Han Zhi Yuan's answer to Shams's Secondary 3 A Maths Singapore question.
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The solution set for the inequality x^2 + cx > 3 is {x : x < -1 or x > d}.
So x(x+c) > 3.
When x = -1, (-1)(c-1) = 3.
=> (c-1) = -3
=> c = -2
When x = d, d(d+c) = 3.
=> d(d-2) = 3
=> d^2 - 2d - 3 = 0
=> (d-3)(d+1) = 0
=> d = 3 or d = -1.
d = -1 is rejected as when x = 0, (0)^2 - 2(0) = 0, which is less than 3.
Hence c = -2 and d = 3.
So x(x+c) > 3.
When x = -1, (-1)(c-1) = 3.
=> (c-1) = -3
=> c = -2
When x = d, d(d+c) = 3.
=> d(d-2) = 3
=> d^2 - 2d - 3 = 0
=> (d-3)(d+1) = 0
=> d = 3 or d = -1.
d = -1 is rejected as when x = 0, (0)^2 - 2(0) = 0, which is less than 3.
Hence c = -2 and d = 3.
Date Posted:
3 years ago
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