A formula to solve sextic degree equation
AA FORMULA TO SOLVE SEXTIC DEGREE EQUATION
Rodrigo José Martinelli Biglia Andrade
Administration DepartmentMethodist University of PiracicabaPiracicaba, SP, Brazil [email protected]
January 7, 2021 A BSTRACT
According to the Abel-Ruffini theorem, equations of degree equal to or greater than 5 cannot, in mostcases, be solved by radicals. Due of this theorem we will present a formula that solves specific casesof sixth degree equations using Martinelli’s polynomial as a base. To better understand how thisformula works, we will solve a sixth degree equation as an example. We will also see that all sixthdegree equations that meet the coefficient criterion have a resolvent of fifth degree that can be splittedinto a second degree and a third degree equation. Throughout the paper we will see a demonstrationof the ratio of the coefficients of a sixth degree equation that can be solved with the formula that willbe presented this paper
First, this paper aims to contribute to the academic content on mathematics and focuses on professionals in themathematic area who wish to increase the range of content they may be teaching in class to their students. Second, thepurpose of this paper is also to present in a didactic way of a formula (the Milanez’s formula) which it is able to solvespecific cases of the sixth degree equation. This formula is deducted by the Martinelli’s polynomial and consequentlywe will have a fifth degree resolvent of the sixth degree equation. Every fifth degree equation that is splittable intoa degree 2 and a degree 3 is the resolvent of a sixth degree equation. The demonstration of the relation of the sixthdegree equation will be showed a posteriori, just in cases where the coefficients are consistent with the relation that willbe presented and proved mathematically it is possible to solve a sixth degree equation where it has a splittable fifthdegree resolvent. Using an example, we will have a better understanding of solving a sixth degree equation according toMilanez’s relation. All equations that obey the Milanez’s relation are solvable by radicals where the roots of the sixthdegree polynomial are the sum of the roots of a polynomial of degree 2 and with a polynomial of degree 3. However, theequation of sixth degree has a resolvent of fifth degree where it is not necessary to use any transformation to eliminatesthe fourth degree term of the fifth degree equation because all resolvent will have the fourth degree term suppressed.It is convenient to demonstrate and explain every detail of the Martinelli’s polynomial so that we can give greaterunderstanding of the development of Milanez’s formula. But, it is also interesting for the reader to know that any of theroots of Martinelli’s polynomial has the unique utility of be used to split an equation of the fifth degree into two ofdegree minor as a degree 2 and a degree 3. So Martinelli’s equality is also useful for solving fifth degree equations. Inthis paper we will present a very simple way to use this feature to be able to deduce a formula for specific cases ofequation of the sixth degree where its roots can be represented by means of radicals. This paper also commits itself to afriendly and didactic content for the reader, making the inviting and interesting reading for any student and teacher inthe mathematic field.
We will see some concepts about Galois’ theory and Abel-Ruffini’s theorem. The theorem, in a succinct way, consistsof the proof that there is no "closed" formula for all equations of degree greater or equal to 5. That is, there is no way a r X i v : . [ m a t h . G M ] J a n PREPRINT - J
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7, 2021to algebraically arrive at a formula for those equations greater or equal to 5. As Zoladek (2000) reported, "A generalalgebraic equation of degree ≥ F ⊆ K ⊆ E ; they are also calledsub-extensions of E / F.).So, according to Abel-Ruffini’s theorem and Galois theory, there are cases where equations of degree 5 or higher aresolvable by radicals or algebraically. To demonstrate the Martinelli’s Polynomial it is necessary to consider the roots of any fifth degree equation. Each rootmust be associated with a single other root. We have then that the combination of the roots of a fifth degree equation isten (due that the polynomial is tenth degree). Let us then consider the following equations: ( x − kx + n )( x + kx + kx + m ) = 0 . (1) ( x − kx + n )( x + kx + lx + m ) = 0 . (2)Equation (1) have a relation with equation (2). If we expand both equations (1) and (2), each corresponding term can besplittable. Let put those terms into a table: Table 1: Equation (1) expanded x x x Term Independent k + n − k = C − k + kn + m = D kn − km = E mn = F Table 2: Equation (2) expanded x x x Term Independent l + n − k = C nk − lk + m = D nl − km = E mn = F Andrade (2019) consider the follow equation to better understand the proof of Martinelli’s Polynomial: x + 3 x + 2 = 0 (3)If we substitute x for the sum of the roots of equation (3) we have 2=0, that would be absurd, but with that idea, we cancreate a second degree equation using the terms C, C , D, D , E e E as shown in Tables 1 and 2 above. Matchingthe equation that will be created with ( − k + n + k − C ) n we will have, on the right side of the equation, the samething E − E , because E is the same as kn − km and − E is the same as − nl + km . The sum of E − E will result kn − nl then l = C − n + k . So we have kn − n ( C − n + k ) that is the same as ( − k + n + k − C ) n .To create a second degree equation where one of the solutions of the equation will be the sum of two solutions ofa fifth degree equation we must follow this logic: We have the general form of the second degree equation that is ax + bx + c = 0 , the coefficients of the second degree equation that will be created will be a = C − C , b = D − D , c = E − E and on the right side of the equation we have to add ( − k + n + k − C ) n because on the left side of theequation will remain E − E .This idea will give us the possibility to know n from equation (2). The goal is to form a tenth degree equation withthe unknown k. So a = C − C e b = D − D according to tables 1 and 2, it follows that a = k + n − k − C , b = − k + kn + m − D , c = nk − km − E and the right side of the equation will be ( − k + n + k − C ) n . Thus,making the substitutions in ak + bk + c = 0 we will have:2 PREPRINT - J
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7, 2021 ( − k + n + k − C ) k + ( − k + nk + m − D ) k + nk − km − E = ( − k + n + k − C ) n. ( − k + nk + k − Ck ) + ( − k + nk + km − Dk ) + nk − km − E = ( − k + n + k − C ) n. − k + 2 nk − Ck − Dk + nk − E = − nk + n + nk − nC. − k − Ck − Dk − E = − nk + n − nC. Putting n on the left side of the equation, we have: n = k + Ck + Dk + E k − n + C . (4)Since k is the sum of two roots of a fifth degree equation, C, D, E and F the coefficients of a fifth degree equation andn is the product of the roots of a second degree equation, so to arrive at Martinelli’s polynomial, we have to replacethe variable n with the use of an algebraic manipulation. This algebraic manipulation consists of taking the equalityreferring to the term D in table (2). Thus: nk − ( k − n + C ) k + Fn = D. (5) n = nk + Cnk − F + Dn k . (6)Now just replace n in equation (4) and get n: n = k + Ck + Dk + E k − n + C . − n + 3 nk + Cn = k + Ck + Dk + E. − nk − Cnk + F − Dn k + 3 nk + Cn = k + Ck + Dk + E.n = 2( k + Ck + Dk + Ek ) − F k + Ck − D . (7)If we have n, then we can create the Martinelli’s polynomial that will be very important to Milanez’s formula. Inequality (4) we can replace the n on the left and the right side of the equation, like this: k + Ck + Dk + Ek ) − F k + Ck − D = k + Ck + Dk + E k − k + Ck + Dk + Ek ) − F k + Ck − D + C .
Arranging the equation on both sides and equaling 0 we arrive at the Martinelli polynomial: (2( k + Ck + Dk + Ek ) − F )(13 k + 6 Ck − Dk + ( − E + C ) k + F − DC ) − ( k + Ck + Dk + E )(5 k + Ck − D ) = 0 . (8) When a fifth degree equation is perfectly separable into a degree 2 equation and another equation of degree 3, then wecan say that the Martinelli’s polynomial is also perfectly separable into two, in this case, one of degree 4 and another ofdegree 6.The following list shows what a splittable equation looks like so that it is algebraically represented, that is, by radicals.3
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7, 2021 ( x − ax + b )( x + ax + cx + d ) = 0 . (9)Expanding equation (9) we arrive at a fifth degree equation of the form: x + ( − a + c + b ) x + ( ab + d − ac ) x + ( bc − ad ) x + bd = 0 . (10)Knowing that C = ( − a + c + b ) , D= ( ab + d − ac ) , E= ( bc − ad ) and F= bd , then we can create a relationship with theMartinelli’s polynomial which, if expanded, will have the following terms: k + 3 Ck + Dk + (3 C − E ) k + (2 DC − F ) k + ( C − D − CE ) k + ( DC − DE − CF ) k + (7 DF − CD − E + EC ) k + (4 EF − F C − D ) k − F + F DC − D E = 0 . (11)If we substitute each letter for the relation of equation (10) then we have a polynomial that can be splittable into a fourthand a sixth degree. Thus: ( k + ( a ) k + ( c − a ) k + ( − a − d ) k + ( ad − a c ))( k + ( − a ) k + ( − a + 3 b + 2 c ) k + ( a − ab − ac + 2 d ) k + ( − a b + c + 3 b − ad ) k + ( − ac − ab + a d + 2 abc − bd + 2 dc ) k + ( adb − adc − b c + d + b + bc )) = 0 . (12)So we can solve specific cases of the sixth degree equation by radicals when the fifth degree resolvent equation isperfectly separable. So, the Milanez’s relation are the coefficients of each term of the sixth degree equation that makesup the Martinelli’s polynomial when it can be separated into two equations (one of degree 4 and one of degree 6). Thus,we have that the relation to obtain a sixth degree equation solvable by radicals when the resolvent is an equation of fifthdegree splittable into one of degree 2 and another of degree 3. So, the Milanez’s relation is: k − ak ( − a + 3 b + 2 c ) k ( a − ab − ac + 2 d ) k ( − a b + c + 3 b − ad ) k ( − ac − ab + a d + 2 abc − bd + 2 dc ) k ( adb − adc − b c + d + b + bc ) To deduce Milanez’s formula is quite simple. If the Martinelli’s polynomial is the sum of two roots of any equationof any fifth degree, then the formula for solving a sixth degree equation where the coefficients are consistent with thegiven Milanez’s relation is: x = (cid:118)(cid:117)(cid:117)(cid:116)(cid:18) − a a + a a a − a a (cid:19) + (cid:115)(cid:18) − a a + a a a − a a (cid:19) + (cid:18) a a − a a (cid:19) + (cid:118)(cid:117)(cid:117)(cid:116)(cid:18) − a a − a a a − a a (cid:19) − (cid:115)(cid:18) − a a + a a a − a a (cid:19) + (cid:18) a a − a a (cid:19) + − a + (cid:112) a − a . (13)4 PREPRINT - J
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7, 2021In equation (13), the coefficients a , a , a , a , a , a are from two equations, one from the third and the other from thesecond degree.Equation (13) is then the formula that provides a root of the sixth degree equation that can be solved by radicalsaccording to the Milanez’s relation. We will see in more details with the resolution of a sixth degree equation. As an example, we will solve a sixth degree equation according to Milanez’s relation. The equation is: x + 2 x + 21 x − x + 51 = 0 . (14)In equation (14) the term a is 0, and based on this we can find the terms b, c and d. For that we will get Milanez’srelation and make the comparisons.If a = 0 then, according to the relation we have: − a = 0 . b + 2 c = 0 .d = 1 .c + 3 b = 21 . − b + 2 c = − . − b c + b + bc = 50 . Now we have a system with two variables that we can solve without problems. It is convenient to choose the simplestequations. b + 2 c = 0 . (15) − b + 2 c = − . (16)If we solve the system of the equation (15) and (16) we will have b = 2 and c = − . Then we find all the variables inour equation according to the Milanez’s relation. So just substitute in equation (9) and we have a fifth degree resolventequation. Let’s see: ( x − x + 2)( x + 0 x − x + 1) = 0 .x − x + x − x + 2 = 0 . (17)Solving the second degree and third degree equations and adding the roots of the second degree equation with each rootof the third degree equation, we will have the roots of the sixth degree equation, that is, we will be using the Milanez’sformula to be able to obtain the roots of equation (14). Thus, according to formula (13) one of the roots of equation (14)is: x = (cid:118)(cid:117)(cid:117)(cid:116) − (0) + (0)( − − (cid:19) + (cid:115)(cid:18) − (0) + (0)( − − (1)2(1) (cid:19) + (cid:18) ( − − (0) (cid:19) + (cid:118)(cid:117)(cid:117)(cid:116)(cid:18) − (0) + (0)( − − (cid:19) − (cid:115)(cid:18) − (0) + (0)( − − (1)2(1) (cid:19) + (cid:18) ( − − (0) (cid:19) + − (0) + (cid:112) (0) − .x = (cid:115) −
12 + (cid:114) −
34 + (cid:115) − − (cid:114) −
34 + √ i. (18)Or approximately . .... + 1 . .....i . The sum of the roots of the second degree equation with the roots of thethird degree equation forms the roots of a sixth degree equation according to the Milanez’s relation. Then the otherroots will be: 5 PREPRINT - J
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7, 2021 x = (cid:115) −
12 + (cid:114) −
34 + (cid:115) − − (cid:114) − − √ i. (19) x = 12 (cid:18) − √ i (cid:19) (cid:115) −
12 + (cid:114) −
34 + 12 (cid:18) − − √ i (cid:19) (cid:115) − − (cid:114) −
34 + √ i. (20) x = 12 (cid:18) − √ i (cid:19) (cid:115) −
12 + (cid:114) −
34 + 12 (cid:18) − − √ i (cid:19) (cid:115) − − (cid:114) − − √ i. (21) x = 12 (cid:18) − − √ i (cid:19) (cid:115) −
12 + (cid:114) −
34 + 12 (cid:18) − √ i (cid:19) (cid:115) −
12 + (cid:114) −
34 + √ i. (22) x = 12 (cid:18) − − √ i (cid:19) (cid:115) −
12 + (cid:114) −
34 + 12 (cid:18) − √ i (cid:19) (cid:115) − − (cid:114) − − √ i. (23)Or if you prefer, we can represent the roots in a trigonometric way, like this: x = 2 cos (cid:18) π (cid:19) + √ i.x = 2 cos (cid:18) π (cid:19) − √ i.x = − (cid:18) π (cid:19) + √ i.x = − (cid:18) π (cid:19) − √ i.x = cos (cid:18) π (cid:19) − √ (cid:18) π (cid:19) + √ i.x = cos (cid:18) π (cid:19) − √ (cid:18) π (cid:19) − √ i. In the case of example equation (14) all roots are complex as seen above.
As seen, Milanez’s relation is fundamental to solving equations of the sixth degree. All sixth degree equations that obeythe coefficients according to Milanez’s relation can be solved by radicals. As shown, Milanez’s formula is based on theMartinelli’s polynomial which has the function of adding two roots of a fifth degree polynomial. Based on Martinelli’spolynomial, it was possible to demonstrate Milanez’s formula. After finding the fifth degree resolvent equation of thesixth degree equation according to Milanez’s relation, it is possible to find the roots of the sixth degree polynomial. It isconcluded that a formula that can solve specific cases of the sixth degree equation was successfully achieved.
References [1] ZOLADEK, H. The topological proof of Abel-Ruffini theorem.
Topological Methods in Nonlinear Analysis .v. 16,n.2, pp. 253-265, 2000.[2] NICKALLS, R. W. D. A new approach to solving the cubic:
Cardan’s solution revealed . The mathematicalGazette, v. 77, n. 480, pp. 354-359, 1993.[3] ANDRADE, R. J. M. B. Resolução de uma equação do quinto grau. C.Q.D - Revista Eletrônica Paulista deMatemática, v. 16, pp. 196-205, dez. 2019. Retrieved from