[請益] 數學一題 不等式
-(x+1)^2 < ax-(a+1) < x^2 恆成立 求a範圍?
ans:
-(x+1)^2 < ax-(ax+1) → x^2+(a+2)x-a > 0 恆成立
所以:
1 > 0
D = (2+a)^2-4.1.(-a) < 0 → -4-2√3 < a < -4+2√3
^^^^^^^^^^^^^^^^^^^^^^
這邊的不等式怎得來的?
D應該=a^2+8a+4 < 0吧.......囧
請幫我看看.....orz..謝謝
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