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344 Answers to Exercises and Solutions to Problems

a0 # 0, the sum of the reciprocals of the squares of the zeros is (a1/ao)2 -

@2/a0).

For a quadratic, a necessary and sufficient condition for real zeros is

that a: 2 4aoa2. Thus to find a counterexample, we choose the oi so that

2aoa2 5 a: < 4aoa2. For example, the coefficients of 2t2 + 3t + 2 satisfy

both conditions, but the zeros are nonreal.

(b) In this case, a; = 4 < 2a4ac = 6, so that not all the zeros are real.

(c) By Rolle’s Theorem, if all the zeros of p(t) = 2&t’ are real, then its

(n - 2)th derivative has at least two real zeros. Hence the discriminant of

(n - 2)(n - 3)... (3)[n(n - l)a,t2 + 2(n - l)a,-lt + 2a,-21

is positive. This yields the desired necessary condition.
For n = 3, the condition becomes a; 2 3ala3 (cf. Exercises 9 (b)). A
cubic polynomial with not all its zeros real which satisfies the condition is
t3 + 5t2 + 7t. A counterexample of degree n is t” + 5t”-’ + 7tnT2.
Remarks. The problem for the case n = 5 was posed in the 1983 USA
Mathematical Olympiad. Observe that it would be unreasonable for a con-
dition which does not involve the constant coefficient to be sufficient for
the zeros to be real; for, if p(t) is any polynomial over R, k could be chosen
sufficiently great that p(t)+k h as at most one real zero. See Exercise 5.2.16.
1.9. (a) If z = 0, then xy = b. If z # 0, then xyz = -c = z3 + az2 + bz
yields the result. Also

(x - Y)” = (x+y)2-4xy=( -a-~)~-4(z~+az+b)

= -[3z2 + 2az - (a2 - 4b)].

(b) (Z - Y)~ 2 0 3 3z2 + 2az - (a” - 46) must be negative j 3t2 +
2at - (a” - 4b) must have real zeros u and v (u 5 v) and z must satisfy
u 5 z < v =+- the discriminant 16(a2 - 3b) must be positive.
(c) Consider the graph of p(t).

A change in the value of c results in a vertical translation of the graph. For
any c, u is no greater than the smallest real zero and v is no less than the
largest. Hence p(u) 5 0 5 p(v).
This can be written as

u3 + au2 + bu < -c < v3 + au2 + bv.