Solutions to Problems; Chapter (^8 389)
Prom the difference of the last two equations, we obtain that
Y(l - Y2> + 4% - 1) + x2(x - 1) = 0.
Subtracting t times (2) from (1) yields
... (2)
Similarly
x(x - 1)(1+ z + 2%) = y(y - 1)(1+ % + yz).... (3)y(y - 1)(1+ y + yx) = %(% - l)(l+ 2 + %X).... (4)It is clear from (3) and (4) that, if x, y, % are positive, then z, y, % are all
equal to 1, all less than 1, or all greater than 1. The last two possibilities
contradict the given equations and the result follows.
- Consider the graphs of the equations y = 6x2, y = 77x - 147 and
y = 77[x] - 147. Since 6x2 - 77x + 147 = (3x - 7)(2x - 21), the first two
curves cross when x = 713 and x = 2112. Prom the graphs, it can be seen
that any solution x of the given equation must satisfy 3 5 x < 2112. Hence
the possible values of [x] are integers between 3 and 10 inclusive. We now
consider the following table:
[xl Y = 77[zl- 147 ~16 [ml
3 84 14 3
4 161 26.8 5
5 238 39.7 6
6 315 52.5 7
7 392 65.3 8
8 469 78.2 8(^9 546 91 9)
10 623 103.8 10
The solutions of the equation are those values of &6 for which [x] =
[m], i.e. a‘, JG@, &i, @Z@.
- Let
Then
f(x) = x4 + kx2 - 2k2(2k + 1)x
= x[x3 + kx - 2k2(2k + l)].(1) f(x) has four distinct real zeros u the discriminant of the cubic
factor is positive _ k3(3k + 1)(36k2 + 24k + 1) < 0.(2) f(k) = -k3(3k + 1).