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332 Answers to Exercises and Solutions to Problems

4.10. Let p(x) = k(x - ar)(x - a2) ... (z - a,). Then p’(al) =

k(a2 - al)(a3 - al) a.. (a, - al). For i = 2,3,... , n,

lai - all 5 lai - bl + lb - a11 < 2lb - ail.

[p(b)1 = k fi lb - ail > k2-“+‘lb - all fi lui - all

i=l i=2

= 2~“+‘lp’(al)(b - al)l.

4.11. Let the left side be f(t) and suppose that a 5 b 5 c. We have that

f(t) = (t - b)[(t - a)(t - c) - e2] - (t - a)d2 - (t - c)f2 - 2def.

The quadratic (t - a)(t - c) - e2 is nonpositive for t = a and t = c and

so has real zeros u and v for which u _< a 5 c < v. Thus (a - u)(c - U) =

(v - a)(v - c) = e2, whence

f(u) = (Ja-;kr f &=iif)2 2 0

and

f(v) = -(Gd& Gf)2 < 0.

Since the leading coefficient off(t) is positive, f(t) has a zero not exceed-

ing u, a zero not less than v and a zero in the interval [u, v]. The only way

in which f(t) might fail to have three real zeros is that u = a = b = c = v.

But, in this case, e = 0 and f(t) = (t - u)[(t - a)2 - d2 - f2] evidently has

three real roots.

Remark. The roots of this equation are the eigenvalues of the real sym-

metric matrix (a f e/f bd/ e d c ), w h ic h must be real from matrix theory.

4.12. By Descartes’ Rule of Signs, there are at most two positive roots. If
f(z) denotes the left side, then

f(x) = (x2 - lOlO) - (Z2 - lore) - (x + 1)

= (X2 - 10” - 0.5)2 - (x + 1.25).

Since f (105) < 0 and f ( lo5 f 10V2) = [10m4 f 2.103 - 0.512 - [lo5 f 10m3 +

1.251 > lo6 - lo5 - 2 > 0, the two roots lie in the open intervals (99999.99,

100000) and (100000, 100000.01). If x < 0, it is straightforward to see that

f(x) > 0, so that there are no negative roots.

The positive roots are solutions of the two equations

x = [lOlo + 0.5 + (x + 1.25)1’2]1’2 (1)

x = [lOlo + 0.5 - (x + 1.25)1/2]1/2. (2)