Unknown

Answers to Exercises; Chapter 6 345

Substituting for u and v yields

(2a3 - gab) - (2a2 - 6b)dG 5 -27~

5 (2a3 - gab) -I- (2a2 - 6b)dG

and this yields the first inequality. The second inequality can be found by

squaring.

(d) The inequality can be rewritten as

(2a3 - 9ab + 27~)~ i 4(a2 - 3b)3

from which it follows that the discriminant 4(a2 - 3b) of the quadratic

p’(t) is positive. Let r and s be the real zeros of p’(t). If r < s, then

3r = -a-(a2- 3b)li2 and 3s = -a + (a2 - 3b)lj2. We have, upon dividing

p(t) by p’(t), that

27p(r) = 3(3r + a)p’(r) - 6(a2 - 3b)r + 3(9c - ab)

= 0 + 2(a2 - 3b)3/2 + (2a3 - 9ab + 27~) > 0

while

27p(s) = 3(3s + a)p’(s) - 6(a2 - 36)s + 3(9c - ab)

= 0-2(a2 - 3b)3/2 + (2a3 - 9ab + 27~) < 0.

Hence p(t) has a real zero less than r, between r and s and greater than s.

If r = s, then a2 - 36 = 2a3 - Sub + 27~ = 0 and p(r) = 0. In this case, p

has a triple zero at r. Thus, all zeros of p(t) are real.

1.10. Let the roots be a, b, c, d with ab = -5. Then cd = 4. We have that

-1 + (a + b)(c + d) = -7

-5(c + d) + 4(a + b) = -23

(a + b) + (c + d) = 1.

Hence a + b = -2 and c + d = 3, so that the equation can be rewritten

(t2 + 2t - 5)(P - 3t + 4) = 0

and thence solved.
1.11. (a) Let a, b, c, d be the roots and suppose that (ab)2 = (cd)2 = s.
Then -p = (a + b) + (c + d) = -r/ah, which yields the result. On the
other hand, if r2 = p2s, let t = z + r/px. The equation can be rewritten
as t2 + pt + (q - 2r/p) = 0. Each solution t of this quadratic leads to an
equation px2 - ptx + r = 0 the product of whose roots for either value oft
is r/p (cf. Problem 1.4.17).