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348 Answers to Exercises and Solutions to Problems

Solutions to Problems

Chapter 6

4.1. (a) If the roots are r - s, r, T + s, then a = -3r, b = 3r2 - s2 and

c= -r(r2 - s2). The result follows by direct substitution.

(b) If the roots are rs-I, r, rs, then a = -r(l + u), b = r2(1 + u) and

c = -r3, where 2~ = s -+ s- l. The result follows by direct substitution.

4.2. Let u = tanA, v =tanB,w= tanC where A, B, C are the angles of

a triangle. Since

O=tan(A+B+C)= U+v’w-Uvw

l-(uv+vw+wu)’

Also,

u+v+w=uvw=p/q.

uw+vw+uv = (u+v)w+uv=(uv-l)w2+uv

= 1 + (1+ w2)(21v - 1)

= 1 + (sec2 C)(sin A sin B - cos A cos B)

cos A cos B

= l-

cos(A + B) 1

cos A cos B cos2 C

=1+

cos A cos B cos C

= 1+1/q.

The result fohows.

4.3. (a) Suppose, if possible, that a is a double zero of z3 - t2 - x - 1.
Then a is a zero of the derivative 3x2 - 2x - 1, and hence a zero of

8x + 10 = (3z2 - 2x - 1)(3x - 1) - 9(x3 - x2 - x - 1).

But then a = -514, which cannot be a zero of the cubic.
(b) The result is readily checked using symmetric functions of the roots
for n = 1 and n = 2. Use induction, based on relations such as the following
for k > 0

bk+3 - $+3 bk+2

• Ck+2 +

bk+l _ Ck+l b” _ Ck

b-c = b-c b -c +b-c-

4.4. The zeros of 17x4 + 36z3 - 14x2-4x + 1 are in harmonic progression iff
the zeros of x4-4x3-14x2+36x+17 are in arithmetic progression. Suppose
that the zeros of the latter polynomial are o - 3d, a - d, a + d, a + 3d. Then
a = 1 and (Q~ - 9d2)(u2 - d2) = 17, whence (d2 - 2)(9d2 + 8) = 0. Checking
the coefficient -14 reveals that 9d2 +8 # 0. Hence d = fi and the roots of
the given equation are the reciprocals of 1 - 3& 1 - 4, 1 + 4, 1 + 34.