Solutions to Problems; Chapter (^4 309)
(Note that u(u + v) = 1 - b, being the sum of the zeros of p(t) - t.)
Hence, two zeros of the quartic p(p(t)) -t are u and v and the remaining
zeros are zeros of at2 + u(b + 1)t + QC + b + 1.
(b) Here, a = 1, b = -3, c = 2, u = 2+fi, v = 2-fiand the quadratic
equation for the remaining two zeros is t2 - 2t = 0. Hence t = 2 f a, 0 or
2.
7.10. For the radical dm to be defined, we require 1x1 2 1. If c < -1,
then x + dm = dm - Ix] < 0 and the left side of the equation is
not defined. Hence x 2 1.
First solution. Let x = (1/2)(u + u-l), where u 1 1. Then &?? =
(1/2)(U - u-1). S’ mce x(x + 1) = (u” + l)(u + 1)2/4u2 and x(z - 1) =
(u” + l)(u - 1)2/4U2, squaring the equation and simplifying yields u4 = 3,
whence x = i(3 ‘I4 + 3-‘i4). This checks out.
Second solution. Squaring the equation yields
cx - JZi)3 = 2x&Zi[x - &Zj
- (x - &ci)” = 2x&F
==+ 2x2 - 1 = 4xJXi * 12x4 - 12x2 - 1 = 0,
from which x2 and eventually x can be determined.
7.11. Let v = cos(nll4). By de Moivre’s Theorem, cos(a/2) = 0 = 64v7 -
112~~ + 56v3 - 7v = v[(~v~)~ - 7(4~~)~ + 14(4v2) - 71. The same equation
is valid for v = cos(3a/14) and v = cos(5n/14), so the three roots of the
given equation must be 4cos2(?r/14), 4cos2(3r/14) and 4cos2(57r/14), of
which the first is the largest (cf. Problem 3.8.29).
7.12. If l-z2 = -x3, then (l-~)(l-x~)~ = (l-x)x’ = (x2-x3)x4 = x4. If
1-x2 = x, then (l-~)(l-x~)~ = (l-x)x2 = x2x2 = x4. The zeros U, v, w
ofx3-x2+1 aredistinct andeachisazerooff(x) = x4-(l-x)(l-~“)~. By
the Factor Theorem, f(x) is divisible by x3 -x2 + 1 = (x - U)(X - v)(x - w).
Similarly, x2 + x - 1 divides j(x). S ince the cubic and quadratic have no
zeros in common, f(x) = k(x3 - x2 + 1)(x2 + x - 1). Checking leading
coefficients of both sides reveals that k = 1.
7.13. The equation can be rewritten
(x2 - 9x - l)iO - loxs(x2 - 9x - 1) + 9x10 = 0.
The quadratic equation x2 - 9x - 1 = x has real roots, and each of these
roots satisfies the given equation.
8.1. Multiply the first equation by y, the second by x, and take the dif-
ference: 0 = (x3 - y3)(2xy - 1). Either x = y or 2xy = 1. If x = y,
then x = y = 0 or 9/8. If y = 1/2x, then the first equation becomes