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Solutions to Problems; Chapter 5 331

We have that f(a) = f(b) = 0 and f’(t) = (n+2)(a-b)t”+‘-(a”+2-bn+2).

If n is even, f’(t) h as exactly one real zero, counting multiplicity, so that

f(t) has at most two. These are already accounted for by the extraneous

roots a and b of f(t) = 0, so that the original equation has no root.

If n is odd, then f’(t) h as exactly two zeros counting multiplicity, so f(t)

has at most three. Since two of these are accounted for by a and b, then

given equation has at most one root.

4.7. Let

f(t) = 2t3 - pt2 + qt - r = (t - al)(t - as)(t - as)

+ (t - a2)(t - a4)(t - as).

Since f (ai) is positive for i = 1,4,5 and negative for i = 2,3,6, there must

be a real zero in each of the intervals (asj-1, asj) for j = 1,2,3.

4.8. By Descartes’ Rule of Signs, there is at most one positive root. De-
noting the left side by f(x), we have that

f (a + b + c) = 2Ca2b + 4abc > 0

and

27f(2(a + b + c)/3) = -[lOXa - 6Ca’b + Gabc] = -{[S(a + b) - c][a - b12

+ [5(b + c) - a][b - cl2 + [5(c + a) - b][c - a]“}.

Without loss of generality, suppose a 2 b 1 c. If a 5 5(b + c), then
f(2(a + b + c)/3) 5 0. On the other hand, if a > 5(b + c), then

lOXa - 6Ca2b + 6abc > [2u3 - 6a2(b + c)] + [2a3 - 6a(b2 + c2]

+ [2a3 - Gbc(b + c)] > 4a2(b + c) + 44a(b2 + c”) + 4bc(b + c) > 0,

so again f (2(u + b + c)/3) < 0. The required result follows.

Remark. Another solution can be found in Crux Math. 11 (1985), 129.
See also the discussion to problem 787 in Crux Math. 10 (1984), 56-58, in
which it is pointed out that, if a, b, c are three sides of a given quadrilateral,
the length E of the fourth side in order to achieve maximum area is given
by the equation of the problem. In fact, the points can be arranged in a
semi-circle whose diameter is the fourth side.

4.9. By the Fundamental Theorem of Algebra, we have that

(x - a1)(2 - a2).. .(x - a,) - 1 = (x - rr)(x - rs) f.. (x - r,),

whence it can be seen that the second equation has n zeros a;.