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254 Answers to Exercises and Solutions to Problems

from which the result follows.
8.6. The quadratics all have the form x2 + ux + b. Each alteration changes
the value of a - b by 1. Initially, a - b = -10; finally, a - b = 10. Hence,
somewhere along the way, a - b = 1, and we have a quadratic with zeros
-1, -b.
8.7. (a + b)2 = (c + d) 2 implies ub = cd, so that -c and -d are the zeros
of t2 - (u + b)t + ab. Hence a + d = b + c = 0. It is easy to check that any
solution of this equation satisfies the given equations.
8.8. Let r be the real root of the given equation. Then (r” + ur + c) +
(br + d)i = 0. S ince the sum of the roots is nonreal, b # 0 and we have
r = -d/b. Sub s t’t 1 u t ing this into r2 + ur + c = 0 yields abd = b2c + d2.
On the other hand, suppose that abd = b2c + d2 and b # 0. Then z2 +
(u+bi)z+(c+di) = (z+d/b)(z+b(c+di)/d) has one real and one nonreal
zero.
8.9. A common zero of x2 + px + q and px2 + qx + 1 is a zero of
x(x2 + px + q) - (px2 + qx + 1) = x3 - 1 = (x - 1)(x2 + x + 1). If the
zero is 1, then 1 + p + q = 0. Otherwise, the zero satisfies x2 + x + 1 = 0,
and it follows that (p - 1)x + (q - 1) = 0. Solving this equation for x and
substituting into x2 + px + q = 0 yields p2 + q2 + 1 = p + q + pq.

8.10. The discriminant of the quadratic reduces to (4q - p2)[(1 - q)” +p2],
the second factor being nonzero by hypothesis.

8.11. Since tan(r cot x) = tan(x/2 - ?r tan x), ?r cot x = 7r/2 - x tan x + nir
for some integer n. Hence

tanx+l/tanx-(2n+1)/2=0.

Solving this equation for tanx yields the result. The quantity under the
radical is (2n + 1)2 - 16, which is negative exactly when n = -2, -1, 0, 1.

8.12. The discriminant is equal to

e(2ak - h+1)2 - (n - 4)d+1.

k=l

If n 5 4, this is always nonnegative. If n > 4, take ai = 1 for 1 5 i 5 n
and a,,+1 = 2 to obtain an equation with nonreal roots.
8.13. Solution 1. Since p(l), ~(-1)~ p(i), p(-i) belong to the unit disc, so
also do (1/2)(p(l) +p(-1)) = 1 + b and (1/2)(p(i) +p(-i)) = -1 + b. A
quick sketch convinces one that b = 0. Since p( 1) = 1 + a and p( -1) = 1 - a
belong to the unit disc, a = 0.
Solution &. Let Q = q + ri, b = s + ti. Then

4 = Ip( + Ip( + Ip( + Ip( = 4 + 4(q2 + r2 + s2 + t2),