bayx (2.8)

The above equation can be considered as a supplement to the system (2.6). After

taking into account equation (2.8), we obtain the system (2.6) and (2.8) which is

over determined and can be solved easily. On the other hand, once 22 ba is a

perfect cube, we can easily infer that the numbers x and y are integers too.

One more remark. If equation (2.8) is written as kyx 22 (k positive integer),

we can guess x and y and then check which values satisfy the system (2.6), thus

leading to a very simple method of solution.

Example 2.1. Find the cubic roots of the expression i469 .

Solution. We let

yixi4693 (2.9)

Taking the cube of both members, we obtain

3)yix(i469 (2.10)

which can be written, after expanding and rearranging


)yyx3(i)yx3x(i469 3223 (2.11)

Equating real and imaginary parts, we obtain from the previous identity the system

}46yyx3,9yx3x{ 3223 (2.12)

The above is a third degree homogeneous system, which can be solved in principle

with respect to x and y. However we cannot find a general solution of this

system.

Considering the equivalence

yixbiayixbia 33 (2.13)

and multiplying the relations

yixi4693 and yixi4693

we obtain

)yix()yix()i469()i469(3


13)46()9(yx3 2222 (2.14)

From the last equation we may infer that

( 4x2 and 9y2 ) or ( 9x2 and 4y2 )

The choice ( 3x and 2y ) satisfies the system (2.12). Therefore one of the

cubic roots of the complex number i469 is i23 .

We thus find that the number i469 has three cubic roots given by

i23 , )i23( and 2)i23( (2.15)

Note. The system of equations (2.12) and (2.14) can also be solved by solving

(2.14) with respect to 2x and substituting into the second equation of (2.12),

obtaining the equation 046y39y4 3 which admits the only integer solution

2y (using say Horner’s algorithm).


Exercise 1. Show that the cubic roots of the number i946 are given by

i32 , )i32( and 2)i32(


i45 , )i45( and 2)i45(


i3 , )i3( and 2)i3(

2.II. Case II. The cubic roots 3

cbia when a, b and c ( 0c ) are integers

and c not square of an integer.

Considering the identity

zyixcbia3

(2.16)

and taking the cube of both members, we find

3)zyix(cbia


z)zyyx3(i)zyx3x(cbia 3223 (2.17)

From the above identity we get the system

cz

bzyyx3

azyx3x32

23

(2.18)

Multiplying the identities

zyixcbia3

and zyixcbia3

we obtain

)zyix()zyix()cbia()cbia(3

which is equivalent to the equation

3 2222 cbazyx (2.19)


As long as cba 22 is a perfect cube, we can solve the system of equations (2.18)

and (2.19).

Example 2.2. Find the cubic roots of the complex number i384405

Solution. Consider the identity

zyixi3844053

(2.20)

Taking the cube of both members of the above identity, we obtain

3)zyix(i384405


z)zyyx3(i)zyx3x(i384405 3223


3z

84zyyx3

405zyx3x32

23

(2.21)


zyixi3844053

and zyixi3844053

we obtain

)zyix()zyix()i384405()i384405(3


57185193384)405(zyx 33 2222 (2.22)

Since 3z , equation (2.22) becomes 57y3x 22 . From the last equation we

can infer that 9x2 and 16y2 . The choice ( 3x and 4y ) satisfies the

first two equations of (2.21). Therefore one of the cubic roots is i343 .

We thus find that the cubic roots of the complex number i384405 are given

by


i343 , )i343( and 2)i343( (2.23)

The cubic roots of the complex conjugate number i384405 are then given

by

i343 , )i343( and 2)i343( (2.24)

Application 2.2. Find the cubic roots of the complex numbers

27

784i15 and

27

784i15 (2.25)

Solution. The above expression can be written as

27

384i405

27

384i15

27

784i15

Therefore

3

384i405

27

784i15

33

Since

3

384i405 i343 , )i343( , 2)i343(

we find that the cubic roots of the expression 27

784i15 are given by

i3

341 ,

i

3

341 and

2i3

341

(2.26)

We also find that the cubic roots of the complex conjugate 27

784i15 are

given by

i3

341 ,

i

3

341 and

2i3

341

(2.27)




zyix66i283

(2.28)


3)zyix(66i28


z)zyyx3(i)zyx3x(66i28 3223


6z

6zyyx3

28zyx3x32

23

(2.29)


zyix66i283

and zyix66i283

we obtain

)zyix()zyix()66i28()66i28(3


106628zyx3 2222 (2.30)

Since 6z , equation (2.30) becomes 10y6x 22 . From the last equation we

can infer that 4x2 and 1y2 . The choice ( 2x and 1y ) satisfies the

first two equations of (2.29). Therefore one of the cubic roots is i62 .

We thus find that the roots of the complex number 66i28 are given by

i62 , )i62( and 2)i62( (2.31)

Application 2.3. Find the cubic roots of the complex numbers

9

62i

27

28 and

9

62i

27

28

Solution. We have


)66i28(27

1

9

62i

27

28 and )66i28(

27

1

9

62i

27

28

Therefore, using the Example 2.3, we conclude that one of the cubic root of

9

62i

27

28 is )i62(

3

1 and one of the cubic root of

9

62i

27

28 is the

number )i62(3

1 .



zyix66i283

(2.28)

Using again the method described earlier, we find that one of the cubic roots of the

number i6628 is 6i2 .

Application 2.4. Suppose that we want to solve the equations

9

62i

27

28m3 and

9

62i

27

28n3 (1)

Since

)66i28(27

1m3 and )66i28(

27

1n3

it suffices to find the cubic roots of the numbers

66i28 and 66i28

respectively.

We have already found that one of the cubic roots of 66i28 is 6i2 and

thus one of the cubic roots of the 66i28 will be 6i2 .

Therefore one of the cubic roots of

)66i28(27

1

9

62i

27

28

will be the number )6i2(3

1 , while one of the cubic roots of


)66i28(27

1

9

62i

27

28

will be the number )6i2(3

1 .



zyixi594763

(2.28)

Using again the method described earlier, we find that one of the cubic roots of the

number i59476 is the number i534 .

We also find that one of the cubic roots of the number i59476 is the

number i534 .

Application 2.5. Suppose that

3

5i

27

476m3 and

3

5i

27

476n3 (1)

Since

)59i476(27

1m3 and )59i476(

27

1n3

One of the cubic roots of the number 59i476 is i534 .

Therefore one of the cubic roots of the number 3

5i

27

476 is )i534(

3

1 .

One of the cubic roots of the number 3

5i

27

476 is )i534(

3

1 .



zyixi527810241753


Using the method described previously we find thet one of the cubic roots of the

number i52781024175 is the number i5185 . We also find that one of

the cubic roots of the complex number i52781024175 is the number

i5185 .


108

5515i

5832

24175m3 and

108

5515i

5832

24175n3

Since

)527810i24175(5832

1m3 and )527810i24175(

5832

1n3

One of the cubic roots of the number 527810i24175 is i5185 .

Therefore one of the cubic roots of the number 108

5515i

5832

24175 is the number

)i5185(18

1 and one of the cubic roots of the number

108

5515i

5832

24175 is

the number )i5185(18

1 .

Exercises.

Exercise 1. Show that the cubic roots of the complex number i23845 are

given by i223 , )i223( and 2)i223( .


given by i332 , )i332( and 2)i332( .


given by i323 , )i323( and 2)i323( .



given by i534 , )i534( and 2)i534( .

Exercise 5. Show that the cubic roots of the complex number

i52781024175 are given by i5185 , )i5185( and

2)i5185( .


given by i53 , )i53( and 2)i53( .

2.III. Case III. The cubic roots 3

ciba when a, b ( 0b ) and c are

integers and b not a square of an integer.

Consider the identity

ziyxciba3

(2.23)


3)ziyx(ciba


)zyzx3(iy)xz3yx(ciba 3223 (2.24)

From the above identity we obtain the system

{ azx3yx 23 , by , czyzx3 32 } (2.25)


ziyxciba3

and ziyxciba3

we obtain

)ziyx()ziyx()ciba()ciba(3


3 2222 cbazyx (2.26)


Solving the system (2.25) and (2.26) we determine the unknown quantities x, y

and z.

Example 2.7. Find the cubic roots of the complex number i2257 .


ziyxi22573

(2.27)


3)ziyx(i2257


)zyzx3(iy)xz3yx(i2257 3223 (2.28)


{ 7zx3yx 23 , 5y , 22zyzx3 32 } (2.29)


ziyxi22573

and ziyxi22573

we obtain

)ziyx()ziyx()i2257()i2257(3


9)22()57(zyx3 2222 (2.30)

Since 5y , equation (2.30) becomes 9zx5 22 from which we can infer that

1x2 and 4z2 . The choice ( 1x and 2z ) satisfies the system (2.29).

Therefore one of the cubic roots of the number i2257 is i25 .

We thus find that the three cubic roots of the number i2257 are given by

i25 , )i25( and 2)i25(


Exercises.


given by i332 , )i332( , 2)i332(


given by i452 , )i452( , 2)i452(

2.IV. Case IV. The cubic roots 3

dciba when a, b ( 0b ), c and d

( 0d ) are integers and b and d not a square of an integer.

Consider the identity

wziyxdciba3

(2.31)

Taking the cube of both members of the previous identity we find

3)wziyx(dciba


w)yzx3wz(iy)wxz3yx(dciba 2323 (2.32)


}dw,cyzx3wz,by,awxz3yx{ 2323 (2.33)

Multiplying

wziyxdciba3

and wziyxdciba3

we obtain

)wziyx)(wziyx()dciba)(dciba(3


3 2222 dcbawzyx (2.34)


Solving the system of equations (2.33) and (2.34) we can determine the unknown

quantities x, y, z and w and thus determine one of the cubic roots of the

complex number.



wziyx312i2343

(2.35)

Taking the cube of both members of the previous identity we find

3)wziyx(312i234


w)yzx3wz(iy)wxz3yx(312i234 2323 (2.36)


}3w,12yzx3wz,2y,34wxz3yx{ 2323 (2.37)

Multiplying

wziyx312i2343

and wziyx312i2343

we obtain

)wziyx()wziyx()312i234)(312i234(3


14)312()234(wzyx3 2222 (2.38)

Since 2y and 3w , the previous equation becomes 14z3x2 22 , from

which we can infer that 1x2 and 4z2 . The choice 1x and 2z satisfies

the equations of the system (2.37). Therefore one of the cubic roots is i322 .

We thus find that the three cubic roots of the complex number i312234 are

given by

i322 , )i322( and 2)i322(


Exercises.

Exercise 1. Show that the cubic roots of the complex number i254384

are given by i2332 , )i2332( , 2)i2332(

Exercise 2. Show that the cubic roots of the complex number i3995122

are given by i3352 , )i3352( , 2)i3352(

Case V. The cubic roots 3

)i1()cba(

Example 2.9. Find the cubic roots of the complex number )i1()31220(

Solution. This number can be written as iaa where 031220a

Since

4sini

4cos)533(24

4sini

4cosa2iaa 2

the primary cubic root of iaa is given by

12sini

12cos)533(24z

30

22

13i

22

13)26(

12sini

12cos)26(

i)32(1

Second Method. We let

i)zuw()zyx()i1()31220(3

Raising to the third power both members of the previous equality, we obtain

3}i)zuw()zyx{()i1()31220(


i)31220()31220(

z)uwy2uxyx(3)wx3x( 2223


z)}uwx2wyyx(3)uzy3yz{( 2223

)}wxw3()uwuyx2yw(z3{i 3222

z)}wyx2uwxu(3)uzyzu3{(i 2232

The previous equation is valid only if

z)uwy2uxyx(3)wx3x(20 2223

)uwx2wyyx(3)uzy3yz(12 2223

3z

)wxw3()uwuyx2yw(z320 3222

)wyx2uwxu(3)uzyzu3(12 2232

The above system is equivalent to

20)uwy2uxyx(9)wx3x( 2223

4)uwx2wyyx()uy3y( 2223

20)wxw3()uwuyx2yw(9 3222

4)wyx2uwxu()uyu3( 2232

3z


i)zuw()zyx()i1()31220(3

and

i)zuw()zyx()i1()31220(3

we obtain

223 2 )zuw()zyx()11()31220(


zwu2zuwzxy2zyx39601664 22223

and this to


z)wuxy(2z)uy(wx348 2222

From the previous equation we obtain the system

3z

8)uy(3wx 2222

2uwyx

From the above system we can infer that

5wx 22 and 1uy 22

and then }4w,1x{ 22 or }1w,4x{ 22 and }1u,0y{ 22 or

}0u,1y{ 22 . The values

( 1x , 0y , 2w , 1u , 3z )

satisfy the system found before (equations (), ). Therefore one of the cubic roots

of the expression )i1()31220( is given by i)32(1 .


)i1(33

5

9

4m3

and )i1(3

3

5

9

4n3

(8)

From the two previous relations we can determine m and n by finding the cubic

roots of the numbers )i1(33

5

9

4

and )i1(33

5

9

4

respectively.

We have

)i1()31220(27

1)i1(3

3

5

9

4

and

)i1()31220(27

1)i1(3

3

5

9

4

One of the cubic roots of the number )i1()31220( is i)32(1

We choose the values


i3

32

3

1m

and i

3

32

3

1n

(9)

that satisfy the relation i9

4nm , which is the first equation of the system (5).

2.VI. Case VI. The cubic roots 3

cba where a, b and c ( 0c ) are

integers, and c is not a square of any integer.

We consider the identity

zyxcba3

(1)

Taking the cube of the previous identity, we find

3)zyx(cba


z)zyyx3()zyx3x(cba 3223 (2)


}cz,bzyyx3,azyx3x{ 3223 (3)

On the other hand, multiplying the identities

zyxcba3

and zyxcba3

we find

)zyx()zyx()cba()cba(3


3 2222 cbazyx (4)

The system of equations (3) and (4) determines completely the quantities x, y and

z.

Example 2.10. Calculate the cubic roots of the number 339156344

Solution. We consider the identity

zyx3391563443

(1)



3)zyx(339156344


z)zyyx3()zyx3x(339156344 3223 (2)


}3z,3915zyyx3,6344zyx3x{ 3223 (3)

On the other hand, multiplying the identities

zyx3391563443

and zyx3391563443

we find

)zyx()zyx()339156344()339156344(3


17933915)6344(zyx3 2222 (4)

Since 3z , the previous equation can be written as 179y3x 22 , from

which there follows that 64x 2 and 81y2 . The choice )9y,8x(

satisfies the system (3). Therefore one of the cubic roots of the number

339156344 is the number 398 .


27

3145

729

6344m3 and

27

3145

729

6344n3 (8)

Since

)339156344(729

1m3 and )339156344(

729

1n3

One of the cubic roots of the number 339156344 is 398

39

8m and 3

9

8n (9)


The two previous values of m and n satisfy the relation 81

179nm , which

is the first equation of the system (5).

Exercises.

Exercise 1. Show that the cubic roots of the irrational number 3117170 are

given by 332 , )332( , 2)332(


given by 523 , )523( , 2)523(

2.VII. Case VII. The cubic roots 3

dcba where a, b, c and d

( 0d,0b ) are integers, and b, d are not a square of any integer.

We consider the identity

wzyxdcba3

(1)


3)wzyx(dcba (2)

and expanding the right member,

w)wzyzx3(y)wzx3yx(dcba 3223 (3)


}dw,by,cwzyzx3,awzx3yx{ 3223 (4)


wzyxdcba3

and wzyxdcba3

we obtain

)wzyx()wzyx()dcba()dcba(3


wzyxdcba 223 22 (5)


we may proceed now once we find that dcba 22 is a perfect cube.

We thus have to solve the system of equations (4) and (5) which can be solved

very easily.

Example 2.11. Calculate the cubic roots of the number 220612

Solution. We consider the identity

wzyx2206123

(1)


3)wzyx(220612


w)wzzyx3(y)wzx3yx(220612 3223 (2)


}2w,6y,20wzyzx3,12wzx3yx{ 3223 (3)


wzyx2206123

and wzyx2206123

we obtain

)wzyx()wzyx()220612()220612(3


wzyx)220()612( 223 22

or (taking into account the values 2w,6y from system (3))

223 z2x664 , from which we get

2zx3 22 (4)

From the last equation we obtain 1x2 and 1z2 . The choice 1x and 1z

satisfies the equations of the system (3). Therefore we find 1x , 6y , 1z

and 2w . We thus obtain that 262206123


Exercises.

Exercise 1. Show that the cubic roots of the irrational number 51482234

are given by 5223 , )5223( , 2)5223(


given by 232 , )232( , 2)232(

Section 3

Solution of the third degree equation

3. Solution of the third degree equation

An equation of the third degree

0dxcxbxa 23 ( 0a )

is first transformed into an equation of the form

0qypy3

the so-called reduced form equation, which does not contain the second order

term. This is done by a linear transformation of the form kyx under an

appropriate choice of the constant k. The reduced equation is then solved using a

number of different methods.

3.1. The general form of a third degree equation.

The general form of an equation of the third degree is given by

0dxcxbxa 23 (3.1)

where the coefficients a, b, c, d are in general complex numbers and 0a .

We can, using a suitable transformation, to reduce equation (3.1) into an equation

of third degree, not containing the second order term.


3.2. Transforming the third degree equation

We try a transformation of the form

kyx (3.2)

where k is a parameter to be determined. Equation (3.1) is then transformed to

0d)ky(c)ky(b)ky(a 23

which is equivalent to (after expanding the binomials)

0d)ky(c)kky2y(b)kky3ky3y(a 223223

The previous equation can be written as

y)ckb2ka3(y)bka3(ya 223

0)dkckbka( 23 (3.3)

We have now the choice of determining k such that the coefficient of 2y to be

zero. In other words we put 0bka3 from which we get

a3

bk (3.4)

This is the choice of the parameter k. Under this choice of k, the coefficients of

equation (3.3) take the form

0bka3 (3.5)

c

a3

bb2

a3

ba3ckb2ka3

22

ca3

b2

a9

ba3

2

2

2

ca3

bc

a3

b2

a3

b 222

(3.6)

d

a3

bc

a3

bb

a3

badkckbka

2323


d

a3

bc

a9

bb

a27

ba

2

2

3

3

da3

bc

a9

b

a27

b

2

3

2

3

da3

bc

a27

b2

2

3

(3.7)

Therefore equation (1.3) becomes

0da3

bc

a27

b2yc

a3

bya

2

323

(3.8)

or

0da3

bc

a27

b2

a

1yc

a3

b

a

1y

2

323

or

0qypy3 (3.9)

where

a

c

a3

bp

2

2

(3.10)

a

d

a3

bc

a27

b2q

23

3

(3.11)

Conclusion. The transformation

a3

byx (3.12)

converts the equation

0dxcxbxa 23

into the equation

0qypy3 .


where

a

c

a3

bp

2

2

and a

d

a3

bc

a27

b2q

23

3

3.3. Solving the reduced third degree equation

We have now to find a method of solving third degree equations of the form

0qypy3 .

Description of the Method. We determine two parameters m and n that satisfy

the equation

3333 nmynm3yqypy (3.13)

This means that m and n should satisfy the system of equations

33 nmq

nm3p (3.14)

Taking the cube of both members of the first equation of the above system, we

obtain

333 nm27p

or

27

pnm

333

This equation, combined with the second equation of the system (3.14), results in

the system

qnm 33 and 27

pnm

333 (3.15)

Therefore the numbers 3m and 3n will be the roots of the quadratic equation

027

ptqt

32 (3.16)

Let D be the discriminant of the equation (3.16). We then have


27

p4q

27

p4qD

32

32

or

27

p

4

q4D

32

(3.17)

Let also d be one of the square roots of D:

2)d2(D

Note that

27

p

4

qd

322

Therefore 3m and 3n will be one of the roots

27

p

4

q

2

q

2

d2q 32

of equation (3.16).

Let

27

p

4

q

2

qm

323 and

27

p

4

q

2

qn

323

We have now to determine m and n by finding the cubic roots of the numbers

27

p

4

q

2

q 32

and 27

p

4

q

2

q 32

respectively. We have formally

3

32

027

p

4

q

2

qm and

332

027

p

4

q

2

qn (3.18)

where 0m and 0n are selected so that to satisfy the equation

3

pnm 00 (3.19)

which is the first equation of the system (3.14).


Since

ynm3)nmy(nmynm3y 333333

)nmnymynmy()nmy( 222 (3.20)

(this is Euler’s identity), equation (3.9), taking into account (3.13) and (3.20),

becomes equivalent to

0)nmnymynmy()nmy( 222 (3.21)

from which we get two equations

0nmy (3.22)

0nmnymynmy 222 (3.23)

Equation (3.22) is a first degree equation with solution

001 nmy (3.24)

while equation (3.23) is a second degree equation, since it can be written as

0)nmnm(y)nm(y 222 (3.25)

The discriminant of the above equation is calculated to be

)nmnm(4)nm( 222

nm4n4m4nnm2m 2222

22 n3nm6m3

)nnm2m(3 22

2)nm(3

One of the square roots of is thus

)nm(3i 00

Therefore the roots of the quadratic equation (3.25) will be

2

)nm(3i)nm( 0000

or


2

)nm(3i)nm(y 0000

2

(3.26)

2

)nm(3i)nm(y 0000

3

(3.27)

Notice that (3.26) and (3.27) can also be written as

02

02 nmy (3.28)

and

002

3 nmy (3.29)

where

2

3i

2

1,

2

3i

2

1ω

Conclusion. The roots of the equation

0qypy3

are given by

001 nmy

2

)nm(3i)nm(y 0000

2

2

)nm(3i)nm(y 0000

3

or

001 nmy

02

02 nmy

002

3 nmy

where 0m and 0n are evaluated by

3

32

027

p

4

q

2

qm and

332

027

p

4

q

2

qn


respectively, subject to

3

pnm 00

and

2

3i

2

1,

2

3i

2

1ω .

Discussion. It is obvious from the equations determining 0m and 0n that the

type of roots depends on the sign of the quantity 27

p

4

qD

32

.

Let us examine in detail the three cases:

I) If 0D , then

027

p

4

q

2

q 32

and 027

p

4

q

2

q 32

In either case 0m and 0n are to be real ( 0m0 and 0n0 ). Therefore 1y is

real while 2y and 3y are complex conjugate.

II) If 0D (the so-called irreducible case) then 0m and 0n are complex

conjugate. In this case 00 nm is real, while 00 nm is purely imaginary

number. Therefore in this case all the roots are real.

III) If 0D then rnm 00 , where r is the cubic root of the real number 2

q ,

3

2

qr which is to be real. Therefore in this case we have the roots r2y1

and ryy 32 , i.e. three real roots, at least two of them equal.

Note. The quantity 27

p

4

qD

32

is usually called by some authors “discriminant”

of the equation 0qypy3 . This is not however the case. The discriminant of

the equation 0qypy3 is

27

p

4

q108

32

.


3.4. Examples.

Example 1. Solve the equation

060x7x6x 23 (1)

Solution. In this case 1a , 6b , 7c and 60d .

Under the substitution 13

6y

a3

byx

, i.e.

2yx (2)

equation (1) takes on the form

060)2y(7)2y(6)2y( 23


030y19y3 (3)

In this case 19p and 30q . Therefore the discriminant of the equation is

evaluated to be

027

784

27

)19(

4

30

27

p

4

qD

3232

(4)

We now determine 0m and 0n by

3330

27

784i15

27

78415D

2

qm (5)

3330

27

784i15

27

78415D

2

qn (6)

subject to

3

19

3

pnm 00 (7)

The quantities 0m and 0n satisfying (5), (6) and (7) are given by (Application

2.2)

i3

341m0 and i

3

341n0 (8)


Since 2nm 00 and i3

38nm 00 , we can determine the roots of

equation (3). We find

2nmy 001

32

i3

383i2

2

)nm(3i)nm(y 0000

2

52

i3

383i2

2

)nm(3i)nm(y 0000

3

The roots of equation (1) are then given by

413

62

a3

byx 11

513

63

a3

byx 22

313

65

a3

byx 33

Note. Equation (3) was considered by Kurosh (Kurosh [45], Example 3, p.230),

without an explicit solution.


01x3xx 23 (1)

Solution. In this case 1a , 1b , 3c and 1d .


1y

a3

byx

, i.e.

3

1yx (2)


013

1y3

3

1y

3

1y

23



027

56y

3

10y3 (3)

In this case we have 3

10p and

27

56q . Therefore the discriminant of

equation (3) is evaluated to be

27

8

27

3

10

4

27

56

27

p

4

qD

32

32

(4)


3330

9

62i

27

28

27

8

27

28D

2

qm (5)

3330

9

62i

27

28

27

8

27

28D

2

qn (6)

subject to

9

10

3

pnm 00 (7)


2.3)

i3

6

3

2m0 and i

3

6

3

2n0 (8)

Since 3

4nm 00 and i

3



3

4nmy 001

23

2

2

i3

623i

3

4

2

)nm(3i)nm(y 0000

2


23

2

2

i3

623i

3

4

2

)nm(3i)nm(y 0000

3


113

1

3

4

a3

byx 11

2113

12

3

2

a3

byx 22

2113

12

3

2

a3

byx 33

Conclusion.

The roots of the equation 01x3xx 23 are given by

1x1 , 21x2 and 21x3

Note. It is obvious that equation (1) can be solved using Horner’s Algorithm.

However we solved the equation using the theory outlined previously.


042x20xx 23 (1)

Solution. In this case we have 1a , 1b , 20c and 42d .


1y

a3

byx

, i.e.

3

1yx (2)

equation (1) takes the form

0423

1y20

3

1y

3

1y

23


027

952y

3

61y3 (3)



61p and

27

952q . Therefore the discriminant of the


9

5

27

3

61

4

27

952

27

p

4

qD

32

32

(4)


3330

3

5i

27

476

9

5

27

476D

2

qm (5)

3330

3

5i

27

476

9

5

27

476D

2

qn (6)

subject to

9

61

3

pnm 00 (7)


2.5)

i53

4m0 and i5

3

4n0 (8)

Since 3

8nm 00 and i52nm 00 , we can determine the roots of


3

8nmy 001

153

4

2

)i52(3i3

8

2

)nm(3i)nm(y 0000

2

153

4

2

)i52(3i3

8

2

)nm(3i)nm(y 0000

3



313

1

3

8

a3

byx 11

15113

115

3

4

a3

byx 22

15113

115

3

4

a3

byx 33

Conclusion.

The roots of the equation 042x20xx 23 are given by

3x1 , 151x2 and 151x3




059x181x8x12 23 (1)

Solution. In this case we have 12a , 8b , 181c and 59d


8y

a3

byx

, i.e.

9

2yx (2)


0599

2y181

9

2y8

9

2y12

23


02916

24175y

108

1645y3 (3)


1645p and

2916


the equation (3) is evaluated to be


11664

1326125

27

108

1645

4

2916

24175

27

p

4

qD

32

32

(4)


3330

108

5515i

5832

24175

11664

1326125

5832

24175D

2

qm (5)

3330

108

5515i

5832

24175

11664

1326125

5832

24175D

2

qn (6)

subject to

324

1645

3

pnm 00 (7)


2.6)

i518

5m0 and i5

18

5n0 (8)

Since 9

5nm 00 and i52nm 00 , we can determine the roots of


9

5nmy 001

1518

5

2

)i52(3i9

5

2

)nm(3i)nm(y 0000

2

1518

5

2

)i52(3i9

5

2

)nm(3i)nm(y 0000

3

The roots of the original equation (1) are then calculated to be

3

1

9

2

9

5

9

2yx 11


152

1

9

215

18

5

9

2yx 22

152

1

9

215

18

5

9

2yx 33

Conclusion.

The roots of the equation 059x181x8x12 23 are given by

3

1x1 , 15

2

1x2 and 15

2

1x3




03x10x4x8 23 (1)



4y

a3

byx

, i.e.

6

1yx (2)


036

1y10

6

1y4

6

1y8

23


027

16y

3

4y3 (3)


4p and

27



027

3

4

4

27

16

27

p

4

qD

32

32

(4)



3

2

27

8D

2

qm 33

0 (5)

3

2

27

8D

2

qn 33

0 (6)

subject to

9

4

3

pnm 00 (7)

Since 3

4nm 00 and 0nm 00 , we can determine the roots of equation

(3). We find

3

4nmy 001

3

2

2

3

4

2

)nm(3i)nm(y 0000

2

3

2

2

3

4

2

)nm(3i)nm(y 0000

3


2

3

6

1

3

4

6

1yx 11

2

1

6

1

3

2

6

1yx 22

2

1

6

1

3

2

6

1yx 33

Conclusion.


2

3x1 and

2

1xx 32 (double root).





026x31x10x3 23 (1)



10y

a3

byx

, i.e.

9

10yx (2)


0269

10y31

9

10y10

9

10y3

23


0729

12688y

27

179y3 (3)


179p and

729



243

21025

27

27

179

4

729

12688

27

p

4

qD

32

32

(4)


3330

27

3145

729

6344

243

21025

729

6344D

2

qm (5)

3330

27

3145

729

6344

243

21025

729

6344D

2

qn (6)

subject to

81

179

3

pnm 00 (7)



2.10)

39

8m0 and 3

9

8n0 (8)

Since 9

16nm 00 and 32nm 00 , we can determine the roots of


9

16nmy 001

i39

8

2

)32(3i9

16

2

)nm(3i)nm(y 0000

2

i39

8

2

)32(3i9

16

2

)nm(3i)nm(y 0000

3


3

2

9

10

9

16

9

10yx 11

i329

10i3

9

8

9

10yx 22

i329

10i3

9

8

9

10yx 33

Conclusion.


3

2x1 , i32x2 and i32x3 .




0)i2(zz)i2(z 23 (1)


Solution. In this case we have 1a , )i2(b , 1c and )i2(d .


)i2(y

a3

byz

, i.e.

3

i2yz

(2)


0)i2(3

i2y

3

i2y)i2(

3

i2y

23


0)i1(27

40y

3

i4y3 (3)


i4p and )i1(

27


the equation (3) is evaluated to be

i27

32

27

3

i4

4

)i1(27

40

27

p

4

qD

32

32

(4)


3330 )i1(

9

34)i1(

27

20i

27

32)i1(

27

20D

2

qm (5)

3330 )i1(

9

34)i1(

27

20i

27

32)i1(

27

20D

2

qn (6)

subject to

i9

4

3

pnm 00 (7)


2.9)

i3

1

3

32m0

and i

3

1

3

32n0

(8)


Since i3

2

3

4nm 00 and

3



i3

2

3

4nmy 001

i3

2

3

2

2

3

323ii

3

2

3

4

2

)nm(3i)nm(y 0000

2

i3

4

3

2

2

3

323ii

3

2

3

4

2

)nm(3i)nm(y 0000

3


i23

i2i

3

2

3

4

3

i2yz 11

i3

i2i

3

2

3

2

3

i2yz 22

i3

i2i

3

4

3

2

3

i2yz 33

Conclusion.

The roots of the equation 0)i2(zz)i2(z 23 are given by

iz1 , i2z2 and iz3 .

Note. It is obvious that equation (1) can be solved by factorization. However we

solved the equation using the theory outlined previously.

The factorization goes as follows:

)1z()i2()zz()i2(zz)i2(z 2323

)i2z()iz()iz()i2z()1z()1z()i2()1z(z 222

From the above factorization you can read off the three roots:

iz1 , i2z2 and iz3


3.5. Another simple method of solving the reduced third degree

equation

We consider the reduced third degree equation

0qypy3 (1)

Under the substitution

zwy (2)


0q)zw(p)zw( 3


0q)zw(p)zw(zw3)zw( 33

or

0qypyzw3)zw( 33

or

0qy)pzw3()zw( 33 (3)

We choose the coefficient of y to be zero

0pzw3 (4)

from which we obtain

3

pzw (5)

Because of (4), equation (3) gives us

qzw 33 (6)

Since

27

p

3

pzw

3333

(7)

equations (6) and (7) tell us that 3w and 3z will be the roots of the quadratic

equation


027

ptqt

32 (8)

We shall then solve the system

( 13 tw , 2

3 tz ) or ( 13 tz , 2

3 tw ) (9)

where 1t , 2t are the roots of the quadratic equation (8).

We have however to select those values of z and w which satisfy equation (5).

Example. Find the roots of the equation

0126y15y3 (1)

Solution. The substitution

zwy (2)

converts equation (1) into the equation

0126)zw(15)zw(zw3)zw( 33

or

0126y15yzw3)zw( 33

or

0126y)15zw3()zw( 33 (3)

We choose the coefficient of y to be zero

015zw3 (4)


5zw (5)

Because of (4), equation (3) gives us

126zw 33 (6)

Since

125zw 33 (7)

we conclude from (6) and (7) that 3w and 3z will be the roots of the quadratic

equation


0125t126t2 (8)

The roots of the above equation are 125t1 and 1t2 .

Therefore we have to solve the equations 125w3 and 1z3 .

The choice 5w and 1z is compatible to equation (5). Therefore one of the

roots of the equation (1) will be 6zwy .

Conclusion. The roots of equation (1) are given by

6y1 , i323y2 , i323y3

Section 4

Solving the reduced third degree equation

using Vieta’s method

4. Solving the reduced third degree equation using

Vieta’s method

The reduced third degree equation

0qypy3 (4.1)

can also be solved using the Vieta method. We introduce the substitution

z

mzy (4.2)

where z is the new variable and m is a parameter to be determined.

Using the above substitution, equation (4.1) becomes

0qz

mzp

z

mz

3



0qz

m

z

1)Am3(mz)pm3(z

3

33

Multiplying through by 3z the above equation becomes

0mz)pm3(mzqz)pm3(z 32346 (4.3)

The choice 0pm3 , i.e.

3

pm (4.4)

transforms equation (4.3) to the equation

027

pzqz

336 (4.5)

which is a sixth-degree equation. This equation can be converted to a second-

degree equation by the substitution

3zw (4.6)

Therefore we obtain

027

pwqw

32 (4.7)


27

p

4

q4

27

p4qD

3232 (4.8)

Therefore the roots of equation (4.7) are given by the formulas

27

p

4

q

2

q

2

27

p

4

q2q

w32

32

2,1

(4.9)

We then have to solve the two equations in order to determine z:

27

p

4

q

2

qz

323 (4.10)

and


27

p

4

q

2

qz

323 (4.11)

Equations (4.10) and (4.11) provide us with six solutions. Either choice of roots of

(4.10) or (4.11) is sufficient to determine the roots of the original equation, i.e. no

matter which equation we consider, we find the same set of solutions of the

original equation.

The three roots of equation (4.1) are determined by the formula

z

1

3

pzy (4.12)

Conclusion. Equation 0qypy3 , under the substitution

z

1

3

pzy

is converted into the equation

027

pzqz

336

The substitution 3zw , transforms the above equation into

027

pwqw

32

The determinant of the above equation is given by

27

p

4

q

2

qw

32

2,1 (4.9)

We then have to solve the two equations in order to determine z:

27

p

4

q

2

qz

323 and

27

p

4

q

2

qz

323 (4.10)

Solving any of the (4.), we find three roots

01 zz , 02 zz and 203 zz

where 0z is a fundamental root either one of (4.).


The roots of equation 0qypy3 are then given by

1

11z

1

3

pzy ,

222

z

1

3

pzy ,

333

z

1

3

pzy


01x3xx 23 (1)

Solution. Under the substitution

3

1yx (2)


027

56y

3

10y3 (3)

The substitution

z

1

9

10zy (4)


0729

1000z

27

56z 36

This equation can be converted to a second-degree equation by the substitution

3zw (5)

Therefore we obtain

0729

1000w

27

56w2 (6)


27

32

729

10004

27

56D

2

Therefore the roots of equation (6) are given by the formulas

9

62i

27

28

2

27

32i

27

56

w 2,1

(7)


We then have to solve the next two equations in order to determine z:

9

62i

27

28z3 (8)

and

9

62i

27

28z3 (9)

We find that the roots of equation (8) are given by

3

6i

3

2 ,

3

6i

3

2,

2

3

6i

3

2

(10)

and those of equation (9) by

3

6i

3

2 ,

3

6i

3

2,

2

3

6i

3

2

(11)

Using (10), for the choice 2

3i

2

1 , we have

3

6i

3

2z1

i6

326

6

232

3

6i

3

2z2

i6

326

6

232

3

6i

3

2z 2

3

We then obtain

3

4

z

1

9

10zy

111 ,

23

2

z

1

9

10zy

222 ,

23

2

z

1

9

10zy

333

The roots of the equation (1) are then evaluated using (2) and the above

expressions:


13

1yx 11 , 21

3

1yx 22 , 21

3

1yx 33

Conclusion.

The roots of the equation are

1x1 , 21x2 and 21x3


042x20xx 23 (1)


3

1yx (2)


027

952y

3

61y3 (3)


z

1

9

61zy (4)

equation (3) becomes

0729

226981z

27

952z 36 (5)

This equation can be converted to a second-degree equation by the substitution

3zw (6)

Therefore we obtain

0729

226981w

27

952w2 (7)


9

20

729

2269814

27

952D

2

and thus


3

5i

27

476

2

3

52i

27

952

w 2,1

We now have to solve the equations

3

5i

27

476z3 and

3

5i

27

476z3 (8)

We find

5i3

4z0 and 5i

3

4z0 (9)

Using either choice

5i3

4z1 ,

5i

3

4z2 , 2

3 5i3

4z

(10)

or

5i3

4z1 ,

5i

3

4z2 , 2

3 5i3

4z

we find, using equation (4), the quantities 1y , 2y , 3y and then using equation (2)

the roots 1x , 2x and 3x .

Conclusion.

The roots of the equation are 3x1 , 151x2 and 151x3


059x181x8x12 23 (1)


9

2y

3

12

8

yx

(2)


02916

24175y

108

1645y3 (3)



z

1

324

1645zy (4)


0324

1645z

2916

24175z

336

(5)

The above equation, using the substitution

3zw (6)

is converted into the quadratic equation

0324

1645w

2916

24175w

32

(7)


232

54

5155

324

16454

2916

24175D

ant thus

108

5515i

5832

24175

2

554

515i

2916

24175

w 2,1

(8)

We then have to solve the equations

108

5515i

5832

24175z3 and

108

5515i

5832

24175z3 (9)

We find

5i18

5z0 and 5i

18

5z0 (10)

Using either choice

5i18

5z1 ,

5i

18

5z2 , 2

2 5i18

5z

(11)

or

5i18

5z1 ,

5i

18

5z2 , 2

2 5i18

5z

(12)




Conclusion.

The roots of the equation are 3

1x1 , 15

2

1x2 and 15

2

1x3


03x10x4x8 23 (1)


6

1yx (2)


027

16y

3

4y3 (3)

The above equation under the substitution

z

1

9

4zy (4)

is converted into

09

4z

27

16z

336

(5)

The substitution

wz3 (6)

converts equation (5) into

09

4w

27

16w

32

(7)


09

44

27

16D

32

and thus equation (7) admits a double root given by


27

8

54

16w (9)

Therefore

27

8z3 and

27

8z3 (10)

Therefore we have the roots

3

2z1 ,

3

2z2 , 2

33

2z (11)

Using equation (4), we find the quantities 1y , 2y , 3y and then using equation (2)


Conclusion.


3x1 and

2

1xx 32 (double root).


026x31x10x3 23 (1)


9

10y

3

3

10

yx

(2)


0269

10y31

9

10y10

9

10y3

23


0729

12688y

27

179y3 (3)


z

1

81

179zy (4)



081

179z

729

12688z

336

(5)

The above equation under the substitution

3zw (6)

transforms into the quadratic equation

081

179w

729

12688w

32

(7)


232

27

3290

243

84100

81

1794

729

12688D

and thus the roots are given by

27

3145

729

6344

2

27

3290

729

12688

w 2.1


27

3145

729

6344z3 and

27

3145

729

6344z3 (8)

We find

39

8z0 and 3

9

8z0 (9)

Using either choice

39

8z1 ,

3

9

8z2 , 2

3 39

8z

or

39

8z1 ,

3

9

8z2 , 2

3 39

8z




Conclusion.


2x1 , i32x2 and i32x3 .


0)i2(zz)i2(z 23 (1)


3

i2yz

(2)


0)i2(3

i2y

3

i2y)i2(

3

i2y

23


0)i1(27

40y

3

i4y3 (3)


z

1

9

i4zy

equation (3) transforms into the equation

0i9

4z

27

)i1(40z

336

(4)


3zw (5)

Equation (4) transforms into the quadratic equation

0i9

4t

27

)i1(40t

32

(6)


i27

128i

9

44

27

)i1(40D

32


and thus find

)i1(33

5

9

4

2

)i1(2

2

3

2

3

8

27

)i1(40

w 2,1

(7)

where we have taken into account that one of the square roots of the imaginary

unit is )i1(2

2 .

We then have to solve the two equations

)i1(33

5

9

4z3

and )i1(3

3

5

9

4z3

(8)

We obtain

}i)32(1{3

1z0 and }i)32(1{

3

1z0 (9)

Using either choice

}i)32(1{3

1z1 , }i)32(1{

3

1z2 , 2

3 }i)32(1{3

1z

or

}i)32(1{3

1z1 , }i)32(1{

3

1z2 , 2

3 }i)32(1{3

1z



Conclusion.

The roots of the equation are iz1 , i2z2 and iz3 .


Section 5

Solving the reduced third degree equation using

Trigonometric Functions


trigonometric functions

5.1. Solving the equation 0qyp3y3 for 0p .


0qyp3y3 ( 0p ) (5.1)


n

zy (5.2)

equation (5.1) takes on the form

0nqznp3z 323 (5.3)

In view of the identity

cos3cos43cos 3

we have

03cos4

1cos

4

3cos3 (5.4)

Comparing (5.3) and (5.4), we obtain

cosz (5.5a)

4

3np3 2 (5.5b)


and

3cos4

1nq 3

(5.5c)

From equation (5.5b) we obtain

p4

1n2

from which we determine n:

p2

1n ( 0p ) (5.6)

Because of (5.6), equation (5.5c) gives us

pp2

q3cos (5.7)

If 0 is an angle such that

pp2

qcos 0 , 00 (5.8)

then from equation (5.7), since

0cos3cos , )2cos(3cos 0 and )4cos(3cos 0

we obtain that

3

0 , 3

2

3

0

and

3

4

3

0

(5.9)

Equations (5.2), (5.6), (5.5a) and (5.9) determine the three roots of the reduced

equation (5.1).

Example. Find the roots of the equation

03x3x6x 23 (1)

Solution. The substitution

2yx (2)


01y3y3 ( 0D ) (3)



n

zy (4)


0nzn3z 323 (5)

We have the identity

03cos4

1cos

4

3cos3 (6)

Comparing (5) and (6), we obtain

cosz (7)

4

3n3 2 (8)

and

3cos4

1n3

(9)

From equation (8) we obtain 4

1n2 , from which we determine n:

2

1n (10)

Equation (9) then becomes

3cos

4

1

2

13

, from which we get

0120cos

2

13cos (11)

Therefore, in view of (5.9), we have

040 ,

00 12040 and 00 24040 (12)

Taking into account (4), (7), (10) and (12), we have

040cos2y ,

0160cos2y , 0280cos2y (13)

Therefore the three roots of the equation (1), in view of (2) and (13), are given by


240cos2x 01 , 2160cos2x 0

2 and 2280cos2x 03

or

240cos2x 01 , 220cos2x 0

2 and 280cos2x 03 (14)

5.2. General solutions for the equation 0qypy3 using

trigonometric functions.

Below we present the general formulas for evaluation of the roots of the reduced

third degree equation using trigonometric functions.

(I) If 027

p

4

qD

32

(irreducible case) the roots of the equation 0qypy3

are given by

cos3

p2y1

)60(cos3

p2y 0

3,2

where

3

p

3

2

q3cos

(II) If 027

p

4

qD

32

and 0p the roots of the equation 0qypy3 are

given by

2cot3

p2y1

}2eccos3i2cot{3

py 3,2

where


3

2tantan

and

3

3

p

q

2tan

, 045|| , 090||

(III) If 027

p

4

qD

32

and 0p the roots of the equation 0qypy3 are

given by

2eccos3

p2y1

}2cot3i2eccos{3

py 3,2

where

3

2tantan

and

3

3

p

q

2sin

, 045|| , 090||

Proof of the above formulas.

We consider the equation

0qypy3 (5.10)

and the quantity

27

p

4

qD

32

(5.11)

(I) Suppose that 0D . It would then be 0p .


n

zy (5.12)


0nqznpz 323 (5.13)


cos3cos43cos 3

we have


03cos4

1cos

4

3cos3 (5.14)


cosz (5.16)

4

3np 2 (5.17)

and

3cos4

1nq 3

(5.18)

From equation (5.17) we obtain

p

3

4

1n2


p

3

2

1n (5.19)

Because of (5.19), equation (5.18) gives us

p

3

p

3

2

q3cos (5.20)

Combining (5.12), (5.16) and (5.19), we can determine one root of the equation

(5.10):

cos

3

p2y1 (5.21)

We also have the identity

)ypyyy()yy(qypy 211

21

3

The roots of the trinomial 211

2 ypyyy of the above relation will determine

the other two roots of the equation (5.10). The discriminant of this trinomial is

evaluated to be

21

21

21 y3p4)yp(4y


2

cos3

p23p4

)cos1(p4cosp4p4 22

2sinp4

Therefore sinp2 . The roots of the trinomial are then given by

sinpcos

3

p

2

sinp2yy 1

3,2

sin

2

3cos

2

1

3

p2

}sin60sincos60{cos3

p2 00

or finally

)60cos(3

p2y 0

3,2

(5.22)

Relations (5.21) and (5.22) provide the three roots of the equation 0qypy3

when 027

p

4

qD

32

.

(II) We suppose that 027

p

4

qD

32

and 0p . One of the roots of the equation

0qypy3 , is known to be given by

3

323

32

00127

p

4

q

2

q

27

p

4

q

2

qnmy (5.23)

We shall express the above expression in terms of trigonometric functions. We

first transform the quantity 27

p

4

q 32

. Since


2323

2

232

3

p

q

21

4

q

27

p

q

41

4

q

27

p

4

q

introducing the function tan by

3

3

p

q

2tan

(5.24)

we obtain

22

2232

sec4

q)tan1(

4

q

27

p

4

q (5.25)

Therefore

sec2

qD (5.26)

and then

2tantan

2

q)sec1(

2

qsec

2

q

2

qD

2

q

2

tan3

p

2tan

3

p

q

2

2

q33

(5.27)

Similarly

2tan

tan

2

q)sec1(

2

qsec

2

q

2

qD

2

q

2tan

1

3

p

2tan

1

3

p

q

2

2

q33

(5.28)

We thus have

3

323

32

00127

p

4

q

2

q

27

p

4

q

2

qnmy


3

33

3

2tan

1

3

p

2tan

3

p

33

2tan

1

2tan

3

p (5.29)

We now introduce the function tan by

3

2tantan

(5.30)

and equation (5.29) gives us

2cot

3

p2

tan

1tan

3

py1 (5.31)

In order to find the other roots, we remark that since

2cot

3

p2

tan

1tan

3

pnm 00

we would have similarly

2csc

3

p2

tan

1tan

3

pnm 00

Therefore, in view of the formulas

2

)nm(3i)nm(y 0000

3,2

we obtain the following formula for the other two roots of the equation

}2csc3i2cot{3

py 3,2

(5.32)

Equations (5.31) and (5.32) give the roots of the equation 0qypy3 when

027

p

4

qD

32

and 0p .


(III) We suppose that 027

p

4

qD

32

and 0p . One of the roots of the equation

0qypy3 , is known to be given by

3

323

32

00127

p

4

q

2

q

27

p

4

q

2

qnmy (5.33)

We shall express the above expression in terms of trigonometric functions. We

first transform the quantity 27

p

4

q 32

. Since

2323

2

232

3

p

q

21

4

q

27

p

q

41

4

q

27

p

4

q

introducing the function sin by

3

3

p

q

2sin

(5.34)

we obtain

22

2232

cos4

q)sin1(

4

q

27

p

4

q (5.35)

Therefore

cos2

qD (5.36)

and then

2tansin

2

q)cos1(

2

qcos

2

q

2

qD

2

q

2

tan3

p

2tan

3

p

q

2

2

q33

(5.37)

Similarly


2tan

sin

2

q)cos1(

2

qcos

2

q

2

qD

2

q

2tan

1

3

p

2tan

1

3

p

q

2

2

q33

(5.38)

We thus have

3

323

32

00127

p

4

q

2

q

27

p

4

q

2

qnmy

3

33

3

2tan

1

3

p

2tan

3

p

33

2tan

1

2tan

3

p (5.39)

We now introduce the function tan by

3

2tantan

(5.40)

and equation (5.39) gives us

2csc

3

p2

tan

1tan

3

py1 (5.41)

In order to find the other roots, we remark that since

2csc

3

p2

tan

1tan

3

pnm 00

we would have similarly

2cot

3

p2

tan

1tan

3

pnm 00

Therefore, in view of the formulas


2

)nm(3i)nm(y 0000

3,2

we obtain the following formula for the other two roots of the equation

}a2cot3i2csc{3

py 3,2

(5.42)

Equations (5.41) and (5.32) give the roots of the equation 0qypy3 when

027

p

4

qD

32

and 0p .

Example 1. Find the roots of the equation 027y27y3

Solution. In this case we have 027p , 27q , 04

2187

27

p

4

qD

32

We have 2

1

27

3

2

27

p

3

2

q3cos

33

. Therefore o1203

and then o40 . Therefore we have the following values for the roots:

o1 40cos6cos

3

p2y

o02 100cos6)60(cos

3

p2y

o03 20cos6)60(cos

3

p2y


Solution. In this case we have 09p , 6q and 01727

p

4

qD

32

We find 33

9

6

2

3

p

q

2tan

33

and then

o60 . We also obtain

6

3 o3

3

130tan

2tantan

. We further calculate


6

6

6

2

62

32

13

3

12

3

11

tan2

tan12cot

and

6

6

6

2

62

32

13

3

12

3

11

tan2

tan12csc

We then obtain the following values of the roots:

)13(932

13322cot

3

p2y

63

6

6

1

6

6

6

6

232

133i

32

133)2csc3i2cot(

3

py

6

6

6

6

332

133i

32

133)2csc3i2cot(

3

py


Solution. In this case we have 09p , 12q and 0927

p

4

qD

32

We find 2

3

3

9

12

2

3

p

q

2sin

33

and then

o60 . We also

obtain 6

3 o3

3

130tan

2tantan

. We further calculate


6

6

6

2

62

32

13

3

12

3

11

tan2

tan12cot

and

6

6

6

2

62

32

13

3

12

3

11

tan2

tan12csc

We then obtain the following values of the roots:

)13(932

13322csc

3

p2y

63

6

6

1

6

6

6

6

232

133i

32

133)2cot3i2csc(

3

py

6

6

6

6

332

133i

32

133)2cot3i2csc(

3

py


Section 6

Solving the reduced third degree equation using

hyperbolic functions


hyperbolic functions


0qyp3y3 (6.1)

when

027

)p3(

4

qD

32

, i.e. 32 p4q (6.2)

6.I. Case I. We first consider the case 0p .


n

zy (6.3)


0nqznp3z 323 (6.4)


usinh3usinh4u3sinh 3

we have

0u3sinh4

1usinh

4

3usinh 3 (6.5)


usinhz (6.6a)


4

3np3 2 (6.6b)

and

u3sinh4

1nq 3 (6.6c)


p4

1n2


p2

1n

( 0p ) (6.7)

Using (6.3), (6.6a) and (6.7), we find that one root of the equation is given by

usinhp2y1 (6.8)


)p3yyyy()yy(qyp3y 211

21

3

the discriminant of the trinomial )p3yyyy( 211

2 is evaluated to be

)p4y(3)p3y(4yD 21

21

21

)]usinh1(p4[3]p4usinh)p(4[3 22

ucosh)p(12 2

Therefore

ucoshp32iD

We thus have obtain the other two roots of the equation, which are the roots of the

quadratic trinomial )p3yyyy( 211

2 :

2

ucoshp32iusinhp2

2

ucoshp32iyy 1

3,2

or

)ucosh3iusinh(py2 (6.9)


)ucosh3iu(sinhpy3 (6.10)


pp2

qu3sinh

(6.11)

From the previous equation we determine usinh and ucosh .

6.II. Case II. We consider next the case 0p and 0q .


n

zy (6.12)


0nqznp3z 323 (6.13)


ucosh3ucosh4u3cosh 3

we have

0u3cosh4

1ucosh

4

3ucosh3 (6.14)


ucoshz (6.15a)

4

3np3 2 (6.15b)

and

u3sinh4

1nq 3 (6.15c)


p4

1n2


p2

1n ( 0p ) (6.16)


From (6.12), (6.15a) and (6.16), we find that one root of the equation is given by

ucoshp2y1 (6.17)


)p3yyyy()yy(qyp3y 211

21

3

the discriminant of the trinomial )p3yyyy( 211

2 is evaluated to be

)p4y(3)p3y(4yD 21

21

21

)]1u(coshp4[3)p4ucoshp4(3 22

usinhp12 2

Therefore

usinhp32iD

We thus have obtain the other two roots of the equation, which are the roots of the

quadratic trinomial )p3yyyy( 211

2 :

2

usinhp32iucoshp2

2

usinhp32iyy 1

3,2

or

)usinh3iucosh(py2 (6.18)

)usinh3iu(coshpy3 (6.19)


pp2

qu3cosh (6.20)

From the previous equation we can determine ucosh and usinh .

Example 1. Find the roots of the equation

012y12y3 (1)

Solution. In this case 04p and 12q . The condition 32 p4q is satisfied

automatically. We have now


4

3

4)4(2

12

pp2

qu3sinh

from which we obtain 4

3

2

ee u3u3

, which upon substitution u3ew we get

the equation 02w3w2 2 which admits the two real roots 2

1w1 and

2w2 . We accept the positive root only and we get 2

1e u3 .

Therefore 3/1u 2e and then

2

22

2

eeusinh

3/13/1uu

(2)

and

2

22

2

eeucosh

3/13/1uu

(3)

The roots of the equation (1) are then given by

usinh4usinhp2y1 (4)

)ucosh3iusinh(2)ucosh3iusinh(py2 (5)

)ucosh3iusinh(2)ucosh3iu(sinhpy3 (6)

where usinh and ucosh are given by (2) and (3) respectively.

Example 2. Find the roots of the equation

012y9y3 (1)

Solution. In this case 03p and 012q . These values of p and q satisfy

the condition 32 p4q . We have now

3

2

332

12

pp2

qu3cosh


from which we obtain 3

2

2

ee u3u3

, which upon substitution u3ew we get

the equation 03w4w3 2 which admits the two real roots 3w1 and

3

1w2 . We can accept either one. For example accepting 3w1 , we get

3e u3 . Therefore 6/1u 3e and then

2

33

2

eeusinh

6/16/1uu

(2)

and

2

33

2

eeucosh

6/16/1uu

(3)

The roots of the equation are then given by

ucosh32ucoshp2y1 (4)

)usinh3iucosh(3)usinh3iucosh(py2 (5)

)usinh3iu(cosh3)usinh3iu(coshpy3 (6)

where usinh and ucosh are given by (2) and (3) respectively.


Section 7

The Tschirnhaus method of solving the reduced

third degree equation

7. The Tschirnhaus method of solving the reduced third

degree equation.


0qypy3 (1)

Introducing the substitution (the so-called Tschirnhaus transformation)

azyby2 (2)

we can convert equation (1) into an equation of the form mz3 , by assigning

appropriate values to the parameters b and a.

The elimination of y between equations (1) and (2) can be done using the

resultant of the two polynomials. The Sylvester Matrix of (1) and (2) is the

following 55 matrix (see Appendix II)

azb100

0azb10

00azb1

qp010

0qp01

(3)

The resultant is the determinant of Sylvester’s matrix:

z)pa4pbpqb3a3(z)p2a3(z 22223

0bqpbpaap2qba3bqqapa 223232 (4)


Requiring coefficients of 2z and z to be zero in (4),

0p2a3 (5)

0pa4pbpqb3a3 222 (6)

we find, solving the simultaneous equations (5) and (6) that

3

p2a (7)

and b to be a root of the quadratic equation

0p3

1bq3bp 22 (8)

The roots of the above equation are given by

p6

Dq9b

(9)

where

23 q81p12D (10)

The values (8) and (9) when substituted in (4), we obtain the equation

mz3 (11)

where

3

2

p486

DDq9Dm

(12)

(I) For 3

p2a and

p6

Dq9b

, quadratic equation (2) will give us two

solutions for each value of the roots of the equation 3

23

p486

DDq9Dmz

.

Therefore we shall obtain six in total roots by considering the equation (2) written

as

03

p2zy

p6

Dq9y2

(13)

where }z,z,z{z 02

00 are the roots of the equation


3

23

p486

DDq9Dmz

(14)

In other words equation (13) is essentially the following set of three equations

03

p2zy

p6

Dq9y 0

2

03

p2zy

p6

Dq9y 0

2

03

p2zy

p6

Dq9y 0

22

where

33

2

0p486

DDq9Dz

and

2

3i

2

1,

2

3i

2

1

(II) For 3

p2a and

p6

Dq9b

, quadratic equation (2) will give us two

solutions for each value of the roots of the equation 3

23

p486

DDq9Dmz

.

Therefore we shall obtain six in total roots by considering the equation (2) written

as

03

p2zy

p6

Dq9y2

(15)

where }z,z,z{z 02

00 are the roots of the equation

3

23

p486

DDq9Dmz

(16)

In other words equation (15) is essentially the following set of three equations

03

p2zy

p6

Dq9y 0

2


03

p2zy

p6

Dq9y 0

2

03

p2zy

p6

Dq9y 0

22

where

33

2

0p486

DDq9Dz

and

2

3i

2

1,

2

3i

2

1

In practice we consider either one of the cases (I) or (II). We then check which

root satisfies the original equation (1).

Despite the elegance of the method (which can be extended into higher than the

third degree equations, like the quintic, etc) it leads to some computational

difficulties compared to the methods considered previously. That is why this

method is usually considered in solving equations above the fourth degree

(quintic, sixtic, …).

Example. We consider the equation

027

56y

3

10y3 (1)

This equation was solved previously in Section 2 (Example 2) using another

method. It was found that this equation admits the roots 3

4y1 , 2

3

2y2

and 23

2y3 . We shall apply Tschirnhaus’s transformation in solving the

above equation (1). In our case we have

3

10p and

27

56q

We then calculate

9

20

3

p2a , 96q81p12D 23


i5

6

15

14

p6

Dq9b

, i

1125

6448

125

64

p486

DDq9Dm

3

2

We also find

i15

64

5

4i

1125

6448

125

64mz 33

0

We then have to find the roots of one of the following sets:

(I)

09

20i

15

64

5

4yi

5

6

15

14y2

(E1)

09

20i

15

64

5

4yi

5

6

15

14y2

(E2)

09

20i

15

64

5

4yi

5

6

15

14y 22

(E3)

(II)

09

20i

15

64

5

4yi

5

6

15

14y2

(E4)

09

20i

15

64

5

4yi

5

6

15

14y2

(E5)

09

20i

15

64

5

4yi

5

6

15

14y 22

(E6)

We find that

the roots of (E1) are 3

4y1 and i

5

6

15

34y2

the roots of (E2) are 23

2y1 and 2

3

2y2 ,

2

3i

2

1

We can check that 3

4y1 and 2

3

2y2 , 2

3

2y3 are the three roots of

equation (1).


Section 8

The Glasser method of solving the

reduced third degree equation

8. The Glasser method.

This method is described in Appendix IV.

The reduced third degree equation 0qzpz3 can be transformed into the

equation 0txx3 under the substitution xpz .

In fact equation 0qzpz3 becomes 0q)xp(p)xp( 3 which

is equivalent to

0)p(

qxp

)p(

px

33

3

0)p(

qx

)p(

px

32

3

0)p(

qx

p

px

3

3

0txx3

where

3)p(

qt

Using the general formulas, we find that the root 1x of the equation

0txx3

is given by


0n

n22

0n

n2

1

)3n2(2

3n

t2

5n3

2

t

)2n2()1n(

t)1n3(

2

t1x

or, using the known series expansion of the hypergeometric function, we have

4

t27;2,

2

3;1,

6

7,

6

5F

8

t3

4

t27;

2

3;

3

2,

3

1F

2

t1x

2

23

22

121

Since

zsin3

1sin

z

3z;

2

3;

3

2,

3

1F 1

12

1zsin3

1cos

z

18z;2,

2

3;1,

6

7,

6

5F 1

23

we find

t

2

33sin

3

1cost

2

33sin

3

1sin

3

1x 11

1

The other two roots are given by

t

2

33sin

3

1cost

2

33sin

3

1sin

3

1x 11

2

t

2

33sin

3

1sin

3

2x 1

3

Note. The Birkeland method.

Birkeland’s method (R. Birkeland: “Comptes Rendus” Vol. 171 (1920) 778-781,

1047-1049, 1370-1372 and Vol. 172 (1920) 309-311 and 1155-1158.

Errata Vol. 172 (1920) 188) is very similar to Glasser’s method. In fact there is a

remarkable overlapping between the two methods.

For the equation

xgx3

the roots are given by


;

2

3;

3

2,

3

1F

33

1;

2

1;

6

1,

6

1Fgx 12121

;

2

3;

3

2,

3

1F

33

1;

2

1;

6

1,

6

1Fgx 12122

;

2

3;

3

2,

3

1F

3

g

3

2x 123

where

3

2

g4

27

The above formulas are true for 1|| . There are similar formulas for 1|| .


Section 9

The method of differential resolvents

9. The method of differential resolvents

The method of differential resolvents, a term coined by R. Harley (Harley [31])

is a general method applied to any equation of n-th degree of the form

0x)1n(ynyn , 2n (1)

where x is a known constant parameter.

According to this method, to every equation of the form (1), we associate an

ordinary differential equation of (n-1) order, formed under some rules.

We follow the next steps:

Step 1. Considering that y is a function of the parameter x, we differentiate

equation (1) with respect to x and we obtain the equation

0)1n(dx

dyn

dx

dyyn 1n )1n(

dx

dy)1y(n 1n

1y

1

n

1n

dx

dy1n

(2)

Step 2. Equation (1) can be written as

1y

xy)1n(y 1n


y

xy)1n(1y 1n

(3)

Step 3. Combining (2) and (3), we obtain another expression of the derivative


xy

y

n

1

dx

dy

(4)

Step 4. From equation (1) we obtain the following relation

y)x1()xy()1n()xy(y 1n1n1n (5)

which can be checked easily. From this relation, dividing by )xy( , we obtain

)x1(xy

y)1n(y

xy

xy 1n1n1n

and from this

)1n(y

xy

xy

x1

1

xy

y 1n1n

1n (6)


2n4n23n2n

1n1n

xyxyxyxy

xy

relation (6) becomes

)}1n(yxyxyxy{x1

1

xy

y 2n3n22n1n

1n

(7)

Step 5. Combining (4) and (7), we obtain the following expression of the

derivative

)}1n(yxyxyxy{x1

1

n

1

dx

dy 2n3n22n1n

1n

(8)

or

)}1n(yxyxyxy{dx

dy)x1(n 2n3n22n1n1n (9)

an expression to be used for the formation of the differential resolvent.

9.1. The differential resolvent for the cubic.

We are now to derive the differential resolvent for the cubic, an ordinary second

order differential equation and then we shall derive the roots of the cubic.

For 3n , we obtain from (1) the cubic


0x2y3y3 (10)

For 3n , we obtain from (9) the following expression for the derivative

)2yxy(dx

dy)x1(3 22 (11)

Differentiate again the above expression with respect to x, we obtain

ydx

dyx

dx

dyy2

dx

yd)x1(3

dx

dyx6

2

22


ydx

dy)y2x5(

dx

yd)x1(3

2

22 (12)

Multiply by )x1(3 2 the previous equation and we obtain

)x1(y3dx

dy)x1(3)y2x5(

dx

yd)x1(9 22

2

222

from which, upon substitution of the bracket by the expression given in (11), we

obtain

)x1(y3)}2yxy({)y2x5(dx

yd)x1(9 22

2

222

which can be simplified into

x6y)x21(yx3dx

yd)x1(9 22

2

222 (13)

We are now in the position to form the differential resolvent: we multiply (11) by

)x( and add the resulting equation into (13). We then obtain the equation

)2yxy(x6y)x21(yx3dx

dy)x1(3

dx

yd)x1(9 2222

2

222

The above equation can be rearranged as

dx

dy)x1(3

dx

yd)x1(9 2

2

222


)x3(2y)x2x1(y)x3( 22 (14)

In order to make 2y vanish, we choose such that 0x3 , i.e. x3 .

Under this choice, equation (14) takes on the form

y)x1(dx

dy)x1(x9

dx

yd)x1(9 22

2

222


0ydx

dyx9

dx

yd)x1(9

2

22 (15)

Equation (15) is the differential resolvent, corresponding to the cubic

0x2y3y3 .

Second Method.

This method is based on the following

Theorem. The differential resolvent corresponding to any equation of the form

0x)1n(ynyn , 3n

has the form

0ymdx

ydm

dx

yd1n2n

2n

11n

1n

where 1m , 2m , … , 1nm are x dependent functions.

For 3n , using the known formula established previously

xy

y

3

1

dx

dy

and taking the second derivative,

3

4

2

2

)xy(

y

9

1

dx

yd

the differential resolvent

0ymdx

dym

dx

yd212

2


takes on the form

0ymxy

y

3

1m

)xy(

y

9

1213

4

The last equation is equivalent to

0)xy(m9)xy(m3y 32

21

3

which upon expanding can be written as

0xm9xm3y)xm9xm2(3y)xm9m(3y)m91( 32

21

221

221

32

Since x2y3y3 , we obtain from the last equation

y)}x1(m9xm21{3y)xm9m(3 221

221

0x2xm3x)2x(m9 21

22

Equating to zero all the coefficients, we obtain the system of simultaneous

equations

0xm9m 21 , 0)x1(m9xm21 221 ,

0x2xm3x)2x(m9 21

22

The above system admits the unique solution

21

x1

xm

and

22x1

1

9

1m

Therefore the differential resolvent takes on the form

0yx1

1

9

1

dx

dy

x1

x

dx

yd222

2

the same expression derived previously using another method.

9.2. Solution of the differential resolvent and evaluation of the roots

Equation (15) is a second order, ordinary differential equation, which can be

solved easily. Under the substitution tsinx , equation (15) transforms into

0y9

1

dt

yd2

2


The general solution of the above equation is given by

3

CxsinsinC

3

CtsinCy 2

1

12

1

The constants are determined by the fact that when 0x , then 3,3,0y .

Therefore

3

CsinC0 2

1 , 3

CsinC3 2

1

,

3

CsinC3 2

1

Combining any two of the above conditions, we find

2C1 and n2C2

where n is zero or any integer. We thus have

3

n2xsinsin2y

1


Section 10

Fourth degree equations.

The Ferrari, Descartes, Euler

and Simpson methods.

The differential resolvent method

10.1. The Ferrari Method

We consider the general form of the equation of the fourth degree

0dxcxbxax 234 (1)

where a, b, c and d are complex numbers in general.

Using the transformation

4

ayx (2)


0ryqypy 24 (3)

where the coefficients p, q and r are given by

2a8

3bp (4)

ca8

1ba

2

1q 3 (5)

da256

3ba

16

1ca

4

1r 42 (6)

Equation (3) can be written as


04

prkpkyqyk2k

2

py

222

22

(7)

where k is an auxiliary parameter chosen so as the expression inside the brackets

to be a perfect square, i.e. to have zero determinant:

04

prkpkk24)q(

222

(8)

Equation (8) is a third degree equation, as we can see through the following series

of equivalent transformations

0kp2kr8kp8k8q 2232

0qk)p2r8(kp8k8 2223

0qk)p2r8(kp8k8 2223

or finally

08

qk

4

prkpk

2223

(9)

Let 0k be one of the roots of equation (9). Then the trinomial inside the brackets

in equation (7) will be a perfect square with a double root given by

0k4

r (10)

Therefore equation (7) can be written as

0k4

ryk2k

2

py

2

00

2

02

(11)

which is a couple of second degree equations.

The four roots of the above equation (11) are the roots of the equation (3).


01x2xx 34 (1)

Solution. The transformation


4

1yx (2)

transforms (1) into the equation

0256

131y

8

15y

8

3y 24 (3)

The above equation can be written as

064

35k

8

3ky

8

15yk2k

16

3y 22

22

(4)

where k is an auxiliary parameter chosen so as the expression inside the brackets

to be a perfect square, i.e. to have zero determinant:

064

35k

8

3kk24

8

15 22

The above equation is equivalent to the equation

0512

225k

64

35k

8

3k 23 (5)

The roots of the above equation are found (using one of the methods explained in

previous sections) to be

8

5, )i1121(

8

1 (6)

For 8

5k , equation (4) takes on the form

0640

131y

4

5

16

7y

222


0640

131y

2

5

16

7y

640

131y

2

5

16

7y 22

(7)

From the above equation it follows that the roots of the original equation are the

roots of the pair of equations


01280

5131

16

7y

2

5y2 (8)

and

01280

5131

16

7y

2

5y2 (9)

The roots of equation (8) are given by

80

8005655

4

5 (10)

while the roots of (9) are given by

80

8005655i

4

5 (11)

Therefore the roots of equation (1) are

80

8005655

4

5 ,

80

8005655

4

5 ,

80

8005655i

4

5 ,

80

8005655i

4

5


01x2x3x4x 234 (1)

Solution. Under the transformation

1yx (2)

equation (1) transforms into

01y3y 24 (3)

This is a bi-quadratic equation solved under the substitution

2yw (4)

Equation (3) thus transforms into the quadratic equation

01w3w2 (5)

with roots given by


2

53w 2,1

(6)

Therefore the roots of the original equation are the solutions of the equations

2

53y2

and 2

53y2

Equation 2

53y2

admits the roots

2

5

2

1y and

2

5

2

1y (7)

Equation 2

53y2

admits the roots

2

5

2

1y and

2

5

2

1y (8)

Therefore the roots of equation (3) are given by

2

5

2

1y and

2

5

2

1y

Therefore the roots of the original equation (1) are given by

2

5

2

3x and

2

5

2

1x

Note. Equation 0rxqypy 24 can be solved using another simplified

approach. Since this equation can be written as

ryqypy 24

adding to both members the quantity 4

kyk

22 to complete the square, we obtain

ryqyp4

kyk

4

kyky 2

22

224

The previous equation is equivalent to

r

4

kyqy)pk(

2

ky

22

22


We transform the right hand member of the above equation into a perfect square,

requiring the discriminant of the quadratic trinomial to be zero:

0r4

k)pk(4)q(0

22

0)qpr4(kr4kpk 223

Let 0k be a root of the last equation (which can be solved according to one of the

known methods). The trinomial

r

4

kyqy)pk(

22 as having a double root

)pk(2

q

0 , can be written as

2

00

)pk(2

qy)pk(

. Therefore we have to

solve the equation

2

00

202

)pk(2

qy)pk(

2

ky

The last equation can be factorized and converted to a pair of second-order

equations

0)pk(2

qypk

2

ky

00

02

and

0)pk(2

qypk

2

ky

00

02

which when solved, will give the four roots of equation 0rxqypy 24 .


0442x58x23x2x 234 (1)


2

1yx (2)



016

7625y80y

2

43y 24 (3)


16

7625y80y

2

43y 24

Adding to both members of this equation the quantity 4

kyk

22 to complete the

square, we obtain

16

7625y80y

2

43

4

kyk

4

kyky 2

22

224


16

7625

4

ky80y

2

43k

2

ky

22

22 (4)

We transform the trinomial

16

7625

4

ky80y

2

43k

22 (5)

into a complete square requiring the discriminant to be zero. Since

016

7625

4

k

2

43k4800

22

we have to solve the third degree equation

08

276675k

4

7625k

2

43k 23 (6)

The roots of the above equation are given by 2

35,

2

85 and

2

93. The choice

2

35k0 , transforms equation (4) into

400y80y44

35y 2

22



22

2 )10y(44

35y

(7)

From the last equation we obtain the pair of equations

)10y(i24

35y2 and )10y(i2

4

35y2

The roots of the first equation are i32

5 and i5

2

5 .

The roots of the second equation are i32

5 and i5

2

5 .

Using the above results and equation (2), we find that the roots of equation (1) are

given by i32 and i53 .


01369x74x457x12x36 234 (1)


12

1yx (2)


0576

793921y150y

2

911y36 24 (3)


576

793921y150y

2

911y36 24

Adding to both members of this equation the quantity 4

kyk6

22 to complete the

square, we obtain

576

793921y150y

2

911

4

kyk6

4

kyk6y36 2

22

224



576

793921

4

ky150y

2

911k6

2

ky6

22

22

(4)

We transform the trinomial

576

793921

4

ky150y

2

911k6

22

into a perfect square requiring its discriminant to be zero. Since

0576

793921

4

k

2

911k641500

22

we have to solve the third degree equation

0288

716782031k

24

793921k

2

911k6 23

The roots of the above equation are

12

889 ,

12

839 and

12

961

The choice 12

839k0 , transforms equation (4) into

4

625y150y36

24

839y6 2

22

which can also be written as

222

2

25y6

24

839y6

(5)

The previous equation can be split into the pair of equations

2

25y6i

24

839y6 2 and

2

25y6i

24

839y6 2 (6)

The roots of the first equation are i212

5 and i3

12

5 .

The roots of the second equation are i212

5 and i3

12

5 .


Taking into account the above roots and equation (2), we find that the roots of

equation (1) are given by i23

1 and i3

2

1 .


0i102x)i91(xi3x)i1(2x 234 (1)


2

i1yx

(2)


0i154

15y)i23(2yi6y 24 (3)


i154

15y)i23(2yi6y 24

Adding the quantity 4

kyk

22 to both members of the above equation, we obtain

4

ki15

4

15y)i23(2y)i6k(

2

ky

22

22

(4)

We require the determinant of the trinomial of the rhs of the above equation to be

zero:

04

ki15

4

15)i6k(4)i23(4

22

0i42340k)i41(15ki6k 23 (5)

The roots of the above equation are given by i61k1 , i64k2 ,

i65k3 . For i61k , equation (4) takes on the form

i125y)i23(2y2

i61y 2

22



22

2 )i23y(2

i61y

From the last equation we obtain the pair of equations

)i23y(i2

i61y2

and )i23y(i

2

i61y2

The first one admits the solutions i2

1

2

3y1 , i

2

1

2

3y2 .

The second one admits the solutions i2

3

2

3y3 , i

2

5

2

3y4 .

Using (2) and the above roots, we find that the roots of the original equation (1)

are given by

i1x1 , i32x2 , 1x3 and 2x4

10.2. The Descartes Method.

We consider the reduced form of the fourth order equation

0ryqypy 24 (1)

and we try to factorize the fourth degree polynomial

)nyky()myky( 22

nmy)mknk(y)knm(y 224 (2)

Comparing the coefficients of the polynomials in equations (1) and (2) we obtain

the system

2knmp , mknkq , nmr (3)

The first two equations of the previous system can be written as

2kpnm and

k

qnm

from which we can express m and n in terms of

k

qkp

2

1m 2

and

k

qkp

2

1n 2

(4)


Substituting the above values to the last equation of the system (3), we derive the

equation

k

qkp

2

1

k

qkp

2

1r 22

The above equation contains only one unknown, the variable k and can be written

as

222

k

q)kp(r4

or equivalently

22222 q)kp(kkr4

242222 q)kkp2p(kkr4

0qk)r4p(kp2k 22246 (5)

The previous equation under the substitution

2kw (6)

takes on the form

0qw)r4p(wp2w 2223 (7)

which is a third degree equation and can be solved accordingly, using one of the

known methods. After determining one root only 0k from (6) and (7), we arrive at

the factorization (using equation (2)):

0)nyky()myky( 02

02 (8)

where m and n are determined by (4):

0

20

k

qkp

2

1m and

0

20

k

qkp

2

1n (9)


0442x58x23x2x 234 (1)



2

1yx (2)


016

7625y80y

2

43y 24 (3)

We try to factorize the fourth degree polynomial

)nyky()myky( 22

nmy)mknk(y)knm(y 224 (4)

Comparing the coefficients of the polynomials in equations (3) and (4) we obtain

the system

2knm

2

43 , mknk80 , nm

16

7625 (5)

The first two equations of the previous system can be written as

2k

2

43nm and

k

80nm

from which we can express m and n in terms of k:

k

80k

2

43

2

1m 2 and

k

80k

2

43

2

1n 2 (6)

Substituting the above values to the last equation of the system (3), we derive the

equation

k

80k

2

43

2

1

k

80k

2

43

2

1

16

7625 22

The above equation contains only one unknown, the variable k and can be written

06400k1444k43k 246 (7)

The previous equation under the substitution

2kw (8)

takes on the form

06400w1444w43w 23 (9)


which is a third degree equation with roots 4,25,64 . The choice 25w ,

gives us 5k and thus we arrive at the factorization (using equation (6)):

0)ny5y()my5y( 22 (10)

where m and n are determined by (6):

4

125

5

8025

2

43

2

1m

and

4

61

5

8025

2

43

2

1n

Therefore equation (10) becomes

04

61y5y

4

125y5y 22

which can be split into the two equations

04

125y5y2 and 0

4

61y5y2

with roots i32

5 and i5

2

5 . Therefore the roots of the original equation are

given by i32 and i53 .

10.3. The Euler Method.


0dxcxbxax 234 (1)


Using the transformation

4

ayx (2)


0rxqypy 24 (3)

where the coefficients p, q and r are given by

2a8

3bp (4)


ca8

1ba

2

1q 3 (5)

da256

3ba

16

1ca

4

1r 42 (6)

We now search for a solution of equation (3) in the form

wvuy (7)

where for the moment any combination of signs is allowed.

Squaring both members of equation (7), we obtain

)]u()w()w()v()v()u[(2)wvu(y2

or

)]u()w()w()v()v()u[(2)wvu(y2 (8)

Squaring again equation (8) we arrive at

224 )wvu()wvu(y2y

)]wvu()w()v()u(2uwwvvu[4


y)w()v()u(8y)wvu(2y 24

0)uwwvvu(4)wvu( 2 (9)

Equating the corresponding coefficients of equations (3) and (9) we arrive at the

system

)wvu(2p (10)

)w()v()u(8q (11)

)uwwvvu(4)wvu(r 2 (12)

The above system can also be written as

2

pwvu (13)

4

r

16

puwwvvu

2

(14)


64

qwvu

2

(15)

Therefore the quantities u, v and w will be the roots of the third degree

equation

064

qz

4

r

16

pz

2

pz

2223

(16)

From this equation we obtain the values of the three variables u, v and w.

The roots of equation (3) will then be given by

wvuy

or, since )v()u(8

q)w(

by (11), the roots will be determined by

)v()u(8

qvuy

(17)

The above formula will provide us with the four roots of equation (3), under the

choice of signs: )( , )( , )( , )( .

The roots of the original equation (1) will be determined using relation (2).


0442x58x23x2x 234 (1)


2

1yx (2)


016

7625y80y

2

43y 24 (3)

We now search for a solution of equation (3) in the form

wvuy (4)

where for the moment any combination of signs is allowed.

Squaring both members of equation (4), we obtain


)]u()w()w()v()v()u[(2)wvu(y2

or

)]u()w()w()v()v()u[(2)wvu(y2 (5)

Squaring again equation (5) we arrive at

224 )wvu()wvu(y2y

)]wvu()w()v()u(2uwwvvu[4


y)w()v()u(8y)wvu(2y 24

0)uwwvvu(4)wvu( 2 (6)

Equating the corresponding coefficients of equations (3) and (6), we arrive at the

system

)wvu(22

43 (7)

)w()v()u(880 (8)

)uwwvvu(4)wvu(16

7625 2 (9)

The above system can also be written as

4

43wvu (10)

4

361uwwvvu (11)

100wvu (12)

Therefore the quantities u, v and w will be the roots of the third degree

equation

0100z4

361z

4

43z 23 (13)


The roots of this equation are given by 1,4

25,16 , which are the values of the

three variables u, v and w.

The roots of equation (3) will then be given by

wvuy

or, since )v()u(

10)w(

by (8), the roots will be determined by

)v()u(

10vuy

(14)

The above formula will provide us with the four roots of equation (3), under the

choice of signs: )( , )( , )( , )( .

We make the choice 16u and 1v .

The choice of signs )( gives us

i52

5

)i()i4(

10ii4

)v()u(

10vuy


i32

5

)i()i4(

10ii4

)v()u(

10vuy


i32

5

)i()i4(

10ii4

)v()u(

10vuy


i52

5

)i()i4(

10ii4

)v()u(

10vuy

Therefore the roots of equation (3) are given by i52

5 and i3

2

5 .

The roots of the original equation are then given by i53 and i32 .


10.4. The Simpson method.


0dxcxbxax 234 (1)


We write the above equation as

0)BAx(mx2

ax 2

22

(2)

Expanding the previous equation we find

0Bmx)AB2ma(xAm24

axax 2222

234

(3)

Comparing the coefficients of (1) and (3) we derive the system

bAm24

a 22

cAB2ma

dBm 22

This is a system of three equations with three unknown. This system is written as

bm24

aA

22 , dmB 22 , cmaAB2 (4)

From the above equations, since from the last equation 222 )cma(BA4 , we

obtain using the first two equations

222

)cma()dm(bm24

a4

which is a third degree equation with respect to m :

0cdadb4m)d4ca(2mb4m8 2223 (5)

Only one root has to be chosen from the three roots of (5) which will ensure the

compatibility of the system of equations (4). After determining m, A and B from

(4) and (5), equation (2) gives us the two equations


)BAx(mx2

ax2

(6)

which when solved, provide the four roots of the original equation (1).


0442x58x23x2x 234 (1)

Solution. We write the above equation as

0)BAx()mxx( 222 (2)

Expanding the previous equation we find

0Bmx)ABm(2x)Am21(x2x 222234 (3)

Comparing the coefficients of (1) and (3) we derive the system

23Am21 2 , 58AB2m2 , 442Bm 22

This is a system of three equations with three unknowns. This system is written as

22m2A2 , 442mB 22 , 29mAB (4)

From the above equations, since from the last equation 222 )29m(BA , we

obtain using the first two equations

22 )29m()442m()22m2(

which is a third degree equation with respect to m:

08883m942m23m2 23 (5)

The roots of the previous equation are 9,2

47,21 . It is only the value

2

47m

which leads to compatible equations of the system of equations (4).

This value of m, determines the values 5A and 2

21B (or the opposite).

Equation (2) then is written as

02

21x5

2

47xx

222

(6)

which is been split into the pair of equations


02

21x5

2

47xx2

, 0

2

21x5

2

47xx2

(7)

The roots of the first equation are i32 , while the second equation has roots

i53 . All these roots are the roots of equation (1).

10.5. The method of differential resolvents.

This method was explained previously in Section 9. For 4n , we have for the

fourth degree equation

0x3y4y4

the following expressions for the derivatives

xy

y

4

1

dx

dy

3

5

2

2

)xy(

y

16

1

dx

yd

5

5

3

3

)xy(

)x7y10(y

64

1

dx

yd

The differential resolvent

0ymdx

dym

dx

ydm

dx

yd322

2

13

3

takes on the form

0ymxy

y

4

1m

)xy(

y

16

1m

)xy(

)x7y10(y

64

1323

5

15

5

The previous equation is equivalent to the equation (removing denominators and

expanding)

513

61 y)10xm8m64(ym4

3232

423

21 y)xm10xm(64y)x7m16xm320xm4(

y)xm320xm64(y)xm640xm96( 43

32

233

22


0xm64xm16 53

42

In the above equation we substitute x3y4y4 and we derive the new equation

32321 y)xm640xm64m16(

23

33

221 y)40m256xm640xm96xm44(

y)x58m64xm1472xm40xm64xm320( 232

13

24

3

0x21xm48xm64xm16xm960xm12 22

53

42

23

31

Equating to zero all the coefficients of y, we obtain the over-determined system of

equations

0xm640xm64m16 2321

040m256xm640xm96xm44 33

32

21

0x58m64xm1472xm40xm64xm320 232

13

24

3

0x21xm48xm64xm16xm960xm12 22

53

42

23

31

The above system admits the unique solution

1x

x

2

9m

3

2

1

, 1x

x

16

43m

32

and 1x

1

32

5m

33

Therefore the differential resolvent has the form

0y1x

1

32

5

dx

dy

1x

x

16

43

dx

yd

1x

x

2

9

dx

yd332

2

3

2

3

3

or

0y5dx

dyx86

dx

ydx144

dx

yd)1x(32

2

22

3

33

The last equation admits the general solution

3

233

23 x;3

4,

3

2;

4

3,

2

1,

4

1FxBx;

3

2,

3

1;

12

5,

6

1,

12

1FAy


3

232 x;

3

5,

3

4;

12

13,

6

5,

12

7FxC

where A, B, C are constants.

Section 11

The Quintic

11.1. Forms of the quintic.

An equation of the fifth degree, usually called the quintic, has the general form

0fxexdxcxbxa 2345 (10.1)

The above form is also called the general quintic. The coefficients are either real

or complex numbers and 0a . Some other forms of the quintic equation, along

with their names, are the following:

The reduced quintic

0dxcxbxax 235 (10.2)

The principal quintic

0cxbxax 25 (10.3)

The Brioschi quintic (containing a single parameter Z )

0ZxZ45xZ10x 2235 (10.4)

The Bring-Jerrard quintic

0bxax5 (10.5)

Equations of fifth degree are not generally solved using formulas which contain a

finite sequence of algebraic operations like addition, subtraction, multiplication,

division and root extraction.


This is the Abel theorem (Stewart [60]) which asserts that we cannot find a

solution of this kind.

Closed-form solutions exist using special functions, like theta (Hermite [32],

Kiepert [40], King [41] and King and Canfield [42], Klein [43]) or

hypergeometric (Birkeland [6], Cockle [11], [12], Harley [31] and Glasser [25])

functions. There is also some classification of solvable quintics (Beversdorff [5],

Malfatti [48], Spearman and Williams [58], [59]) which are solved in the sense

referred previously (i.e. their solution contains a finite sequence of algebraic

operations like addition, subtraction, multiplication, division and root extraction).

11.2. Transformation of the General Quintic into the Principal Quintic.

We consider the general quintic

0NxMxCxBxAx 2345 (1)

and the quadratic Tschirnhaus transformation

bxaxy 2 (2)

and eliminate x between the two polynomial equations.

This can be done using the resultant of the two polynomials. The Sylvester Matrix

between the two polynomials is a 77 matrix given by

yba10000

0yba1000

00yba100

000yba10

0000yba1

NMCBA10

0NMCBA1

The determinant of the above matrix gives us the resultant of the two polynomials

which when equated to zero, gives a fifth degree equation:

0PyPyPyPyPy 542

33

24

15


where the coefficients 51 P,,P are polynomials given by

B2Ab5aAP 21

22222 BM2AC2b)B8A4(a)ABC3(baA4b10aBP

22222333 b)AB2(6a)ACM4(baA6baB3b10aCP

b)M2BAC2(3a)AM3N5CB(ba)C3AB(3 2

AN2CDB2 2

32333444 b)B2A(4a)AMN5(Aab4bCa2b5aMP

22222 a)AN4BM(ba)ABC3(3ba)M4AC(2bBa3

a)CMBN3(ab)N5BCAM3(2b)BAC2M2(3 22

CN2Mb)CBM2AN2(2 22

424322344555 b)AB2(aANbaBbCabaAbaMbaNP

32233 aBNba)CAM4(ab)C3AB(ba)N5AM(

2232 ab)AM3BCE5(ba)BMAN4(b)M2BAC2(

MNaab)BN3CM(b)BM2CAN2(aNC 222

22 Nb)MCN2(

We determine a and b such that 0P1 and 0P2 . Solving 0P1 with

respect to b, we find

)B2AaA(5

1b 2

Equation 0P2 , because of the above value of b, gives the equation

)B2AaA(

5

1aA4)B2AaA(

5

110aB 2

222

0BM2CA2)B2AaA(5

1)B8A4(a)BAC3( 222


which is equivalent to the following second order equation

a)C15AB13A4(a)A2B5( 322

0M10B3AC10A2BA8 242

The previous equation determines the value of the coefficient a. We may choose

any one of the two roots of the above equation. Using the value of b given by (),

we find that 3P , 4P and 5P can be expressed in terms of the coefficients A, B, C,

M and N and the coefficient a whose value is given by any root of equation ().

Conclusion. The general quintic (1) under the Tschirnhaus transformation (2)

and for the choice of a and b given by any root of equation () and ()

respectively, takes on the form

0PyPyPy 542

35

where 3P , 4P and 5P are given by (), () and () respectively, where the values of

the variables a and b are to be substituted by any root of equation () and ()

respectively.

11.3. Transformation of the principal quintic to the Bring-Jerrard

quintic.

We consider the principal quintic

0PyLyKy 25

and the quartic Tschirhaus transformation

dycybyayz 234

We shall eliminate the variable y from these two equations. The Sylvester matrix

is the following 99 matrix


zdcba10000

0zdcba1000

00zdcba100

000zdcba10

0000zdcba1

PLK001000

0PLK00100

00PLK0010

000PLK001

The resultant is the determinant of the Sylvester matrix, which when equated to

zero, gives

0SzSzSzSzSz 542

33

24

15

where

d5aK3L4S1

Ld16KLa5Pab5L6Kad12d10S 222

)Kb3La4P5(cKP4aK3bK3Lb2 2222

There are similar expressions for the other coefficient functions 3S , 4S and 5S but

there is no need to write them down.

We set 0S1 , which when solved with respect to d, gives

5

aK3L4d

We then put to zero the coefficient of c in the expression for 2S :

0Kb3La4P5

From the previous expression we solve with respect to b:

K3

La4P5b

When the values of d and b expressed by () and () respectively substituted in the

expression for 0S2 , we obtain the equation


a)KP375LK27PL400(a)K27KLP300L160( 232243

0LK18PK45LP250 2232

The above equation provides the value of the coefficient a (any root is

acceptable).

The coefficients 3S and 4S can then be expressed (after substituting b and d

into the original expressions) as

322

13

03 VcVcVcVS

432

23

14

04 WcWcWcWcWS

542

33

24

15

05 ZcZcZcZcZcZS

where the various V’s , W’s and Z’s are polynomials of K, L, P and a.

We now have the following two choices:

Choice 1. 0S3 leads us to the Bring-Jerrard quintic:

0SzSz 545

Equation 0S3 provides us with the value of c.

Choice 2. 0S4 leads us to the Euler-Jerrard quintic:

0SzSz 52

35

Equation 0S4 provides us with the value of c.

In performing the above calculations the reader can take advantage of any of the

known Computer Algebra Systems (e.g. “Maple” or “Mathematica”).

Example 1. Transform the general quintic

030x5x10x4x2x 2345 (1)

into the principal quintic.

Solution. We consider the quadratic Tschirnhaus transformation

bxaxy 2 (2)


and eliminate x between the two polynomial equations. The Sylvester matrix of

the polynomial equations is the following 77 matrix

yba10000

0yba1000

00yba100

000yba10

0000yba1

305104210

030510421

The determinant of the above matrix gives us the resultant of the two polynomials

which when equated to zero, gives a fifth degree equation:

0PyPyPyPyPy 542

33

24

15

where the coefficients 51 P,,P are polynomials depending on a and b.

We have

12b5a2)b,a(P1

66a22b48ab8a4b10)b,a(P 222

We do not write down explicitly the rest polynomials 43 P,P and 5P . Their

explicit expressions have appeared before.

Since we want to have 0)b,a(P1 and 0)b,a(P2 , we should solve the system

012b5a2 and 066a22b48ab8a4b10 22

Solving the first equation with respect to b and substituting to the second

equation, we arrive at the system

5

12a2b

and 03aa2 2

The second of the above equations admits the roots 1a and 2

3a .

We make the choice 1a . Therefore 2b . We are then being able to calculate

the rest polynomials. We find


50)2,1(P3 , 380)2,1(P4 and 1728)2,1(P5

Therefore the principal quintic has the form

01728y380y50y 25

Most of the above calculations have been performed using “Maple”.

Example 2. Transform the principal quintic

010y15y10y 25 (1)

into the Bring-Jerrard quintic and the Euler-Jerrard quintic.

Solution. We consider the quartic Tschirhaus transformation

dycybyayz 234 (2)

We shall eliminate the variable y from the last two equations. The Sylvester

matrix is the following 99 matrix

zdcba10000

0zdcba1000

00zdcba100

000zdcba10

0000zdcba1

101510001000

010151000100

001015100010

000101510001

The resultant is the determinant of the Sylvester matrix, which when equated to

zero, gives

0SzSzSzSzSz 542

33

24

15 (3)

where

60d5a30S1 (4)

222 b30ab50a750ad120d240d10S

)b30a6050(c1750b300a300 2 (5)


There are similar expressions for the other coefficient functions 3S , 4S and 5S but

there is no need to write them down at the moment.

Equating 1S to zero and solving with respect to d, we obtain

a612d (6)

Equating to zero the coefficient of c in the expression for 2S and solving with

respect to b, we obtain

a23

5b (7)

Substituting the above values of d and b into the equation 0S2 , we obtain the

following quadratic equation with respect to a:

02a7a3 2 (8)

The above equation admits the roots )737(6

1a . The choice )737(

6

1a

gives us the following expressions for the coefficients 3S , 4S and 5S :

c)73482539875(9

1c)11957385(

3

1c10S 23

3

73333527

647975 (9)

2344 c)73234557395(

3

1c)23973(

3

10c15S

)7314875045150020015(162

1c)6369857349795(

27

10 (10)

3455 c)732392837(

3

10c)731(15c10S

c)7312099725162134375(81

1c)532873732345(

27

10 2

)299122061773285513187(243

1 (11)

Taking 0S3 , we can determine c (any of the three roots) and then equation (3)

becomes


0SzSz 545 (12)

where 4S and 5S are given by (10) and (11) respectively, where c has been

substituted by any root of the equation 0S3 . Equation (12) is a Bring-Jerrard

quintic.

Taking 0S4 , we can determine c (any of the four roots) and then equation (3)

becomes

0SzSz 52

35 (13)

where 3S and 5S are given by (9) and (11) respectively, where c has been

substituted by any root of the equation 0S4 . Equation (13) is an Euler-Jerrard

quintic. Most of the above calculations have been performed using “Maple”.

11.4. Solution of the Bring-Jerrard quintic by Hermite’s method.

The Bring-Jerrard quintic has been solved by Hermite using the Modular

Functions. See C. Hermite: Oeuvres Vol.2 (1908) pp. 5-12 and H. T. Davis:

“Introduction to Nonlinear Differential and Integral Equations”, Dover 1962.

We first have the following Lemma:

Lemma. The Bring-Jerrard quintic 0qzpz5 can be transformed into the

equation 0mxx5 under the substitution xpz 4 .

Proof. In fact equation 0qzpz5 becomes

0qp)xp()xp( 454


0)p(

qxp

)p(

px

5454

45

0)p(

qx

)p(

px

5444

5


0)p(

qx

p

px

54

5

0mxx5

where

54 )p(

qm

We first define the functions

4 k)( and

4 k)( (1)

and then the function

5

164

5

16

5)5()(

5

163

5

162 (2)

It is then proved that the quantities

)( , )16( , )162( , )163( and )164( (3)

are the five roots of the quintic equation

0)](1[)()(52)()(52 81636 5164345 (4)

The substitution

x)()(52 44 5 (5)

reduces equation (4) into the equation

0mxx5 (6)

where

kk5

)k1(2

)()(5

)](1[2m

4 5

2

424 5

8

(7)

When m is given, then we conclude, using (7), that k can be determined from

the equation


0)k1(km5)k1(4 22522 (8)

To solve equation (8), we define m2

5A

4 5

and determine the angle from the

equation

2A

4sin (9)

The modulus k then has one of the following values

4

tank

, 4

2tan

,

4tan

,

4

3tan

(10)

Choosing any of the above values for the modulus, the roots of equation (6) are

given by

)(Bx1 , )16(Bx2 , )162(Bx3 ,

)163(Bx4 and )164(Bx5 (11)

where the coefficient B is given by

kk52

1

)()(52

1B

4 344 3

(12)

In compact form the roots are given by the formula

)()(52

)16k(x

44 3k

, 4,3,2,1,0k

Note. For numerical evaluation of the roots of the equation (6), we may use the

following expansion

)Q9Q8Q9Q8QQQ1(Q40)( 8765328 3 (13)

where

5 qQ , ieq ( q Jacobi’s nome) (14)

11.5. Solution of the Bring-Jerrard quintic by Glasser’s method.

The Bring-Jerrard quintic has also been solved recently by Glasser (M.L. Glasser:

“The Quadratic Formula Made Hard”, arXiv:math/9411224v1, Nov. 1994) by


hypergeometric functions, using a theorem due to Lagrange. Details are given in

Appendix IV. According to the general procedure introduced in Appendix IV, we

find that the roots of the equation 0txx5 are given by

)t(Ftx 21

)t(Ft32

5)t(Ft

32

5)t(Ft

4

1)t(Fx 4

33

2212

)t(Ft32

5)t(Ft

32

5)t(Ft

4

1)t(Fx 4

33

2213

)t(Ft32

5)t(Fti

32

5)t(Ft

4

1)t(Fix 4

33

2214

)t(Ft32

5)t(Fti

32

5)t(Ft

4

1)t(Fix 4

33

2215

where

4

341

t256

3125;

4

3,

4

2,

4

1

;20

11,

20

7,

20

3,

20

1

F)t(F ,

4

342

t256

3125;

4

5,

4

3,

4

2

;5

4,

5

3,

5

2,

5

1

F)t(F

4

343

t256

3125;

4

6,

4

5,

4

3

;20

21,

20

17,

20

13,

20

9

F)t(F ,

4

344

t256

3125;

4

7,

4

6,

4

5

;10

13,

10

11,

10

9,

10

7

F)t(F

11.6. Solution of the Bring-Jerrard quintic by the Cockle-Harley

method.

The Bring-Jerrard quintic has also been solved a long time ago by Cockle and

Harley (Cockle [11], [12], Harley [31]) using the method of differential

resolvents. Their solution is also expressed in terms of hypergeometric

functions.

Consider the Bring-Jerrard quintic:

0axx)x(f 5


and determine a function )a( such that

0))a((f , i.e. 0a)a()a( 5

0))a((fda

d , i.e. 0]a)a()a([

da

d 5

0))a((fda

d2

2

, i.e. 0]a)a()a([da

d 5

2

2

0))a((fda

d3

3

, i.e. 0]a)a()a([da

d 5

3

3

0))a((fda

d4

4

, i.e. 0]a)a()a([da

d 5

4

4

In other words we determine a function )a( such that

0a)a()a( 5

01da

d

da

d)a(5 4

0da

d

da

d)a(5

da

d)a(20

2

2

2

24

23

0da

d

da

d)a(5

da

d

da

d)a(60

da

d)a(60

3

3

3

34

2

23

32

2

2

23

2

222

4

da

d)a(60

da

d

da

d)a(360

da

d)a(120

0da

d

da

d)a(5

da

d

da

d)a(80

4

4

4

44

3

33

Eliminating )a( from the above equations we obtain the equation

2

22

3

33

4

44

da

d

77

a4875

da

d

231

a6250

da

d

1155

a3125256

0da

d

77

a2125


The general solution of the above equation is

4

34

5 32

4

345

1

a3125

256;

5

6,

5

1,

5

3

;10

9,

20

13,

5

2,

20

3

FaC

a3125

256;

5

4,

5

3,

5

2

;10

7,

20

9,

5

1,

20

1

FaC)a(

4

34

5 114

4

34

5 73

a3125

256;

5

8,

5

7,

5

6

;10

13,

20

21,

5

4,

20

11

FaC

a3125

256;

5

7,

5

6,

5

4

;10

11,

20

17,

5

3,

20

7

FaC

From the above solution we can determine the roots of the Bring-Jerrard quintic as

given by the formulas of the preceding section.

11.7. Solution of the Bring-Jerrard quintic by the Birkeland method.

The Bring-Jerrard quintic has also been solved by Birkeland:

R. Birkeland: “Über die Auflösung algebraischer Gleichungen durch

hypergeometrische Functionen” Mathematische Zeitschrift 26 (1927) 565-578

See also R. Birkeland: “Comptes Rendus” Vol. 171 (1920) 778-781, 1047-1049,

1370-1372 and Vol. 172 (1920) 309-311 and 1155-1158.

Errata Vol. 172 (1920) 188

It is remarkable that Lagrange’s theorem use by Glasser has also been used

previously by Birkeland.

The solutions of the equation

xgx5

can be expressed in terms of generalized hypergeometric functions )(F34 with

variable

5

4

4

5

g4

5

If 1|| the roots are given by


)(F2

5)(Fi

2

5)(F

4

1)(Fix 352

2

5101

)(F2

5)(F

2

5)(F

4

1)(Fx 3

3

522

5102

)(F2

5)(Fi

2

5)(F

4

1)(Fix 3

3

522

5103

)(F2

5)(F

2

5)(F

4

1)(Fx 3

3

522

5104

)(Fx 15

where

;4

3,

4

2,

4

1

;20

11,

20

7,

20

3,

20

1

F)(F 340 ,

;4

5,

4

3,

4

2

;5

4,

5

3,

5

2,

5

1

F)(F 341

;4

6,

4

5,

4

3

;20

21,

20

17,

20

13,

20

9

F)(F 342 ,

;4

7,

4

6,

4

5

;10

13,

10

11,

10

9,

10

7

F)(F 343

If 1|| the roots are given by

1

125

1

25

1

5

1y 3

5

114

25

73

15

32

05

1

1

1

125

1

25

1

5

1y 3

5

113

25

7

15

34

05

1

22

1

125

1

25

1

5

1y 3

5

112

25

74

15

3

05

1

33

1

125

1

25

1

5

1y 3

5

11

25

72

15

33

05

1

44


1

125

11

25

11

5

11y 3

5

11

25

7

15

3

05

1

5

where is a primitive root of 15 and

1;

5

2,

5

3,

5

4

;5

1,

20

9,

20

7,

20

1

F1

340 ,

1;

5

3,

5

4,

5

6

;5

2,

20

13,

10

9,

20

3

F1

341

1;

5

4,

5

6,

5

7

;5

3,

20

17,

20

11,

20

7

F1

342 ,

1;

5

6,

5

7,

5

6

;5

4,

20

21,

10

13,

20

11

F1

343

11.8. Solution of the Bring-Jerrard quintic by the Malfatti method.

We consider the Bring-Jerrard quintic (Malfatti [48], Bewersdorff [5])

0x5x5 ( 0 ) (1)

and suppose that the roots are given by

)nqpm(x k4k3k2k1k , 4,3,2,1,0k (2)

where

5

2sini

5

2cos

(3)

Expanding the product

0)xx()xx()xx()xx()xx( 54321 (4)

we derive the equation

2222235 x)qnpmpnqm(5x)pqmn(5x

x)qpnmqpnmpnqmqnpm(5 22223333

0)pnqmqnpm()qpnm(5qpnm 22225555 (5)

Comparing the coefficients of (1) and (5), we obtain


0qpnm (6)

0qnpmpnqm 2222 (7)

qpnmqpnmpnqmqnpm 22223333 (8)

)pnqmqnpm)(qpnm(5qpnm 22225555 (9)

We define the quantities y, r, and w by

nmqpy (10)

)qnpm(pnqmr 2222 (11)

qnpm 33 (12)

33 pnqmw (13)

The last equalities appearing in (10) and (11) follow because of (6) and (7).

Because of (12), (13) and (10), we obtain from (8)

2y3w (14)

Because of (10) and (11), we obtain from (9)

yr20qpnm 5555 (15)

We can easily check the following identities:

255 yry)nm(r (16)

and

255 yry)qp(wr (17)

Because of the last two identities, equation (15) becomes

yyr22)w(r 2 (18)

We can also check the following two identities:

4y4w (19)

and

y)w(r2 (20)

Therefore, using the above two identities, we find


622222224 y16y)w(yw4y)w(y)w(r (21)

Eliminating w between (14) and (21), we obtain the equation

22464 yy6y25r (22)

Eliminating w between (14) and (18), we obtain the equation

y)y25(r 2 (23)

Raising to the fourth power the above equation and using (22) to eliminate r, we

obtain the equation

44422242 y)y25()y6y25(y (24)

It is now apparent that 0y , since otherwise we would have 0 . Therefore


2442224 y)y25()y6y25( (25)


2y25z (26)


z)z()25z6z( 4422

which can further be written as

z)256()5z15z5z( 5423223 (27)

which is actually the (bicubic) resolvent of the original equation.

The bicubic resolvent cannot in general be solved in radicals. In case we can find a

value for z, it would then easily find the roots of the equation as follows:

From (26) we obtain

z5

1y (28)

From (23) we obtain

2y25

yr (29)


Solving the system of (14) and (20)

y

rw,y3w

22

we find

y2

ryy3 23 (30)

and

y2

ryy3w

23 (31)

From (10) and (16) we get the system

y

)y(rnm,ynm

255555 (32)

from which we conclude that 5m and 5n are the two roots of the equation

0yty

)y(rt 5

22

(33)

We then obtain that

5

222

55 y4y

)y(r

y

)y(r

2

1n,m (34)

and therefore

5 5

222

yy2

)y(r

y2

)y(rn,m

(35)

From (10) and (17) we get the system

y

)yw(rqp,yqp

255555 (36)

from which we conclude that 5p and

5q are the two roots of the equation


0yty

)yw(rt 5

22

(37)

We then obtain that

5

222

55 y4y

)yw(r

y

)yw(r

2

1q,p (38)

and therefore

5 5

222

yy2

)yw(r

y2

)yw(rq,p

(39)

Since we have determined m, n, p and q, we can find the roots of the equation

(1), using equations (2).

Example. Calculate the roots of the equation 012x5x5

Solution. In this case we have 1 and 12 . Equation (27) then becomes

z20480)5z15z5z( 223

One of the roots of the previous equation is 5z .

We then have

5

5

5

zy ,

5

52

y25

yr

2

10

5463

y2

ryy3 23

,

10

5463

y2

ryy3w

23

5

2

5 5

222

125

5

10

6554

10

6554y

y2

)y(r

y2

)y(rn,m

5

2

5 5

222

125

5

10

6554

10

6554y

y2

)yw(r

y2

)yw(rq,p

The roots are then given by

)nqpm(x k4k3k2k1k , 4,3,2,1,0k


where

5

2sini

5

2cos

11.9. The Kiepert Algorithm

The Kiepert algorithm (Kiepert [40], King [41]) is a general algorithm which

expresses the roots of the general quintic in terms of its coefficients. The steps of

the algorithm are the following:

Step 1. The general quintic 0ExDxCxBxAx 2345 is transformed

into the principal quintic 0czb5za5z 25 through a Tschirnhaus

transformation

Step 2. The principal quintic is transformed into the (single-parameter) Brioschi

quintic 0ZyZ45yZ10y 2235 through another Tschirnhaus

transformation

Step 3. The Brioschi quintic is transformed into the Jacobi sextic

05

sg12

s10

s22

236

which admits exact solutions through theta functions.

Details of the algorithm (using the symmetries of the Icosahedron) and some

solved examples will be published in an expanded version of this report.

11.9.1 Transforming the Brioschi quintic to the principal quintic

We shall transform the Brioschi quintic

0ZwZ45wZ10w 2235 (1)

into the principal quintic

0cbY5aY5Y 25 (2)

using the transformation

3wZ

wY

21

(3)

Introducing the notation


3wZ 21 (4)

we obtain

)3(Zw 2 (5)

Writing (1) as

2224 Z)Z45wZ10w(w

and using (5), we obtain

22222 Z)Z45)3(Z10)3(Z{w


1)244(w 2 (6)

Squaring the previous relation and using (5), we have

1)244()3(Z 22 (7)


0V405 345 (8)

where

Z

11728V (9)

We now obtain from (6)

244

1w

2 (9)


22

2

)244(

244w

or using (7),

)244()3(Z

)3(Z

1

244w 2

2



)7212(Zw 23 (10)

We then apply to (8) the transformation (3), written as, because of (4),

wY (11)

Using the usual notation for power sums of roots, we have for equation (8):

5s1 , 55s2 , 0ss 21 , V

120s 3 ,

V

20s 4 (12)

For the transformation (11), we obtain

0)60ss(ZY 12 (13)

0)5s(Z2Y 12 (14)

The above two relations mean that there are no terms in 4Y and 3Y in

transforming (1) into (2).

Using

)3(Z3)721(Z3Y 3232333

)162(Z 323terms in 1 and 2

and since a15Y3 , we obtain

ZZ728Va 3223 (15)

Similarly, since

)3(Z6)7212(Z4Y 4322433444

)96(Z)2161081(Z4 43244323

terms in 1 and

2

and b20Y4 , we obtain the equation

243224 Z27ZZ18Vb (16)

We similarly obtain the equation

2524235 ZZ45Z10Vc (17)


Equations (15), (16) and (17) provide us with the transformation of reducing

equation (1) into equation (2).

11.9.2 Transforming the Brioschi quintic to the principal quintic

We are now to solve the system

ZZ728Va 3223 (1)

243224 Z27ZZ18Vb (2)

2524235 ZZ45Z10Vc (3)

Z

11728V (4)

with respect to , , V and Z.

We do that using the following steps (Dickson [17]):

Step 1. Multiply (2) by and add to (3). We then obtain the relation

)ZZ728(ZV)cb( 32232

which can be written, because of (1), as VaZV)cb( 2 from which we have

aZcb 2 (5)

Step 2. Multiply (3) by and (2) by Z2 and subtract. We then derive the

equation

362422462 Z27Z27Z9V)bZc(

The right member of the previous equation is a perfect cube. We thus have the

equation

3222 )Z3(V)bZc( (6)

Step 3. Multiply (1) by and (2) by 8 and add. We then divide by the

resulting equation

23223 Z216Z9Z216

V)b8a(

We then square both members of the above equation and divide by Z:


Z

)Z216Z9Z216(

Z

V)b8a( 223223

2

22

(7)

Step 4. Square (1) and multiply by 27:

2322322 )ZZ728(27aV27 (8)

We then subtract from equation (8), equation (7):

23223

2

2222 )ZZ728(27

Z

V)b8a(aV27

Z

)Z216Z9Z216( 223223


2422424

62

2

22 Z46656Z155529

ZV

Z

)b8a(a27

2663642 Z271728Z46656Z27

which can also take the form

Z

11728Z9

Z

11728V

Z

)b8a(a27 2462

2

22

Z

11728Z27

Z

11728Z27 36242

or, using (4),

VZ27VZ27VZ9VVZ

)b8a(a27 362422462

2

22

or, in a more simplified form,

V)Z3(VZ

)b8a(a27 3232

2

22

(9)

Using equation (6), equation (9) can be expressed as

222

2

22 V)bZc(V

Z

)b8a(a27


or

bZcZ

)b8a(a27 2

2

22

(10)

Step 5. Eliminate the quantity Z2 between (5) and (10) and we obtain the

equation

a

)cb(bc

cb

)b8a(aa27

22

The above is a quadratic equation with respect to and can be written as

0cbca27ba64)cb2caba11()abcba( 2322223234 (11)

Step 6. We find Z2 from (5),

a

cbZ2 (12)

We then determine V from (6), using (12),

ba

cbc

a

cb3

bZc

)Z3(V

32

2

322

or

cba)bca(a

)c3b3a(V

222

32

(13)

Finally we determine Z from (4):

V1728

1Z

(14)

11.10. Solvable Quintics.

A quintic of the form

0bxax5

is solvable in terms of radicals if a and b have the form


1c

)c43(e5a

2

4

and

1c

)c211(e4b

2

5

where c and e are rational numbers, with 0c , 0e and 1 .

This is a result due to Spearman and Williams ([58] and [59]).

In this case the roots are given by

)qpnm(ex k4k3k2kk , 5,4,3,2,1k

where

5

2sini

5

2cos

The quantities m, n, p and q are given by

52

2

m

, 5

2

2

n

, 5

2

2

p

, 5

2

2

q

where

,

,

and

1c2

Example. The quintic 012x5x5 , corresponds to 1e , 2c and 1 .

We find, according to the above formulas, that 5 and

555 , 555

555 , 555

We also obtain

541010)52(102 , 541010)52(102

541010)52(102 , 541010)52(102

The quantities m, n, p and q are then evaluated to be


552

2

}5410)52{(5

2m

,

552

2

}5410)52{(5

2n

,

552

2

}5410)52{(5

2p

,

552

2

}5410)52{(5

2q

The roots are then given by

qpnmx k4k3k2kk , 5,4,3,2,1k

where

5

2sini

5

2cos

Note 1. It is not always possible to identify a solvable quintic. Given a and b, we

can in some cases determine c and e by solving the system

1c

)c43(e5a

2

4

and

1c

)c211(e4b

2

5

(1)

(I) For 1 , the system (1) of simultaneous equations can be solved with respect

to e and c in terms of a and b. In fact, dividing (1) by (2), we find

11c2

3c4

b5

ea4

(2)

Solving with respect to c we find from the previous equation

b20ea8

b15ea44c

(3)

Substituting the above value of c into the first equation of (1), we obtain

b5ae2

ae50e51

b20ea8

b15ea44a 4

2



0b5eab8ea16eb160ea64 22256 (4)

Equations (4) and (3) determine e and c, given a and b, in case where 1 .

Considering the equation 012x5x5 , i.e. 5a and 12b , equation (4)

takes on the form 09e6e5e24e4 256 which admits obviously one

solution 1e . From (3) we then obtain 2c .

(II) For 1 , the system (1) of simultaneous equations can be solved with

respect to e and c in terms of a and b. In fact, dividing (1) by (2), we find

c211

c43

b5

ea4

(5)

Solving with respect to c we find from the previous equation

b20ea8

b15ea44c

(6)

Substituting the above value of c into the first equation of (1), we obtain

b5ae2

ae50e51

b20ea8

b15ea44a 4

2


0b5eab8ea16eb160ea64 22256 (7)

Equations (7) and (6) determine e and c, given a and b, in case where 1 .

Note 2. G. P. Young in 1885 (Young [72]) has shown that all the irreducible

solvable quintics have the form

01n

)3n4)(1n2(m4x

1n

)3n4(m5x

2

5

2

45

where m and m are any rational numbers.


Section 12

Exercises

12. Exercises.

12.1. Third degree equations

Exercise 1. Solve the third degree equation

0126x59x4x 23

Answer. 2x1 , 7x2 , 9x3


0150x25x6x 23

Answer. 5x1 , 5x2 , 6x3


020x57x11x6 23

Answer. 3

1x1 ,

2

5x 2 , 4x3


018x51x11x10 23

Answer. 2

3x1 ,

5

2x2 , 3x3


028x)21(4x)21(2x 23

Answer. 22x1 , 51x2 , 51x3



027x9x3x 23

Answer. 3xx 21 , 3x3


020x16xx 23

Answer. 2xx 21 , 5x3


030x28x9x 23

Answer. 3x1 , i3x2 , i3x3


039xxx 23



012x)36(2x)132(x 23



012x)223(2x)21(2x 23



050x)i92(5x)i910(x2 23

Answer. i2

5x1 , i2x2 , 5x3


0i148x)i316(x)i1511(x3 23


Answer. 3

2x1 , i21x2 , i32x3


0ix)i34(x)i21(2x6 23

Answer. i2

1

2

1x1 , i

6

1

6

1x2 , 1x3

12.2. Fourth degree equations

Exercise 15. Solve the fourth degree equation

01394x138x5x2x 234

Answer. i35x 2,1 , i54x 4,3


021x26x18x6x 234

Answer. i21x 2,1 , i32x 4,3


052x28x7x2x 234

Answer. 2xx 21 , i23x 4,3


020x22x2x5x 234

Answer. 2x1 , 1x2 , i3x 4,3


018x39x46xx6 234

Answer. 2x1 , 3x2 , 2

3x3 ,

3

1x4


Appendix I.

The solutions of the equation az3 when the argument of the complex number a

is not a known angle, can be determined by a pocket calculator. However there are

cases where we want to have an exact solution. In this case we try an algebraic

way of solving the equation.

Suppose that we want to find the cubic root of the complex number bia . We let

yixbia3 (I.1)

Raising to the third power of both members of the above equation, we obtain

bia)yix( 3 (I.2)

where a and b are known real numbers. We have to determine the real numbers

x and y. Since

)yyx3(i)yx3x()yix( 32233 (I.3)

we have the equation

bia)yyx3(i)yx3x( 3223 (I.4)

from which we obtain the system (equating the real and imaginary parts)

byyx3

ayx3x32

23

(I.5)

Since this is a homogeneous (third degree) system, we consider the substitution

xλy (I.6)

The equations of the system take the form

b)λλ3(x

a)λ31(x

bxλxλ3

axλ3x33

23

333

323

(I.7)

Dividing the two equations we obtain the equation

a

b

λ31

λλ3

2

3

(I.8)

which is equivalent to the third degree equation


0μλ3λμ3λ 23 (I.9)

where

a

bμ (I.10)


μρλ (I.11)

equation (I.9) takes on the form

0)1(2)1(3 223 (I.12)

Under the further substitution (Vieta substitution)

z

μ1zρ

2 (I.13)

the previous equation becomes

0133z)1(2z 246326 (I.14)

The substitution

3zw (I.15)

converts equation (I.14)) to the equation

0133w)1(2w 24622 (I.16)

which is a quadratic equation. The discriminant of this equation is

22 )1(4D (I.17)

Therefore the roots are given by

)μ1(i)μ1(μw 222,1 (I.18)

or

)i()1(w 22,1 (I.19)


)i()1(wz 22,1

3 (I.20)


In order to have exact solution, the quantity )i()1( 2 should be a perfect

cube.

The reader has already noticed the complexity of the calculations which may lead

to nowhere. Equations (I.9) or (I.20) might be harder than solving the original

qubic. Our algebraic method of solutions thus avoids those vicious circle

situations. Using the procedure of this Appendix, there is no way of calculating the

cubic roots of complex numbers considered in Section 2.

Appendix II. The Sylvester Matrix.

The Sylvester Matrix of two polynomials

m

0k

kkxa)x(A and

n

0k

kk xb)x(B

is a )nm()nm( matrix M defined by

01n

011nn

011nn

0m

011mm

011mm

bbb

bbbb

bbbb

aa

aaaa

aaaa

M

All the elements of the above matrix not appearing explicitly are zero.

The determinant of the Sylvester matrix is called the resultant of the two

polynomials )x(A and )x(B :

)Mdet())x(B),x(A(res

Example 1. The Sylvester Matrix of the polynomials

1x6x5x2)x(A 23 and 4x2x3)x(B 2

is the 55 matrix M given by


42300

04230

00423

16520

01652

M

The resultant of the polynomials is

499)Mdet())x(B),x(A(res

Maple support:

> with(LinearAlgebra):

> A:=2*x^3-5*x^2+6*x-1; := A 2 x3 5 x2 6 x 1

> B:=3*x^2-2*x+4; := B 3 x2 2 x 4

> M:=SylvesterMatrix(A,B,x);

:= M

2 -5 6 -1 0

0 2 -5 6 -1

3 -2 4 0 0

0 3 -2 4 0

0 0 3 -2 4

> Determinant(M); 499

> resultant(A,B,x); 499

Example 2. The Sylvester Matrix of the polynomials

1xy3xy5xy2)x(A 3232 and y3xy2xy5)x(B 223

considered as polynomials of x, is the 55 matrix M given by

y3y2y500

0y3y2y50

00y3y2y5

1y3y5y20

01y3y5y2

M

23

23

23

32

32


The resultant of the polynomials is

)Mdet())y,x(B),y,x(A(resx

781091113 y837y645y162y125y150y675

Maple support:

> with(LinearAlgebra):

> A:=2*y^2*x^3-5*y*x^2-3*y^3*x+1; := A 2 y2 x3 5 y x2 3 y3 x 1

> B:=5*y^3*x^2-2*y^2*x-3*y;

:= B 5 y3 x2 2 y2 x 3 y

> M:=SylvesterMatrix(A,B,x);

:= M

2 y2 5 y 3 y3 1 0

0 2 y2 5 y 3 y3 1

5 y3 2 y2 3 y 0 0

0 5 y3 2 y2 3 y 0

0 0 5 y3 2 y2 3 y

> Determinant(M); 837 y7 654 y8 162 y10 675 y13 150 y11 125 y9

> resultant(A,B,x); y3 ( ) 675 y10 150 y8 162 y7 125 y6 654 y5 837 y4

Appendix III. Vieta’s formulas, symmetric polynomials

and power sums.

A given polynomial

n1n1n

1n axaxax)x(f

(III.1)

of degree n admits, according to the basic theorem of the algebra, n roots

n21 ,,,

and has the following factorization

)x()x()x()x(f n21 (III.2)

Upon expanding the right hand side of the previous identity, and comparing the

derived expression with the original expression, we obtain


)(a n211

n1n32n131212a

)(a n1n2n4213213

…………………………………………………………………..

)()1(a n32n2n211n211n

1n

n211n

n )1(a

The above formulas can also be written in short-hand notation as

11a

212a

3213a

……………………………….

1n211n

1n )1(a

n211n

n )1(a

and are called Vieta’s formulas.

For 3n for example, we have

)(a 3211 , 3231212a , 3213a

Symmetric Polynomials. We define the polynomials

n21 ,,,

called fundamental symmetric polynomials, by

n211 xxx

n1n32n131212 xxxxxxxxxx

n1n2n4213213 xxxxxxxxx

…………………………………………………………

n32n2n211n211n xxxxxxxxxx

n21n xxx


defined out of the n variables n21 x,,x,x .

We have the following fundamental theorem on symmetric polynomials:

Any symmetric polynomial in the unknowns n21 x,,x,x over the field P is a

polynomial in the elementary symmetric polynomials n21 ,,, with

coefficients in P.

Power Sums. In many applications we have to calculate expressions like power

sums of the form

kn

k2

k1k xxxs , ,3,2,1k

In all these cases this kind of expressions can be expressed in terms of the

fundamental symmetric polynomials n21 ,,, .

It is obvious that 11s . Considering the obvious relations

21k

1k11k xxss

322k

121k

122k xxxxxs

……………………………………………………………………

1ii2ik

1i21ik

1iik xxxxxxxs

( 2ki2 )

……………………………………………………………………

k1k2211k1 kxxxs

and taking the sum with alternating signs and transposing all the terms to one side,

we obtain the formula

0k)1(s)1(sss kk

1k11k

22k11kk

( nk ) (III.3)

For nk we consider the relations

21k

1k11k xxss

322k

121k

122k xxxxxs

……………………………………………………………………

1ii2ik

1i21ik

1iik xxxxxxxs

( 1ni2 )


……………………………………………………………………

n21nk

1nnk xxxs

From the previous relations we get the following formula

0s)1(sss nnkn

22k11kk ( nk ) (III.4)

The last two formulas (III.3) and (III.4) are due to Newton.

Example 1. We shall calculate 2s , 3s and 4s in terms of symmetric polynomials

for 3n .

For 1k , we have 11s .

For 2k we have from (III.3) 02ss 2112 and since 11s , we obtain

2212 2s

For 3k we have from (III.3) 03sss 321123 and using the previous

expression for 2s , we obtain

3213132112

213 333)2(s

For 4k we have from (III.4) 0ssss 3122134 and using the

previous expressions for 2s and 3s , we obtain

3122211321

314 )2()33(s

or

22312

21

414 244s

Example 2. We shall calculate 2s , 3s , 4s and 5s in terms of symmetric

polynomials for 5n .

For 1k , we have 11s .

For 2k we obtain 2212 2s

For 3k we have 321313 33s

For 4k we have 422312

21

414 4244s


For 5k we have )(5s 532411223

212

31

515

For the quintic

0axaxaxaxax 012

23

34

45

or

0xxxxx 542

33

24

15

we have the recursive formula

j5

1n

1jjnn5n asans

, 0a j for 0j

Using this formula we obtain

41 as , 3242 a2as , 243

343 a3aa3as ,

12342

243

444 a4a2aa4aa4as

)aaaaaaaaaaa(5as 03241423

242

343

544

Appendix IV. The Discriminant of a Polynomial.

The discriminant of a polynomial

n1n1n

1n

0 axaxaxa)x(f (IV.1)

with roots n21 ,,, is defined by

1jin

2ji

2

)1n(n

)()1(D (IV.2)

It is proved that the discriminant can be expressed in terms of the resultant of

)x(f and its derivative )x(f :

Da)1())x(f),x(f(R 02

)1n(n

(IV.3)

There is also another formula expressed as determinant of the symmetric

polynomials n21 ,,, of the roots n21 ,,, :


2n21nn1n

1n432

n321

1n21

2n20

ssss

ssss

ssss

sssn

aD

Example 1. Let cxbxa)x(f 2 . Then bxa2)x(f . The Sylvester

matrix of )x(f and )x(f is the following 33 matrix

ba20

0ba2

cba

M

The determinant of the Sylvester matrix is the resultant:

22 baca4)Mdet())x(f),x(f(R

Since 2n , we have 12

)1n(n

and thus because of (IV.3),

Da)1(baca4 22

from which we get

ca4b)baca4(a

1D 222

which is the known formula for the discriminant of the quadratic trinomial.

Example 2. Let qxpx)x(f 3 . Then px3)x(f 2 . The Sylvester matrix

of )x(f and )x(f is the following 55 matrix

p0300

0p030

00p03

qp010

0qp01

M

The determinant of the Sylvester matrix is the resultant:

23 q27p4)Mdet())x(f),x(f(R


Since 3n , we have 32

)1n(n

and thus because of (IV.3),

D)1(q27p4 23

from which we get

27

p

4

q108)q27p4(D

3223

Example 3. Let cxbxax)x(f 23 be the complete cubic polynomial.

We then have

432

321

21

sss

sss

ss3

D

We have

as 11

b2a2s 22

212

c3ab3a33s 3321

313

and

22422312

21

414 b2ca4ba4a244s

Therefore

234

21

3232142 s3ssssss2ss3D

23322 c27abc18ca4b4ba

For 0a , we find 23 c27b4D thus recovering the result we found

previously in Example 2.

Appendix V. Glasser’s Method.

Glasser’s method is based on a theorem proved by Lagrange. We first state

Lagrange’s Theorem (see E. T. Whittaker and G. N. Watson: “A Course of

Modern Analysis”, Ch.7, section 7.32):


Let )z(f and )z( be functions of z analytic on and inside a contour C

surrounding a point a and let t be such that the inequality

|az||)z(t|

is satisfied at all points z on the perimeter of C; then the equation

)(ta

Regarded as an equation in , has one root in the interior of C; and further any

function of analytic on and inside C can be expanded as a power series in t

by the formula

])}a(){a(f[da

d

!n

t)a(f)(f n

1n

1n

1n

n

The theorem belongs to the Theory of Analytic Functions.

Consider now the equation

0txx N ( ,5,4,3,2N ) (V.1)


1N

1

x

(V.2)

equation (V.1) becomes

0t1N

1

1N

N

from which, multiplying by 1N

N

, we derive the equation

0t1 1N

N

(V.3)


)(te i2 (V.4)

where

1N

N

)( (V.5)


According to Lagrange’s theorem, for any function analytic in a neighborhood of a

root of (V.4), we have the expansion

i2ea

n

1n

1n

1n

ni2 ])}a(){a(f[

da

d

!n

t)e(f)(f

(V.6)

We apply the above expansion for the function

1N

1

)(f

(V.7)

We have

1N

N)1n(

n a1N

1)}a(){a(f

(V.8)

We thus have

i2ea

1N

N)1n(

1n

1n

1n

ni2 a

da

d

!n

t

1N

1)e(f)(f

(V.9)

Using the formula

kpp

k

k

x)1kp(

)1p()x(

dx

d

we have

1N

1n

1N

N)1n(

1n

1n

a

11N

1n

11N

N)1n(

ada

d

Therefore equation (IV.9) becomes

i2ea

1N

1n

1n

ni2 a

11N

1n

11N

N)1n(

!n

t

1N

1)e(f)(f

Changing the summation from 1n to 0n (i.e. n1n ), we obtain from the

above equation


i2ea

1N

n

0n

1ni2 a

11N

n

11N

Nn

!)1n(

t

1N

1)e(f)(f


i2ea

1N

n

0n

ni2 a

11N

n

11N

Nn

)2n(

t

1N

t)e(f)(f

or finally

0n

n

1N

i2

i2 et

11N

n)2n(

11N

Nn

1N

t)e(f)(f (V.10)

Therefore the roots are given by

0n

n

1N

ik2

1N

ik2

k et

11N

n)2n(

11N

Nn

1N

tex (V.11)

1N,,3,2,1k

while Nx is evaluated using the formula

0xN

1kk

(V.12)

In case of two roots (quadratic trinomial) we use the formula 1xx 21 .

The series (V.11) can be simplified further using Gauss multiplication theorem.

The summation of series gives us the generalized hypergeometric functions.

Defining )q(n by


2N

0k

1N

0k1N

qN

q

1N

in2

n

1N

2kq1

1N

q

N

1k

1N

q

N1N

te)q(

1N

0k

1N

qN

q

1N

in2

1N

kq

N

1k

1N

q

N1N

te (V.13)

we have the following formulas for the roots of the equation 0txx N :

2N

0qN1Nn2

1N

in2

n )(F)q()1N(2

N

)1N(

tex (V.14)

1N,,3,2,1n

2N

0qN1Nm

1N

1m2N )(F)q(

)1N(2

N

)1N(

tx (V.15)

where

N

1N

1N

in2

N1NN1N

N1N

te

;1N

1Nq,

1N

Nq,,

1N

2q

;1,1N

1Nq,,

)1N(N

1NqN

F)(F

(V.16)

is the generalized hypergeometric function (see for example E. D. Rainville:

“Special Functions”, Chelsea 1971).


Bibliography

[1] M. Abramowitz and I. Stegun: “Handbook of Mathematical Functions”

Dover 1972

[2] V. Adamchik and D. J. Jeffrey: “Polynomial Transformations of

Tschirnhaus,Bring and Jerrard”

ACM SIGSAM Bulletin, Vol 37, No 3, September 2003

[3] R. S. Ball: “Note on the algebraical solution of biquadratic equations”

Quart. J. of Pure and Applied Math. 7 (1866) 6-9 and 358-369

[4] B. Berndt, B. K. Spearman and K. S. Williams: “Commentary on an

unpublished lecture by G.N. Watson on solving the quintic”

Math. Intell. 24 (2002) 15-33

[5] J. Bewersdorff: “Galois theory for beginners: A historical perspective”

AMS 2006

[6] R. Birkeland: “Über die Auflösung algebraischer Gleichungen durch

hypergeometrische Functionen”

Mathematische Zeitschrift 26 (1927) 565-578

[7] E. S. Bring: “Meletemata quaedam mathematematica circa

transformationem Aequationum algebraicarum” Lund 1786

[8] E. S. Bring: Quart. J. Math. 6 (1684)

[9] F. Cajori: “An Introduction to the Modern Theory of Equations”

Macmillan, 1919

[10] A. Cayley: “On Tschirnhausen’s Transformation”

Phil. Trans. Roy. Society London 151 (1861) 561

[11] J. Cockle: “Sketch of a Theory of Transcendental Roots”

The London, Edinburgh and Dublin Philosophical Magazine and Journal of

Science 20 (1860) 145-148

[12] J. Cockle: “Note on Transcendental Roots”


[13] N. B. Conkwright: “Introduction to the Theory of Equations”

Ginn and Co. Boston 1941

[14] H. T. Davis: “Introduction to Nonlinear Differential and Integral

Equations”. Dover 1962

[15] E. Dehn: “Algebraic Equations”. Columbia University Press, 1930

[16] L.E. Dickson: “First Course in the Theory of Equations” Wiley, N.Y. 1922

[17] L.E. Dickson: “Modern Algebraic Theories”. B.H. Sanborn & Co. 1926

[18] L.E. Dickson: “Algebraic Invariants”

Wiley, N.Y. 1914 and Chapman and Hall, London 1914

[19] L.E. Dickson: “Linear Groups with an Extension of the Galois Field

Theory”. Teubner, Leipzig 1901

[20] R. J. Drociuk: “On the complete solution to the most general fifth degree

polynomial” arXiv:math.GM/0005026, May 2000

[21] D. S. Dummit: “Solving Solvable Quintics”.

Mathematics of computation 57 (1991) 387-401 and 59 (1992) 309

[22] D. S. Dummit and R. M. Foote: “Abstract Algebra”

Prentice-Hall, N.Y. 1990

[23] S. Gabelli: “Teoria delle Equazioni e Teoria di Galois”

Springer-Verlag Italia, Milano 2008

[24] R. Garver: “On the transformation which leads from the Brioschi quintic

to a general principal quintic”. Presentation to the AMS, 1930

[25] M. L. Glasser: “The quadratic formula made hard”

arXiv:math/9411224v1, 16 Nov 1994

[26] O. E. Glenn: “A Treatise on the Theory of Invariants”

Ginn and Company, Boston 1915

[27] P. Gordan: “Über die Auflösung der Gleichungen vom Fünften Grade”

Math. Annalen 13 (1878) 375-404


[28] H. Hall and S. Knight: “Higher Algebra” ,Macmillan and Co. 1955

[29] W. R. Hamilton: “Inquiry into the validity of a method recently proposed

By George B. Jerrard, esq., for transforming and resolving equations of

elevated degrees”. British Association Report, Bristol 1836, pp 295–348.

Report available online:

http://www.maths.tcd.ie/pub/HistMath/People/Hamilton/Jerrard/

[30] W. R. Hamilton: “Theorems respecting algebraic elimination”

Philosophical Magazine (3rd series):

vol. viii (1836), pp. 538–543, vol. ix (1836), pp. 28-32.

[31] R. Harley: “On the Theory of Transcendental Solution of Algebraic

Equations”. Quart. J. Pure Appl. Math 5 (1862) 337-361

[32] C. Hermite: Oeuvres Tome 2, pp.5-12, Gauthier-Vilars 1905

[33] W. Heymann: “Studien über die Transformation und Integration der

Differential und Differenzengleichungen” . Leipzig 1891

[34] W. Heymann: Zeitschrift für Mathematik und Physik 39 (1894) 162-182

[35] D. Hilbert: “Theory of Algebraic Invariants”. CUP 1993

[36] G. B. Jerrard: “An essay on the resolution of equations”

Taylor and Francis, London 1859

[37] G. B. Jerrard: “On the possibility of solving equations of any degree

however elevated” Phil. Mag. Ser. 4, 3 (1852) 457-460

[38] R. S. Johnson: “The solution of cubic and quartic equations”

Notes, School of Mathematics and Statistics, Newcastle University

[39] A. Kaw: “Exact Solution to Cubic Equations”

School of Engineering, USF, July 2009

[40] L. Kiepert: “Auflösung der Gleichungen Fünften Grades”

J. für Math 87 (1878) 114-133

[41] B. R. King: “Beyond the Quartic Equation” Birkhäuser 1996


[42] B. R. King and E. R. Canfield: “An algorithm for calculating the roots

of a general quintic equation from its coefficients”,

J. Math. Phys 32(1991) 823

[43] F. Klein: “Lectures on the Icosahedron”. Dover 1956

[44] G. A. Korn and T. M. Korn: “Mathematical Handbook for Scientists and

Engineers”, McGraw-Hill, 1961

[45] A. Kurosh: “Higher Algebra” Mir Publishers, Moscow 1984

[46] D. Lazard: “Solving quintics in radicals”

Olav Arnfinn Laudal and Ragni Piene (eds)

“The Legacy of Niels Henrik Abel” pp. 207-225

[47] S. L. Loney: “Plane Trigonometry”, Parts I and II

Cambridge University Press 1964

[48] G. Malfatti: “De aequationibus quadrato-cubicis dissertation analitica”

Atti dell’ Accademia delle scienze di Siena, Vol. IV (1771) 129-185

[49] S. Mukai: “An Introduction to invariants and moduli”

Cambridge University Press 2003

[50] B. Netto: “Elementare Algebra”. Teubner 1904

[51] B. Netto: “Vorlesungen über Algebra”.

Vol.1, Teubner 1896, Vol.2, Teubner 1900

[52] B. Netto: “The Theory of Substitutions and its Applications to Algebra”

F. N. Cole, translator. The Inland Press, 1892

[53] O. Perron: “Algebra”. Vol.1 and Vol.2, Third Edition, de Gruyter,

Berlin 1951

[54] G. Pierre et B. Sébastien: “Résolution de l’équation du cincuième degré”

Preprint

[55] C. Runge: “Über die Auflösung der Gleichungen vom der form

0xux5 ”. Acta Math. 7 (1886) 173-186

[56] J.-A. Serret: “Cours d’Algèbre Supérieure”, Gauthier-Villars 1866, Vol. 1


[57] J. Shurman: “Geometry of the Quintic”, Wiley 1997

[58] B. Spearman and K. S. Williams: “Characterization of solvable quintics

bxax5 ” American Mathematical Monthly 101 (1994) 986-992

[59] B. Spearman and K. S. Williams: “Conditions for the Insolubility of the

quintic equation bxax5 ” Preprint

[60] I. Stewart: “Galois Theory” Chapman and Hall, London 1973

[61] J. Stillwell: “Elements of Algebra”. Springer 1994

[62] J. Swallow: “Exploratory Galois Theory”. CUP 2004

[63] J-P. Tignol: “Galois’ Theory of Algebraic Equations”.

World Scientific 2001

[56] I. Todhunter: “An Elementary Treatise on the Theory of Equations”

MacMillan 1875

[64] E. W. von Tschirnhaus: “A method of removing all intermediate terms

from a given equation”

ACM SIGSAM Bulletin, Vol 37, No 1, March 2003

Translation of the original paper:

“Methodus auferendi omnes terminus intermedios ex data equatione”

Acta Eruditorium 2 (1683) 204-207

[65] H. W. Turnbull: “Theory of Equations”.

Oliver and Boyd, Fourth Edition, 1947

[66] J. V. Uspensky: “Theory of Equations”. McGraw Hill 1948

[67] J. Villanueva: “The Cubic and Quartic Equations in Intermediate Algebra

Courses” Notes, Florida Memorial College

[68] H. Weber: “Lehrbuch der Algebra” Vols. I, II, III.

Braunschweig 1898, 1896, 1908

[69] Wikipedia: “Bring radical” available online

[70] Wikipedia: “Quintic function” available online


[71] Wolfram Research: “Solving the Quintic”, poster available online

http://library.wolfram.com/examples/quintic/

[72] G. P. Young: “Solvable Quintic Equations with Commensurable

Coefficients”, American Journal of Mathematics 10 (1888) 99-130

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