http://i987.photobucket.com/albums/ae355/delaynopemore/1-9.jpg
http://i987.photobucket.com/albums/ae355/delaynopemore/687b0ede.jpg
唔明白點解可以出到 1+4K>0 and ...............<0
呢2個conditions, please explain in detail
1條al問題,, quadratic??
2012-03-05 8:18 am
回答 (2)
2012-03-05 4:45 pm
✔ 最佳答案
For a quadratic function ax^2 + bx + c to be > 0 for ALL values of x, that means the function must be above the x - axis for all values of x. So 'a' must be positive (open upwards). If 'a' is negative, it will be open downwards and will cut the x - axis. In this case a = 1 + 4k, so 1 + 4k > 0 AND delta < 0.
2012-03-05 6:47 pm
For a quadratic function f(t) = (1 + 4k)t^2 - 4kt - 2k, if f(t) > 0 for all real value t, then the curve is a parabola open upward, so (1 + 4k) > 0 and the discriminant is smaller than 0.
ie (1 + 4k) > 0 and (-4k)^2 - 4(1 + 4k)(-2k) < 0
ie (1 + 4k) > 0 and (-4k)^2 - 4(1 + 4k)(-2k) < 0
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120305000051KK00011