Solutions to Problems; Chapter (^8 379)
When p = -l/4, then a = b. Hence there is a unique solution if and only
ifp=-1/4orO<p<2.
- Let a, b, c be the zeros of the cubic polynomial. Then, a, b, c are the
lengths of the sides of a triangle if and only if a, b, c are positive real
numbers for which a + b - c, a + c - b and b + c - a are all positive.
Suppose that these conditions hold. Then the discriminant 18uvw +
u2v2 - 27w2 - 4v3 - 4u3w is nonnegative and the numbers u, v, w,
4uv - u3 - 8w = (a + b - C)(U + c - b)(b + c - u)are all positive.
On the other hand, let these conditions hold. The discriminant condition
guarantees that all the zeros are real. Since the polynomial is negative for
nonpositive I, a, b, c must all be positive. Since two of a + b - c, a + c - b,
b+c-a are positive in any case, the condition that their product is positive
ensures that all three are positive.
- First solution. We look at a number of cases.
(1) a = b = c = 0. The system is trivially solvable.
(2) a = b = 0, c # 0. The system is not solvable, since it becomes
z = dm = 0, which is impossible.
(3) a = 0, bc # 0. The system becomes
If the system is solvable, then b2y2 = c2z2 and b2(1 - y2) = c2(1 - z2),
which implies b2 = c2. On the other hand, suppose that b2 = c2. Then
(i) if b = c, the system is satisfied by (x, y, Z) = (p, 1, -l), where p is
arbitrary;(ii) if b = -c, the system is satisfied by (x, y, Z) = (p, q, q) where p, q are
arbitrary.Thus, in Case (3)’ there is a solution iff lb/ = ICI.
(4) abc # 0: a, b, c all positive. Then
Since ax + by + cz = 0, x, y, z cannot have all the same sign, so one of a,
b, c is the sum of the other two. On the other hand, if, say, a = b + c, then
(2, y, z) = (-1, 1’1) is a solution.
(5) abc # 0: a, b, c all negative. As in (4), it can be shown that there is
a solution iff one of a, b, c is the sum of the other two.
(6) a, b, c # 0: not all of a, b, c have the same sign. Let there be a solution
(x, y, 2) = (cos u, cos v, cos w) where 0 5 u, v, w 5 r. Then
acosu+bcosv+ccosw=O