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246 Answers to Exercises and Solutions to Problems

Hence, p(t) has a single term, and it can be seen that the polynomials

commuting with t2 are precisely the powers of t.

2.1. (b) 3, 4;

Cc) 2u

-b+,/j?=-&ii and A-@-==

2u

2.3. m, (1 - m2)/2m.

2.5. 1, -l/3, 0.

2.7. The zeros of 6x2 - 5x - 4 are -l/2 and 413. Thus 6x2 - 5x - 4 =

(2x + 1)(3x - 4) is negative for -l/2 < x < 413.

2.8. The discriminant of x2 + (1 - 3k)x + (2 - k) is 9k2 - 2k - 7 =

(9k+7)(k- l), w h’ ic h is nonnegative for k 2 1 and k 5 -7/g. These are

the only values of k for which the equation is solvable.

2.9. The range of the function is the set of k for which

(k - 1)x2 + (3k - 1)x + (2k + 1) = 0

is solvable. The discriminant, equal to (k - 1)2 + 4, is always positive so

that the equation is solvable for all k.

2.10. Since m + n = 516 and mn = -l/2, we have that the sum of the

roots of the desired quadratic is (m+n)-(m2+n2) = (m+n)-(m+n)2+

2mn = -31/36 and the product of the roots is mn - (m3 + n”) + (mn)2 =

mn - (m + n)3 + 3mn(m + n) + (mn)2 = -449/216. Thus, a quadratic

polynomial with the desired roots is 216t2 + 186t - 449.

2.12. (a) Suppose r is a common nonrational zero of p(t) and q(t). Choose
integers a and b for which (up + bq)(t) is either linear or a constant. If
(up+ bq)(t) were linear, it would have rational coefficients but a nonrational
zero, which cannot occur. Hence it is a constant. Since (up+ bq)(r) = 0, we
must have that (up + bq)(t) is th e zero polynomial, from which the result
follows.
(b) t2 - t; t2 + t.
2.13. Suppose x = (b- d)/(u - c ) sa t’ IS fi es x2-ux+b = 0. Since (a-c)x =
(b - d), --ax + b = -cx + d, so that x2 - cx + d = x2 - ax + b =^0 as
required.
2.14. (a) The quadratic polynomial is always nonnegative and can vanish
only if t = -bi/ui for each i simultaneously. Hence it has either coincident
real zeros or two nonreal zeros.
(b) The desired inequality is a rewriting of the condition that the dis-
criminant of the polynomial in (a) is nonpositive. Equality occurs if and
only if the discriminant vanishes, which occurs when there is a single real
zero. The condition for equality is that bi/ui are equal for all i.