2 Chapter 1. Groups I gh = N (for g2 Mg + N = 0). In particular, we may assume that g + h = u and gh = 1 3 q. This forces 3gh + q = 0, and Equation (1) now gives g3 + h3 = −r. After cubing gh = 1 3 q, we obtain the pair of equations g3 + h3 = −r g3h3 = 1 27 q3. Thus, there is a quadratic equation in g3: g6 + rg3 1 27 q3 = 0. The quadratic formula gives g3 = 1 2 −r + r2 + 4 27 q3 = 1 2 −r + R [note that h3 is also a root of this quadratic, so that h3 = 1 2 ( −r R ) ]. There are three cube roots of g3, namely, g, ωg, and ω2g. Because of the constraint gh = −q/3, each of these has a “mate:” g and h = −q/(3g) ωg and ω2h = −q/(3ωg): ω2g and ωh = −q/(3ω2g) (since ω3 = 1). Example 1.1. If f(x) = x3 15x 126, then q = −15, r = −126, R = 15376, and R = 124. Hence, g3 = 125, so that g = 5. Thus, h = −q/(3g) = 1. Therefore, the roots of f(x) are 6, + ω2 = −3 + 2i 3, 5ω2 + ω = −3 2i 3. Alternatively, having found one root to be 6, the other two roots can be found as the roots of the quadratic f(x)/(x 6) = x2 + 6x + 21. Example 1.2. The cubic formula is not very useful because it often gives the roots in unrecognizable form. For example, let f(x) = (x 1)(x 2)(x + 3) = x3 7x + 6 the roots of f(x) are, obviously, 1, 2, and −3. The cubic formula gives g + h = 3 1 2 −6 + −400 27 + 3 1 2 −6 −400 27 . It is not at all obvious that g + h is a real number, let alone an integer. There is another version of the cubic formula, due to Vi` ete, which gives the roots in terms of trigonometric functions instead of radicals (FCAA, pp. 360–362). Before the cubic formula, mathematicians had no difficulty in ignoring negative numbers or square roots of negative numbers when dealing with quadratic equa- tions. For example, consider the problem of finding the sides x and y of a rectangle having area A and perimeter p. The equations xy = A and 2x + 2y = p give the quadratic 2x2 px + 2A. The quadratic formula gives x = 1 4 p ± p2 16A and y = A/x. If p2 16A 0, the problem is solved. If p2 16A 0, one did not invent fantastic rectangles whose sides involve square roots of negative numbers instead, one merely said that there is no rectangle whose area and perimeter are so related. But the cubic formula does not allow us to discard “imaginary” roots,
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