✔ 最佳答案
1. Let the roots be a and a - 3
Sum of roots = -2k/4 = -k/2 = a + a - 3 = 2a - 3 => k = 6 - 4a
Product of roots = -5/4 = a(a - 3)
a(a - 3) = -5/4
a^2 - 3a + 5/4 = 0
4a^2 - 12a + 5 = 0
(2a - 5)(2a - 1) = 0
a = 5/2 or a = 1/2
When a = 5/2 => k = 6 - 10 = -4
When a = 1/2 => k = 6 - 2 = 4
2. x^2 + 1 = k(x - 1)
x^2 - kx + (1 + k) = 0
Sum of roots = P + Q = k
Product of roots PQ = 1 + k
(P - Q)^2
= P^2 - 2PQ + Q^2
= P^2 + 2PQ + Q^2 - 4PQ
= (P + Q)^2 - 4PQ
= k^2 - 4(1 + k) = 8
k^2 - 4k - 12 = 0
(k - 6)(x + 2) = 0
k = 6 or k = -2
3. y=-x^2 + 4x + k
when y = 0, x = a or x = c
So a and c are the roots of -x^2 + 4x + k = 0
Sum of roots a + c = 4
Product of roots ac = -k
(a - c)^2
= a^2 - 2ac + c^2
= a^2 + 2ac + c^2 - 4ac
= (a + c)^2 - 4ac
= 16 + 4k
PQ = | a - c | = 6
(a - c)^2 = 36
16 + 4k = 36
4k = 20
k = 5
