一元二次式

1.
Let p and q (p > q) be the roots of 4x2 + 2kx - 5 = 0
p + q = -2k/4 = -k/2
pq = -5/4

The difference of the roots = 3
p - q = 3
(p - q)2 = 9
p2 - 2pq + q2 = 9
(p2 + 2pq + q2) - 4pq = 9
(p + q)2 - 4(pq) = 9
(-k/2)2 - 4(-5/4) = 9
(k2/4) + 5 = 9
k2/4 = 4
k2 = 16
k = 4 or k = -4


2.
x2 + 1 = k(x - 1)
x2 + 1 = kx - k
x2 - kx + (k + 1) = 0

P and Q are the roots of x2 - kx + (k + 1)
P + Q = k
PQ = k + 1

(P - Q)2 = 8
P2 - 2PQ + Q2 = 8
(P2 + 2PQ + Q2) - 4PQ = 8
(P + Q)2 - 4(PQ) = 8
k2 - 4(k + 1) = 8
k2 - 4k - 4 = 8
k2 - 4k - 12 = 0
(k + 2)(k - 6)
k = -2 or k = 6


3.
(a, 0) and (c, 0) lie on the graph y = -x2 + 4x + k
When y = 0, then x = a or x = c
Then, a and c are the roots of -x2 + 4x + k = 0
a + c = -4/(-1) = 4
ac = k/(-1) = -k

(a - c)2
= a2 - 2ac + c2
= (a2 + 2ac + c2) - 4ac
= (a + c)2 - 4(ac)
= (4)2 - 4(-k)
= 16 + 4k

PQ = 6
√[(a - c)2 - (0 - 0)2] = 6
√(a - c)2 = 6
√(16 + 4k) = 6
16 + 4k = 36
4k = 20
k = 5