Solutions to Problems; Chapter 8 375
(iii)x+y+%=l;y=z.Theny=z=$(l-x)andx2+2y2=zimply
0=3x2--442+1=(32-1)(x - 1). Hence, (x, y, Z) = (l,O,O) or (l/3, l/3,
I/3).
(iv) x + y + z = 1; y + z = 2x - 1. Then x = 2/3, and y and z are the
zeros of 9t2 - 3t + 1. Hence (z, y, Z) = (2/3,-w/3, -w2/3) or (2/3, -w2/3,
-w/3).
Remark. An alternative approach starts with multiplying the second
equation by y and the third by Z, and subtracting to get 0 = y3 - z3 =
(y - z)(yā + yz + zā) w h ence y = Z, y = wz or y = w2%. The reader should
follow up these three possibilities.
- Q(x) - P(x) = (x - u)F(x) where deg F 5 2, F(x) 1 0 for x 2 u and
F(x) < 0 for z 5 u. Hence F(z) = (x - u)G(x) where degG(x) 5 1 and
G(x) > 0 for each x. Thus G(z) must be zero or constant. Since the case
that G(x) = 0 is straightforward, we can suppose that G(x) = a and
Q(x) = P(x) + (x - U)~U where a > 0.Similarly
Q(x) = R(x) - (x - u)2b where b > 0.
Thus the result follows with k = b/(a + b).
The result does not hold when P, Q, R are of degree 4. For example, let
P(x) = x4-x2, Q(x) = x4, R(x) = 2x4+x2 (so u = 0). If Q = kP+(l-k)R,
then taking x = 1 yields 3k = 2 while taking x = 2 yields 6k = 5, which
are inconsistent.
A straight line can intersect the graph of the quartic curve in exactly
four points if and only if the curve is convex somewhere between each of
the outer pair of intersection points and concave somewhere in between
the inner pair (sketch a diagram). This occurs exactly when there are two
inflection points, i.e. the second derivative of the quartic polynomial has
two distinct real zeros. Thus P(a, b) = 3a2 - 8b.
2b = a + c =+ 2b3 = b(ab + bc) = b(3 - ac) = 3b - 10
+ O=*2b3-3b+10=(b+2)(2b2-4b+5).Hence b = -2, p = -(a + b + c) = -3b = 6 and the equation is
0 = x3 + 6x2 + 3x - 10 = (x - 1)(x + 2)(x + 5).- First solution. Suppose, if possible, that p(p(x)) = q(q(x)) has a real
solution x = 20. Let u = p(w), v = q(w). By hypothesis, u # v. We have
that P(U) = q(v), while p(v) = p(q(w)) = q(p(w)) = q(u). Hence the
polynomial p(x) - q(x) h as opposite signs for x = u and x = v, and so
must vanish between u and v. But this contradicts the hypothesis.