The cubic and quartic equations Dr. Mario Duran Camejo Adjunct Faculty Department of Liberal Arts and Sciences - Mathematics Miami Dade College, Hialeah Campus E-mail:
[email protected] Abstract This paper discusses the analytic method of solution of the quartic and cubic equations in historical perspective. Additionally, an alterative method to the graphic Khayyám’s solution for the cubic is also provided.
Keywords: Cubic equation; History of mathematics and mathematicians; Quartic equation. 2010 Mathematics Subject Classification: 00A05, 00A06, 01A05.
1. Introduction The analytic method of solution of the quartic and cubic equations is known since the Renaissance although a geometrical solution of the cubic equation was developed by the Persian poet and mathematician Omar Khayyám during the Middle Ages. Nonetheless, because of the attractiveness of its implications it is a recurrent theme in mathematics. Although the quartic equation can be solved by factorization, it is not possible to factorize a cubic equation because the constraints imposed to the coefficients lead again to a cubic equation and the problem becomes circular. The reason is that a quartic equation may be decomposed into two polynomials but the cubic has to be decomposed in a binomial and a polynomial and the coefficient of the binomial is one of the cubic roots. An alternative decomposition of the cubic would lead to infinite series rather than polynomials. In fact, some solutions of a cubic equation may be expressed by means of hypergeometric functions.
2. Cubic Equation Two, already classical, algebraic methods were developed for solving the cubic equation. The first one in chronological order is due to Scipione del Ferro and Niccolò Fontana Tartaglia but named after Girolamo Cardano who, later on, published the method. The other one was devised by François Viète. Both methods require to transform the whole cubic into a depressed equation. Cardano's solution assimilates the depressed cubic to a quadratic equation whereas Viète's
assimilates it to a biquadratic. As the transformation into a biquadratic actually doubles the number of solutions, both solutions from Viète's algebraic method are identical. The algebraic methods do really provide only one solution, and the other two have to be found multiplying the first one by the two complex cubic roots of 1, namely, − 1⁄2 ± √3 𝑖 ⁄2. Additionally, Viète also found a trigonometric solution assimilating the depressed cubic equation to the formula for the cosine of the triple angle.
2.1 Khayám’s geometric solution Khayyám uses the intersection of two curves, a parabola and a circle, to get the solution of a depressed cubic. The solution is based in transforming the depressed cubic 𝑥 + 𝑚𝑥 + 𝑙 = 0 into the incomplete depressed quartic 𝑥 + 𝑚𝑥 + 𝑙𝑥 = 0, multiplying the cubic by 𝑥. On the other hand, the 𝑥 coordinate of the point of intersection of the parabola 𝑦 = 𝑎𝑥 and the circle (𝑥 + 𝑅) + 𝑦 = 𝑅 , is given by (𝑥 + 𝑅) + (𝑎𝑥 ) = 𝑅 𝑥
+
𝑥 2𝑥 𝑅 + =0 𝑎 𝑎
Now, the solution of the incomplete depressed quartic can be assimilated to 𝑥 making 𝑎 = 1⁄√𝑚 and 𝑅 = 𝑙 ⁄(2𝑚). As the intersection point is unique (because the origin is a trivial solution), and following what we have seen in the introduction above, the other two roots have to be obtained multiplying respectively by − 1⁄2 ± √3 𝑖 ⁄2.
2.2 Alternative geometric solution An interesting geometric solution would also be given by the intersection of the parabola 𝑦 = 𝑥 + 𝑘 and the equilateral hyperbola 𝑥𝑦 = −𝑎 (for 𝑎 > 0 and 𝑘 > 0), which lead to the equation 𝑦 + 𝑘𝑦 + 𝑎 = 0. When compared to the depressed cubic, we have 𝑘 = 𝑚, and 𝑎 = 𝑙. Unlike the solution due to Khayyám, this intersection would give us the three unequal real roots of the cubic (in case they exist).
2.3 Cardano’s algebraic solution The solution is obtained by making the substitution 𝑥 = 𝑢 + 𝑣 into the depressed cubic 𝑥 + 𝑚𝑥 + 𝑙 = 0, which leads to 𝑢 + 𝑣 + (3𝑢𝑣 + 𝑚)(𝑢 + 𝑣) + 𝑙 = 0. Then, imposing the condition 3𝑢𝑣 + 𝑚 = 0, it is possible to arrive to the two equations 𝑢 𝑣 = − 𝑚 ⁄27, and −(𝑢 + 𝑣 ) = 𝑙. This is equivalent to consider 𝑢 and 𝑣 the solutions of the quadratic equation 𝑧 + 𝑙𝑧 − 𝑚 ⁄27 = (𝑧 − 𝑢 )(𝑧 − 𝑣 ) = 0, so 𝑢 = √𝑧 , 𝑣 = √𝑧 , and x is obtained summing up 𝑢 and 𝑣. It is interesting to have also a look to the new approach of Cardano’s solution, given in [1].
2.4 Viète’s algebraic solution The solution is obtained by substituting 𝑥, also in the depressed cubic 𝑥 + 𝑚𝑥 + 𝑙 = 0, by 𝑧 − 𝑚⁄(3𝑧), which leads to the equation 𝑧 + 𝑙𝑧 − 𝑚 ⁄27 = 0, biquadratic in 𝑧 , and, then,
making the new substitution 𝑧 = 𝑣, the two values 𝑣 and 𝑣 may be obtained. Thus, 𝑥 = √𝑣 − 𝑚⁄(3 √𝑣 ), which has to be multiplied by − 1⁄2 ± √3 𝑖 ⁄2, respectively, to obtain the whole set of three solutions. The other value, 𝑣 , will yield the same set as 𝑣 .
2.5 Viète’s trigonometric solution Starting with the depressed cubic, 𝑥 + 𝑚𝑥 + 𝑙 = 0, let us set 𝑥 = 𝑢 cos 𝜃, to make the equation coincide with the identity 4(cos 𝜃) − 3 cos 𝜃 − cos(3𝜃) = 0. Then, we get 𝑢 (cos 𝜃) + 𝑚𝑢 cos 𝜃 + 𝑙 = 0 Now, we choose 𝑢 = 4 and divide the equation by 𝑢 ⁄4, which gives 4(cos 𝜃) +
4𝑚 4𝑙 cos 𝜃 + =0 𝑢 𝑢
Next, we make 4𝑚⁄𝑢 = −3 and 4𝑙 ⁄𝑢 = cos(3𝜃). These two conditions lead to 𝑢=2 −
cos(3𝜃) =
𝑚 3
3𝑙 3 − 2𝑚 𝑚
and the solutions are
𝑥 = 2 −
𝑚 1 cos cos 3 3
3𝑙 3 2𝜋𝑘 − + , 2𝑚 𝑚 3
𝑘 = 0,1,2
Many years later, René Descartes made a geometric approach to Viète’s trigonometric solution [2].
3. Quartic equation I'm bringing here a slight variation of the method due to René Descartes and improved later by Leonhard Euler. The point is that the solution can be reached in a simpler way if we transform the depressed quartic into a difference of squares instead of a product of two quadratic polynomials 𝑦 + 𝑝𝑦 + 𝑞𝑦 + 𝑟 = 0 (𝑦 + 𝐴) − (𝐵𝑦 + 𝐶) = 0, which gives the two quadratic equations
𝑦
+ 𝐵𝑦 + (𝐴 + 𝐶) = 0, where 𝑦
+ 𝑦
𝑦
− 𝐵𝑦 + (𝐴 − 𝐶) = 0, where 𝑦
+ 𝑦
Thus, (𝑦
+ 𝑦 )(𝑦
=𝐵 =𝐵
+ 𝑦 ) = −𝐵
The resolvent cubic is 𝐴 −
𝑝 1 𝐴 − 𝑟𝐴 − (𝑞 − 4𝑝𝑟) = 0, 2 8
and the other coefficients are 𝐵=
2𝐴 − 𝑝 ; 2
𝐶=
𝑞 2 2𝐴 − 𝑝
The rest of the formulation coincides with Euler's.
4. Concluding Remarks and Directions for Future Research The aim of this work has been to bring together in a common frame two close related topics of algebra hoping that by doing this both will gain in perspective and clarity. Additionally, I have also provided a curious alternative to an ancient and, possibly, already forgotten method due to a great preclassical mathematician. Although it is a quite hard task to add new insights on an old subject like the solution of algebraic equations, the reward may be expected on the side of making it more accesible and suggestive to the young students of mathematics.
Acknowledgements I would like to acknowledge in the first place the Editorial Committee of Polygon for accepting this work for publication. I also thank very sincerely Professor Rene Barrientos and Dr. Mohammad Shakil for their warm and open support. Finally, a mention is deserved by the Math Dept of Miami Dade College, Hialeah Campus, which, giving me the opportunity to teach, has made all this work possible.
References: [1] R. W. D. Nickalls, “A new approach to solving the cubic: Cardan’s solution revealed”, The Mathematical Gazette (1993), 77 (November), 354–359.
[2] R. W. D. Nickalls, “Viète, Descartes, and the cubic equation”, The Mathematical Gazette (2006), 90 (July, No. 518), 203–208.
