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298 Answers to Exercises and Solutions to Problems

then by Exercise 4.2, x : y : z = (br - cq) : (cp - ar) : (aq - bp). If there

is a solution of the form (x, y, z) = ( u2, u, l), then (a) follows, and y2 = xz

shows that the condition (b) is necessary.

If (b) holds, then for every solution of system (+), y2 = xz. If there is a

solution with nonzero Z, then there is a solution with z = 1; if y = u, then

x = u2 and the two quadratic equations have a common root. On the other

hand, if it turns out that z = 0, then y = 0. Since up # 0, x = 0, so 0 is a

common root of the two quadratic equations.

1.5. (a) 12(b - 4a) = (a - b)2.

1.6. (b) Since x-l : y-l : 2-i = (qr - p2) : (pr - q2) : (pq - r2), we

obtain a(qr - p2) + b(pr - q2) + c(pq - r”) = 0.

1.8. x, y, z are the zeros of t3 - 12t2 + 41t - 42 and are therefore 2, 3, 7 in

some order.

1.9. (a) 3, 5, 7. (b) 1, -2, 3.

1.10. (x,y) = (u,u) or (-4 v , v ) w h

or (-1 f ifi)/14.

ere u = 1 or (-2 f i&)/7 and v = 2

1.11. (x, y,z) = (6,8,2).
2.1. (g) x = 7 satisfies (d), but not (a), (b), (c).
(h) Consider an equation of the type U = V. This implies U2 = V2.
However, if U2 = V2, then either U = V or U = -V. Hence, U2 = V2 is
valid under a wider range of circumstances than is U = V. We have that
(a) is equivalent to the given equation; (a) 3 (b); (b) _ (c); (c) 3 (d).
(i) m - &FZ = 1.

2.2. The equation becomes cy2 + aky + (ad -,bc) = 0. If this equation has a
nonnegative solution y, then there is exactly one corresponding solution x
to the given equation. Any negative or nonreal solution y yields no solution
2.
If c = 0, there is a unique solution y = -d/k and a solution x when
d/k < 0.
If c # 0, there are no solutions 3: if a2k2 < 4c(ad - bc) or if a2k2 2
4c(ad - bc) 1 0 and ak/c > 0. If a2k2 > 4c(ad - bc) > 0 and ak/c < 0,
then there are two nonnegative solutions y and two solutions x. If a2k2 =
4c(ad - bc) and ak/c 5 0, or if a2k2 > 4c(ad - bc) and 4c(ad - bc) < 0,
then there is one nonnegative solution y and one solution x.
Discussion of problems of this type can be found in Goro Nagase, Exis-
tence of real roots of a radical equation, Math. Teacher 80 (1987), 369-370.

2.3. (a) y = m leads to y2 -y-6 = 0. The solution y = 3 corresponds
to x = 5; y = -2 is extraneous.
(b) y = m leads to y2 - 4y + 3 = 0, and (x, y) = (8,3), (0,l).
(c)y=-1 ea d s t o y2 + 4y +^3 = 0. There are no solutions x.