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secondary 3 | A Maths
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Date Posted: 3 years ago
Views: 191
x² - x + 4 > k(x-1)
x² - (k+1)x + (4+k) > 0
a = 1
b = -(k+1)
c = 4+k
b²-4ac > 0
= (k+1)² - 4(4+k)
= k² -2k -15
= (k-5)(k+3) > 0
-3 < k < 5
x² - (k+1)x + (4+k) > 0
a = 1
b = -(k+1)
c = 4+k
b²-4ac > 0
= (k+1)² - 4(4+k)
= k² -2k -15
= (k-5)(k+3) > 0
-3 < k < 5
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