中四數學 - 不等式 (快)

p + q + k = 1 ... (1)
(p + k)(q + k) = 1 ... (2)
From (1),
p + q + k = 1
p + q = 1 - k ... (3)
From (2),
(p + k)(q + k) = 1
pq + (p + q)k + k2 = 1
pq = 1 - (p + q)k - k2
pq = 1 - (1- k)k - k2 【From (3)】
pq = 1 - k + k2 - k2
pq = 1 - k .... (4)
Using (3) and (4), the quadratic equation with roots p and q is
x2 - (1-k)x + (1-k) = 0 .... (*)
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(b)hence, find the range of values of k.
由於p、q和k均為實數, 所以其判別式為(*)≧ 0
[-(1-k)]2 - 4(1)(1-k) ≧ 0
1 - 2k + k2 - 4 + 4k ≧ 0
k2 + 2k - 3 ≧ 0
(k + 3)(k - 1) ≧ 0
k ≦ -3 or k ≧ 1
k可取值的範圍是 k ≦ -3 or k ≧ 1

勁呀