Answers to Exercises; Chapter 4 2992.4. Both equations lead to 0 = (x - 2)(4x - 7).
(a) 2, 7/4; (b) no solutions.
2.5. (b) Let u = 14 + x, v = 14 - x. Apply (a) to obtain w = 4 and
64 = 28 + 12qm. Th’ is simplifies to 27 = 196 - x2, whence x = f13.
Both are valid solutions.
2.6. Set y = x2 + 18x + 45 and obtain y - 15 = 2fi. There are two roots
y = 9 and y = 25. The first is extraneous; the second leads to a quadratic
equation which can be solved for x.
2.7. (a) Since dm > x for all real x, the left side of the equation is
always strictly positive.
(b) Suppose that the equation is solvable. Thenx2 + b = (1 - x)~ = 1 - 2x + x2so that 2x = 1 - b. Testing this out, we find that1 - x + I/= = (1/2)[1+ b + dm]= (1/2)[1+b+(l+b)]=l+b#O whenb>-1
{ (1/2)[1+ b - (l+ b)] =^0 when b^5 -1.
Thus, there is no solution when b > -1 and x = (1 - b)/2 is a solution
when b < -1.
3.2. (a) See Exercises 1.2.4, 1.4.4, 1.4.11, 1.4.12, 1.4.13.
(b) Let the reciprocal polynomial have the variable t. If it has even degree
2k, the substitution x = t + t-’ leads to a polynomial equation of degree k
in x. If k 5 4, then, by (a), the number of solutions, counting multiplicity
is, k. Each of the solutions gives rise to two values of t, namely the roots
of t2 - xt + 1 = 0. If the reciprocal polynomial has odd degree, it is the
product oft + 1 and a polynomial of even degree, and the result follows.
3.3. (a) The solutions for various n are: 2 : 2, -2; 3 : -1,-l, 2; 4 :
-2,0,0,2;6:-2,-l,-1,1,1,2;8:-2,-fi,-fi,O,O,&,&2.
(c) If tl and t2 are the roots of t2 - yt + 1 = 0, then, for each n 1 3,
tl = yty-’ - tyv2, from which the result follows.
(f) The result is true for n = 1,2. Suppose it holds up to n = k 1 2.
Then
2cos(k + i)e = y(2 cos ke) - (2 cos(k - i)e)
= Yfk (2 cos 0) - fk-l (2 case) = fk+1(2COSo).3.5. The four equations are a = w + v - u2; b = z + uw - uv; c = uz + VW;
d = vz. Eliminating w yields the system b = z + au + u3 - 2uv; c =
uz +av + u2v - v2; d = vz. Eliminating z yields the system c = bu + 3u2v -