Download - 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.
S. Antoniou: Cubic, Quartic and Quintic 3
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
S. Antoniou: Cubic, Quartic and Quintic 4
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)
S. Antoniou: Cubic, Quartic and Quintic 5
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
S. Antoniou: Cubic, Quartic and Quintic 6
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
S. Antoniou: Cubic, Quartic and Quintic 7
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(
S. Antoniou: Cubic, Quartic and Quintic 8
despite the fact that the argument of the number 66i28 cannot be
determined explicitly, since equation 28
66xtan ,
x
2 cannot be solved
analytically.
Example 3. The number i2257 has three cubic roots given by
i25 , )i25( and 2)i25(
despite the fact that the argument of the number i2257 cannot be
determined explicitly, since equation 57
22xtan , 0x
2
cannot be solved
analytically.
Example 4. The number i312234 has three cubic roots given by
i322 , )i322( and 2)i322(
despite the fact that the argument of the number i312234 cannot be
determined explicitly, since equation 234
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
S. Antoniou: Cubic, Quartic and Quintic 9
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
S. Antoniou: Cubic, Quartic and Quintic 10
}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
S. Antoniou: Cubic, Quartic and Quintic 11
)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
which is equivalent to
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).
S. Antoniou: Cubic, Quartic and Quintic 12
Exercise 1. Show that the cubic roots of the number i946 are given by
i32 , )i32( and 2)i32(
Exercise 2. Show that the cubic roots of the number i236115 are given by
i45 , )i45( and 2)i45(
Exercise 3. Show that the cubic roots of the number i2618 are given by
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
which can be written as
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)
S. Antoniou: Cubic, Quartic and Quintic 13
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
which can be written as
z)zyyx3(i)zyx3x(i384405 3223
From the above identity we get the system
3z
84zyyx3
405zyx3x32
23
(2.21)
Multiplying the identities
zyixi3844053
and zyixi3844053
we obtain
)zyix()zyix()i384405()i384405(3
which is equivalent to the equation
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
S. Antoniou: Cubic, Quartic and Quintic 14
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)
Example 2.3. Find the cubic roots of the complex number i6628
Solution. Consider the identity
S. Antoniou: Cubic, Quartic and Quintic 15
zyix66i283
(2.28)
Taking the cube of both members of the above identity, we obtain
3)zyix(66i28
which can be written as
z)zyyx3(i)zyx3x(66i28 3223
From the above identity we get the system
6z
6zyyx3
28zyx3x32
23
(2.29)
Multiplying the identities
zyix66i283
and zyix66i283
we obtain
)zyix()zyix()66i28()66i28(3
which is equivalent to the equation
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
S. Antoniou: Cubic, Quartic and Quintic 16
)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 .
Example 2.4. Find the cubic roots of the complex number i6628
Solution. Consider the identity
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
S. Antoniou: Cubic, Quartic and Quintic 17
)66i28(27
1
9
62i
27
28
will be the number )6i2(3
1 .
Example 2.5. Find the cubic roots of the complex number i59476
Solution. Consider the identity
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 .
Example 2.6. Find the cubic roots of the complex number i52781024175
Solution. Consider the identity
zyixi527810241753
S. Antoniou: Cubic, Quartic and Quintic 18
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 .
Application 2.6. Suppose that
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( .
Exercise 2. Show that the cubic roots of the complex number i345154 are
given by i332 , )i332( and 2)i332( .
Exercise 3. Show that the cubic roots of the complex number i33081 are
given by i323 , )i323( and 2)i323( .
S. Antoniou: Cubic, Quartic and Quintic 19
Exercise 4. Show that the cubic roots of the complex number i59476 are
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( .
Exercise 6. Show that the cubic roots of the complex number i52218 are
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)
Taking the cube of both members of the above identity, we obtain
3)ziyx(ciba
which can be written as
)zyzx3(iy)xz3yx(ciba 3223 (2.24)
From the above identity we obtain the system
{ azx3yx 23 , by , czyzx3 32 } (2.25)
Multiplying the identities
ziyxciba3
and ziyxciba3
we obtain
)ziyx()ziyx()ciba()ciba(3
which is equivalent to
3 2222 cbazyx (2.26)
S. Antoniou: Cubic, Quartic and Quintic 20
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 .
Solution. Consider the identity
ziyxi22573
(2.27)
Taking the cube of both members of the above identity, we obtain
3)ziyx(i2257
which can be written as
)zyzx3(iy)xz3yx(i2257 3223 (2.28)
From the above identity we obtain the system
{ 7zx3yx 23 , 5y , 22zyzx3 32 } (2.29)
Multiplying the identities
ziyxi22573
and ziyxi22573
we obtain
)ziyx()ziyx()i2257()i2257(3
which is equivalent to
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(
S. Antoniou: Cubic, Quartic and Quintic 21
Exercises.
Exercise 1. Show that the cubic roots of the complex number i81330 are
given by i332 , )i332( , 2)i332(
Exercise 2. Show that the cubic roots of the complex number i176556 are
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
which can be written as
w)yzx3wz(iy)wxz3yx(dciba 2323 (2.32)
From the above identity we obtain the system
}dw,cyzx3wz,by,awxz3yx{ 2323 (2.33)
Multiplying
wziyxdciba3
and wziyxdciba3
we obtain
)wziyx)(wziyx()dciba)(dciba(3
which is equivalent to
3 2222 dcbawzyx (2.34)
S. Antoniou: Cubic, Quartic and Quintic 22
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.
Example 2.8. Find the cubic roots of the complex number i312234
Solution. Consider the identity
wziyx312i2343
(2.35)
Taking the cube of both members of the previous identity we find
3)wziyx(312i234
which can be written as
w)yzx3wz(iy)wxz3yx(312i234 2323 (2.36)
From the above identity we obtain the system
}3w,12yzx3wz,2y,34wxz3yx{ 2323 (2.37)
Multiplying
wziyx312i2343
and wziyx312i2343
we obtain
)wziyx()wziyx()312i234)(312i234(3
which is equivalent to
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(
S. Antoniou: Cubic, Quartic and Quintic 23
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(
which is equivalent to
i)31220()31220(
z)uwy2uxyx(3)wx3x( 2223
S. Antoniou: Cubic, Quartic and Quintic 24
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
Multiplying the identities
i)zuw()zyx()i1()31220(3
and
i)zuw()zyx()i1()31220(3
we obtain
223 2 )zuw()zyx()11()31220(
which is equivalent to
zwu2zuwzxy2zyx39601664 22223
and this to
S. Antoniou: Cubic, Quartic and Quintic 25
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 .
Application 2.9. Suppose that
)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
S. Antoniou: Cubic, Quartic and Quintic 26
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
which can be written as
z)zyyx3()zyx3x(cba 3223 (2)
From the above identity we obtain the system
}cz,bzyyx3,azyx3x{ 3223 (3)
On the other hand, multiplying the identities
zyxcba3
and zyxcba3
we find
)zyx()zyx()cba()cba(3
which is equivalent to
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)
S. Antoniou: Cubic, Quartic and Quintic 27
Taking the cube of the previous identity, we find
3)zyx(339156344
which can be written as
z)zyyx3()zyx3x(339156344 3223 (2)
From the above identity we obtain the system
}3z,3915zyyx3,6344zyx3x{ 3223 (3)
On the other hand, multiplying the identities
zyx3391563443
and zyx3391563443
we find
)zyx()zyx()339156344()339156344(3
which is equivalent to
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 .
Application 2.10. Suppose that
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)
S. Antoniou: Cubic, Quartic and Quintic 28
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(
Exercise 2. Show that the cubic roots of the irrational number 594207 are
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)
Taking the cube of the previous identity, we find
3)wzyx(dcba (2)
and expanding the right member,
w)wzyzx3(y)wzx3yx(dcba 3223 (3)
From the above identity we obtain the system
}dw,by,cwzyzx3,awzx3yx{ 3223 (4)
Multiplying the identities
wzyxdcba3
and wzyxdcba3
we obtain
)wzyx()wzyx()dcba()dcba(3
which can be written as
wzyxdcba 223 22 (5)
S. Antoniou: Cubic, Quartic and Quintic 29
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)
Taking the cube of the previous identity, we find
3)wzyx(220612
which can be written as
w)wzzyx3(y)wzx3yx(220612 3223 (2)
From the above identity we obtain the system
}2w,6y,20wzyzx3,12wzx3yx{ 3223 (3)
Multiplying the identities
wzyx2206123
and wzyx2206123
we obtain
)wzyx()wzyx()220612()220612(3
which can be written as
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
S. Antoniou: Cubic, Quartic and Quintic 30
Exercises.
Exercise 1. Show that the cubic roots of the irrational number 51482234
are given by 5223 , )5223( , 2)5223(
Exercise 2. Show that the cubic roots of the irrational number 238336 are
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.
S. Antoniou: Cubic, Quartic and Quintic 31
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
S. Antoniou: Cubic, Quartic and Quintic 32
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 .
S. Antoniou: Cubic, Quartic and Quintic 33
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
S. Antoniou: Cubic, Quartic and Quintic 34
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).
S. Antoniou: Cubic, Quartic and Quintic 35
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
S. Antoniou: Cubic, Quartic and Quintic 36
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
S. Antoniou: Cubic, Quartic and Quintic 37
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
.
S. Antoniou: Cubic, Quartic and Quintic 38
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
which is equivalent to
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)
S. Antoniou: Cubic, Quartic and Quintic 39
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.
Example 2. Solve the equation
01x3xx 23 (1)
Solution. In this case 1a , 1b , 3c and 1d .
Under the substitution 13
1y
a3
byx
, i.e.
3
1yx (2)
equation (1) takes on the form
013
1y3
3
1y
3
1y
23
S. Antoniou: Cubic, Quartic and Quintic 40
which is equivalent to
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)
We now determine 0m and 0n by
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)
The quantities 0m and 0n satisfying (5), (6) and (7) are given by (Application
2.3)
i3
6
3
2m0 and i
3
6
3
2n0 (8)
Since 3
4nm 00 and i
3
62nm 00 , we can determine the roots of
equation (3). We find
3
4nmy 001
23
2
2
i3
623i
3
4
2
)nm(3i)nm(y 0000
2
S. Antoniou: Cubic, Quartic and Quintic 41
23
2
2
i3
623i
3
4
2
)nm(3i)nm(y 0000
3
The roots of equation (1) are then given by
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.
Example 3. Solve the equation
042x20xx 23 (1)
Solution. In this case we have 1a , 1b , 20c and 42d .
Under the substitution 13
1y
a3
byx
, i.e.
3
1yx (2)
equation (1) takes the form
0423
1y20
3
1y
3
1y
23
which is equivalent to
027
952y
3
61y3 (3)
S. Antoniou: Cubic, Quartic and Quintic 42
In this case we have 3
61p and
27
952q . Therefore the discriminant of the
equation (3) is evaluated to be
9
5
27
3
61
4
27
952
27
p
4
qD
32
32
(4)
We now determine 0m and 0n by
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)
The quantities 0m and 0n satisfying (5), (6) and (7) are given by (Application
2.5)
i53
4m0 and i5
3
4n0 (8)
Since 3
8nm 00 and i52nm 00 , we can determine the roots of
equation (3). We find
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
The roots of equation (1) are then given by
S. Antoniou: Cubic, Quartic and Quintic 43
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
Note. It is obvious that equation (1) can be solved using Horner’s Algorithm.
However we solved the equation using the theory outlined previously.
Example 4. Solve the equation
059x181x8x12 23 (1)
Solution. In this case we have 12a , 8b , 181c and 59d
Under the substitution 123
8y
a3
byx
, i.e.
9
2yx (2)
equation (1) takes the form
0599
2y181
9
2y8
9
2y12
23
which is equivalent to
02916
24175y
108
1645y3 (3)
In this case we have 108
1645p and
2916
24175q . Therefore the discriminant of
the equation (3) is evaluated to be
S. Antoniou: Cubic, Quartic and Quintic 44
11664
1326125
27
108
1645
4
2916
24175
27
p
4
qD
32
32
(4)
We now determine 0m and 0n by
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)
The quantities 0m and 0n satisfying (5), (6) and (7) are given by (Application
2.6)
i518
5m0 and i5
18
5n0 (8)
Since 9
5nm 00 and i52nm 00 , we can determine the roots of
equation (3). We find
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
S. Antoniou: Cubic, Quartic and Quintic 45
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
Note. It is obvious that equation (1) can be solved using Horner’s Algorithm.
However we solved the equation using the theory outlined previously.
Example 5. Solve the equation
03x10x4x8 23 (1)
Solution. In this case we have 8a , 4b , 10c and 3d .
Under the substitution 83
4y
a3
byx
, i.e.
6
1yx (2)
equation (1) takes the form
036
1y10
6
1y4
6
1y8
23
which is equivalent to
027
16y
3
4y3 (3)
In this case we have 3
4p and
27
16q . Therefore the discriminant of the
equation (3) is evaluated to be
027
3
4
4
27
16
27
p
4
qD
32
32
(4)
S. Antoniou: Cubic, Quartic and Quintic 46
We now determine 0m and 0n by
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
The roots of the original equation (1) are then calculated to be
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.
The roots of the equation 03x10x4x8 23 are given by
2
3x1 and
2
1xx 32 (double root).
S. Antoniou: Cubic, Quartic and Quintic 47
Note. It is obvious that equation (1) can be solved using Horner’s Algorithm.
However we solved the equation using the theory outlined previously.
Example 6. Solve the equation
026x31x10x3 23 (1)
Solution. In this case we have 3a , 10b , 31c and 26d .
Under the substitution 33
10y
a3
byx
, i.e.
9
10yx (2)
equation (1) takes the form
0269
10y31
9
10y10
9
10y3
23
which is equivalent to
0729
12688y
27
179y3 (3)
In this case we have 27
179p and
729
12688q . Therefore the discriminant of the
equation (3) is evaluated to be
243
21025
27
27
179
4
729
12688
27
p
4
qD
32
32
(4)
We now determine 0m and 0n by
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)
S. Antoniou: Cubic, Quartic and Quintic 48
The quantities 0m and 0n satisfying (5), (6) and (7) are given by (Application
2.10)
39
8m0 and 3
9
8n0 (8)
Since 9
16nm 00 and 32nm 00 , we can determine the roots of
equation (3). We find
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
The roots of the original equation (1) are then calculated to be
3
2
9
10
9
16
9
10yx 11
i329
10i3
9
8
9
10yx 22
i329
10i3
9
8
9
10yx 33
Conclusion.
The roots of the equation 026x31x10x3 23 are given by
3
2x1 , i32x2 and i32x3 .
Note. It is obvious that equation (1) can be solved using Horner’s Algorithm.
However we solved the equation using the theory outlined previously.
Example 7. Solve the equation
0)i2(zz)i2(z 23 (1)
S. Antoniou: Cubic, Quartic and Quintic 49
Solution. In this case we have 1a , )i2(b , 1c and )i2(d .
Under the substitution 13
)i2(y
a3
byz
, i.e.
3
i2yz
(2)
equation (1) takes the form
0)i2(3
i2y
3
i2y)i2(
3
i2y
23
which is equivalent to
0)i1(27
40y
3
i4y3 (3)
In this case we have 3
i4p and )i1(
27
40q . Therefore the discriminant of
the equation (3) is evaluated to be
i27
32
27
3
i4
4
)i1(27
40
27
p
4
qD
32
32
(4)
We now determine 0m and 0n by
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)
The quantities 0m and 0n satisfying (5), (6) and (7) are given by (Application
2.9)
i3
1
3
32m0
and i
3
1
3
32n0
(8)
S. Antoniou: Cubic, Quartic and Quintic 50
Since i3
2
3
4nm 00 and
3
32nm 00 , we can determine the roots of
equation (3). We find
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
The roots of the original equation (1) are then calculated to be
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
S. Antoniou: Cubic, Quartic and Quintic 51
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)
equation (1) takes on the form
0q)zw(p)zw( 3
which can be written as
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
S. Antoniou: Cubic, Quartic and Quintic 52
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)
from which we obtain
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
S. Antoniou: Cubic, Quartic and Quintic 53
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
S. Antoniou: Cubic, Quartic and Quintic 54
which can be written as
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)
The discriminant of the above equation is calculated to be
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
S. Antoniou: Cubic, Quartic and Quintic 55
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.).
S. Antoniou: Cubic, Quartic and Quintic 56
The roots of equation 0qypy3 are then given by
1
11z
1
3
pzy ,
222
z
1
3
pzy ,
333
z
1
3
pzy
Example 1. Solve the equation
01x3xx 23 (1)
Solution. Under the substitution
3
1yx (2)
equation (1) takes the form
027
56y
3
10y3 (3)
The substitution
z
1
9
10zy (4)
converts equation (3) into the equation
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)
The discriminant of the above equation is calculated to be
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)
S. Antoniou: Cubic, Quartic and Quintic 57
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:
S. Antoniou: Cubic, Quartic and Quintic 58
13
1yx 11 , 21
3
1yx 22 , 21
3
1yx 33
Conclusion.
The roots of the equation are
1x1 , 21x2 and 21x3
Example 2. Solve the equation
042x20xx 23 (1)
Solution. Under the substitution
3
1yx (2)
equation (1) takes the form
027
952y
3
61y3 (3)
Under the substitution
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)
The discriminant of the above equation is calculated to be
9
20
729
2269814
27
952D
2
and thus
S. Antoniou: Cubic, Quartic and Quintic 59
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
Example 3. Solve the equation
059x181x8x12 23 (1)
Solution. Under the substitution
9
2y
3
12
8
yx
(2)
equation (1) takes the form
02916
24175y
108
1645y3 (3)
Under the substitution
S. Antoniou: Cubic, Quartic and Quintic 60
z
1
324
1645zy (4)
equation (3) takes on the form
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)
The discriminant of the above equation is calculated to be
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)
S. Antoniou: Cubic, Quartic and Quintic 61
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 3
1x1 , 15
2
1x2 and 15
2
1x3
Example 4. Solve the equation
03x10x4x8 23 (1)
Solution. Under the substitution
6
1yx (2)
equation (1) takes the form
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)
The discriminant of the above equation is calculated to be
09
44
27
16D
32
and thus equation (7) admits a double root given by
S. Antoniou: Cubic, Quartic and Quintic 62
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)
the roots 1x , 2x and 3x .
Conclusion.
The roots of the equation are 2
3x1 and
2
1xx 32 (double root).
Example 5. Solve the equation
026x31x10x3 23 (1)
Solution. Under the substitution
9
10y
3
3
10
yx
(2)
equation (1) takes the form
0269
10y31
9
10y10
9
10y3
23
which is equivalent to
0729
12688y
27
179y3 (3)
Under the substitution
z
1
81
179zy (4)
equation (3) takes on the form
S. Antoniou: Cubic, Quartic and Quintic 63
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)
The discriminant of the above equation is calculated to be
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
We then have to solve the equations
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
we find, using equation (4), the quantities 1y , 2y , 3y and then using equation (2)
the roots 1x , 2x and 3x .
S. Antoniou: Cubic, Quartic and Quintic 64
Conclusion.
The roots of the equation are 3
2x1 , i32x2 and i32x3 .
Example 6. Solve the equation
0)i2(zz)i2(z 23 (1)
Solution. Under the substitution
3
i2yz
(2)
equation (1) takes the form
0)i2(3
i2y
3
i2y)i2(
3
i2y
23
which is equivalent to
0)i1(27
40y
3
i4y3 (3)
Under the substitution
z
1
9
i4zy
equation (3) transforms into the equation
0i9
4z
27
)i1(40z
336
(4)
Under the substitution
3zw (5)
Equation (4) transforms into the quadratic equation
0i9
4t
27
)i1(40t
32
(6)
The discriminant of the above equation is calculated to be
i27
128i
9
44
27
)i1(40D
32
S. Antoniou: Cubic, Quartic and Quintic 65
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
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 iz1 , i2z2 and iz3 .
S. Antoniou: Cubic, Quartic and Quintic 66
Section 5
Solving the reduced third degree equation using
Trigonometric Functions
5. Solving the reduced third degree equation using
trigonometric functions
5.1. Solving the equation 0qyp3y3 for 0p .
We consider the reduced third degree equation
0qyp3y3 ( 0p ) (5.1)
Under the substitution
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)
S. Antoniou: Cubic, Quartic and Quintic 67
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)
converts equation (1) into the equation
01y3y3 ( 0D ) (3)
S. Antoniou: Cubic, Quartic and Quintic 68
Under the substitution
n
zy (4)
equation (3) takes on the form
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
S. Antoniou: Cubic, Quartic and Quintic 69
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
S. Antoniou: Cubic, Quartic and Quintic 70
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 .
Under the substitution
n
zy (5.12)
equation (5.10) takes on the form
0nqznpz 323 (5.13)
In view of the identity
cos3cos43cos 3
we have
S. Antoniou: Cubic, Quartic and Quintic 71
03cos4
1cos
4
3cos3 (5.14)
Comparing (5.13) and (5.14), we obtain
cosz (5.16)
4
3np 2 (5.17)
and
3cos4
1nq 3
(5.18)
From equation (5.17) we obtain
p
3
4
1n2
from which we determine n:
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
S. Antoniou: Cubic, Quartic and Quintic 72
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
S. Antoniou: Cubic, Quartic and Quintic 73
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
S. Antoniou: Cubic, Quartic and Quintic 74
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 .
S. Antoniou: Cubic, Quartic and Quintic 75
(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
S. Antoniou: Cubic, Quartic and Quintic 76
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
S. Antoniou: Cubic, Quartic and Quintic 77
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
Example 2. Find the roots of the equation 06y9y3
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
S. Antoniou: Cubic, Quartic and Quintic 78
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
Example 3. Find the roots of the equation 012y9y3
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
S. Antoniou: Cubic, Quartic and Quintic 79
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
S. Antoniou: Cubic, Quartic and Quintic 80
Section 6
Solving the reduced third degree equation using
hyperbolic functions
6. Solving the reduced third degree equation using
hyperbolic functions
We consider the reduced third degree equation
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 .
Under the substitution
n
zy (6.3)
equation (6.1) takes on the form
0nqznp3z 323 (6.4)
In view of the identity
usinh3usinh4u3sinh 3
we have
0u3sinh4
1usinh
4
3usinh 3 (6.5)
Comparing (6.4) and (6.5), we obtain
usinhz (6.6a)
S. Antoniou: Cubic, Quartic and Quintic 81
4
3np3 2 (6.6b)
and
u3sinh4
1nq 3 (6.6c)
From equation (6.6b) we obtain
p4
1n2
from which we determine n:
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)
On the other hand, since
)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)
S. Antoniou: Cubic, Quartic and Quintic 82
)ucosh3iu(sinhpy3 (6.10)
Because of (6.7), equation (6.6c) gives us
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 .
Under the substitution
n
zy (6.12)
equation (6.1) takes on the form
0nqznp3z 323 (6.13)
In view of the identity
ucosh3ucosh4u3cosh 3
we have
0u3cosh4
1ucosh
4
3ucosh3 (6.14)
Comparing (6.13) and (6.14), we obtain
ucoshz (6.15a)
4
3np3 2 (6.15b)
and
u3sinh4
1nq 3 (6.15c)
From equation (6.13b) we obtain
p4
1n2
from which we determine n:
p2
1n ( 0p ) (6.16)
S. Antoniou: Cubic, Quartic and Quintic 83
From (6.12), (6.15a) and (6.16), we find that one root of the equation is given by
ucoshp2y1 (6.17)
On the other hand, since
)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)
Because of (6.16), equation (6.15c) gives us
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
S. Antoniou: Cubic, Quartic and Quintic 84
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
S. Antoniou: Cubic, Quartic and Quintic 85
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.
S. Antoniou: Cubic, Quartic and Quintic 86
Section 7
The Tschirnhaus method of solving the reduced
third degree equation
7. The Tschirnhaus method of solving the reduced third
degree equation.
We consider 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)
S. Antoniou: Cubic, Quartic and Quintic 87
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
S. Antoniou: Cubic, Quartic and Quintic 88
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
S. Antoniou: Cubic, Quartic and Quintic 89
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
S. Antoniou: Cubic, Quartic and Quintic 90
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).
S. Antoniou: Cubic, Quartic and Quintic 91
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
S. Antoniou: Cubic, Quartic and Quintic 92
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
S. Antoniou: Cubic, Quartic and Quintic 93
;
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|| .
S. Antoniou: Cubic, Quartic and Quintic 94
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
from which we obtain
y
xy)1n(1y 1n
(3)
Step 3. Combining (2) and (3), we obtain another expression of the derivative
S. Antoniou: Cubic, Quartic and Quintic 95
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)
In view of the identity
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
S. Antoniou: Cubic, Quartic and Quintic 96
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
which is equivalent to
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
S. Antoniou: Cubic, Quartic and Quintic 97
)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
from which we obtain
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
S. Antoniou: Cubic, Quartic and Quintic 98
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
S. Antoniou: Cubic, Quartic and Quintic 99
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
S. Antoniou: Cubic, Quartic and Quintic 100
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)
equation (1) transforms into the equation
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
S. Antoniou: Cubic, Quartic and Quintic 101
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).
Example 1. Solve the equation
01x2xx 34 (1)
Solution. The transformation
S. Antoniou: Cubic, Quartic and Quintic 102
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
which is equivalent to
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
S. Antoniou: Cubic, Quartic and Quintic 103
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
Example 2. Solve the equation
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
S. Antoniou: Cubic, Quartic and Quintic 104
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
S. Antoniou: Cubic, Quartic and Quintic 105
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 .
Example 3. Solve the equation
0442x58x23x2x 234 (1)
Solution. Under the substitution
2
1yx (2)
equation (1) takes on the form
S. Antoniou: Cubic, Quartic and Quintic 106
016
7625y80y
2
43y 24 (3)
The previous equation can be written as
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
The previous equation is equivalent to
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
which can be written as
S. Antoniou: Cubic, Quartic and Quintic 107
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 .
Example 4. Solve the equation
01369x74x457x12x36 234 (1)
Solution. Under the substitution
12
1yx (2)
equation (1) takes on the form
0576
793921y150y
2
911y36 24 (3)
The previous equation can be written as
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
The previous equation is equivalent to
S. Antoniou: Cubic, Quartic and Quintic 108
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 .
S. Antoniou: Cubic, Quartic and Quintic 109
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 .
Example 5. Solve the equation
0i102x)i91(xi3x)i1(2x 234 (1)
Solution. Under the substitution
2
i1yx
(2)
equation (1) takes on the form
0i154
15y)i23(2yi6y 24 (3)
The above equation can be written as
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
which is equivalent to
S. Antoniou: Cubic, Quartic and Quintic 110
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)
S. Antoniou: Cubic, Quartic and Quintic 111
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)
Example 1. Solve the equation
0442x58x23x2x 234 (1)
Solution. Under the substitution
S. Antoniou: Cubic, Quartic and Quintic 112
2
1yx (2)
equation (1) takes on the form
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)
S. Antoniou: Cubic, Quartic and Quintic 113
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.
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)
equation (1) transforms into the equation
0rxqypy 24 (3)
where the coefficients p, q and r are given by
2a8
3bp (4)
S. Antoniou: Cubic, Quartic and Quintic 114
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
which is equivalent to
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)
S. Antoniou: Cubic, Quartic and Quintic 115
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).
Example 1. Solve the equation
0442x58x23x2x 234 (1)
Solution. Under the substitution
2
1yx (2)
equation (1) takes on the form
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
S. Antoniou: Cubic, Quartic and Quintic 116
)]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
which is equivalent to
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)
S. Antoniou: Cubic, Quartic and Quintic 117
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
The choice of signs )( gives us
i32
5
)i()i4(
10ii4
)v()u(
10vuy
The choice of signs )( gives us
i32
5
)i()i4(
10ii4
)v()u(
10vuy
The choice of signs )( gives us
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 .
S. Antoniou: Cubic, Quartic and Quintic 118
10.4. The Simpson 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.
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
S. Antoniou: Cubic, Quartic and Quintic 119
)BAx(mx2
ax2
(6)
which when solved, provide the four roots of the original equation (1).
Example 1. Solve the equation
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
S. Antoniou: Cubic, Quartic and Quintic 120
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
S. Antoniou: Cubic, Quartic and Quintic 121
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
S. Antoniou: Cubic, Quartic and Quintic 122
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.
S. Antoniou: Cubic, Quartic and Quintic 123
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
S. Antoniou: Cubic, Quartic and Quintic 124
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
S. Antoniou: Cubic, Quartic and Quintic 125
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
S. Antoniou: Cubic, Quartic and Quintic 126
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
S. Antoniou: Cubic, Quartic and Quintic 127
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)
S. Antoniou: Cubic, Quartic and Quintic 128
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
S. Antoniou: Cubic, Quartic and Quintic 129
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)
S. Antoniou: Cubic, Quartic and Quintic 130
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
S. Antoniou: Cubic, Quartic and Quintic 131
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
which is equivalent to
0)p(
qxp
)p(
px
5454
45
0)p(
qx
)p(
px
5444
5
S. Antoniou: Cubic, Quartic and Quintic 132
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
S. Antoniou: Cubic, Quartic and Quintic 133
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
S. Antoniou: Cubic, Quartic and Quintic 134
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
S. Antoniou: Cubic, Quartic and Quintic 135
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
S. Antoniou: Cubic, Quartic and Quintic 136
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
S. Antoniou: Cubic, Quartic and Quintic 137
)(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
S. Antoniou: Cubic, Quartic and Quintic 138
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
S. Antoniou: Cubic, Quartic and Quintic 139
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
S. Antoniou: Cubic, Quartic and Quintic 140
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
equation (24) takes on the form
2442224 y)y25()y6y25( (25)
Under the substitution
2y25z (26)
equation (25) takes on the form
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)
S. Antoniou: Cubic, Quartic and Quintic 141
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
S. Antoniou: Cubic, Quartic and Quintic 142
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
S. Antoniou: Cubic, Quartic and Quintic 143
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
S. Antoniou: Cubic, Quartic and Quintic 144
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
which is equivalent to
1)244(w 2 (6)
Squaring the previous relation and using (5), we have
1)244()3(Z 22 (7)
which is equivalent to
0V405 345 (8)
where
Z
11728V (9)
We now obtain from (6)
244
1w
2 (9)
which can be written as
22
2
)244(
244w
or using (7),
)244()3(Z
)3(Z
1
244w 2
2
which can be written as
S. Antoniou: Cubic, Quartic and Quintic 145
)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)
S. Antoniou: Cubic, Quartic and Quintic 146
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:
S. Antoniou: Cubic, Quartic and Quintic 147
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
The previous equation can be written as
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
S. Antoniou: Cubic, Quartic and Quintic 148
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
S. Antoniou: Cubic, Quartic and Quintic 149
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
S. Antoniou: Cubic, Quartic and Quintic 150
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
S. Antoniou: Cubic, Quartic and Quintic 151
The last equation is equivalent to
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
The last equation is equivalent to
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.
S. Antoniou: Cubic, Quartic and Quintic 152
Section 12
Exercises
12. Exercises.
12.1. Third degree equations
Exercise 1. Solve the third degree equation
0126x59x4x 23
Answer. 2x1 , 7x2 , 9x3
Exercise 2. Solve the third degree equation
0150x25x6x 23
Answer. 5x1 , 5x2 , 6x3
Exercise 3. Solve the third degree equation
020x57x11x6 23
Answer. 3
1x1 ,
2
5x 2 , 4x3
Exercise 4. Solve the third degree equation
018x51x11x10 23
Answer. 2
3x1 ,
5
2x2 , 3x3
Exercise 5. Solve the third degree equation
028x)21(4x)21(2x 23
Answer. 22x1 , 51x2 , 51x3
S. Antoniou: Cubic, Quartic and Quintic 153
Exercise 6. Solve the third degree equation
027x9x3x 23
Answer. 3xx 21 , 3x3
Exercise 7. Solve the third degree equation
020x16xx 23
Answer. 2xx 21 , 5x3
Exercise 8. Solve the third degree equation
030x28x9x 23
Answer. 3x1 , i3x2 , i3x3
Exercise 9. Solve the third degree equation
039xxx 23
Answer. 3x1 , i32x2 , i32x3
Exercise 10. Solve the third degree equation
012x)36(2x)132(x 23
Answer. 1x1 , i33x2 , i33x3
Exercise 11. Solve the third degree equation
012x)223(2x)21(2x 23
Answer. 2x1 , i22x2 , i22x3
Exercise 12. Solve the third degree equation
050x)i92(5x)i910(x2 23
Answer. i2
5x1 , i2x2 , 5x3
Exercise 13. Solve the third degree equation
0i148x)i316(x)i1511(x3 23
S. Antoniou: Cubic, Quartic and Quintic 154
Answer. 3
2x1 , i21x2 , i32x3
Exercise 14. Solve the third degree equation
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
Exercise 16. Solve the fourth degree equation
021x26x18x6x 234
Answer. i21x 2,1 , i32x 4,3
Exercise 17. Solve the fourth degree equation
052x28x7x2x 234
Answer. 2xx 21 , i23x 4,3
Exercise 18. Solve the fourth degree equation
020x22x2x5x 234
Answer. 2x1 , 1x2 , i3x 4,3
Exercise 19. Solve the fourth degree equation
018x39x46xx6 234
Answer. 2x1 , 3x2 , 2
3x3 ,
3
1x4
S. Antoniou: Cubic, Quartic and Quintic 155
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
S. Antoniou: Cubic, Quartic and Quintic 156
0μλ3λμ3λ 23 (I.9)
where
a
bμ (I.10)
Under the substitution
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)
We then have to solve the equations
)i()1(wz 22,1
3 (I.20)
S. Antoniou: Cubic, Quartic and Quintic 157
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
S. Antoniou: Cubic, Quartic and Quintic 158
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
S. Antoniou: Cubic, Quartic and Quintic 159
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
S. Antoniou: Cubic, Quartic and Quintic 160
)(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
S. Antoniou: Cubic, Quartic and Quintic 161
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 )
S. Antoniou: Cubic, Quartic and Quintic 162
……………………………………………………………………
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
S. Antoniou: Cubic, Quartic and Quintic 163
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 ,,, :
S. Antoniou: Cubic, Quartic and Quintic 164
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
S. Antoniou: Cubic, Quartic and Quintic 165
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):
S. Antoniou: Cubic, Quartic and Quintic 166
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)
Under the substitution
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)
The above equation can be written as
)(te i2 (V.4)
where
1N
N
)( (V.5)
S. Antoniou: Cubic, Quartic and Quintic 167
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
S. Antoniou: Cubic, Quartic and Quintic 168
i2ea
1N
n
0n
1ni2 a
11N
n
11N
Nn
!)1n(
t
1N
1)e(f)(f
which can be written as
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
S. Antoniou: Cubic, Quartic and Quintic 169
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).
S. Antoniou: Cubic, Quartic and Quintic 170
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