※ 引述《debdeb ()》之铭言： : Find the range of values of the constant a for which there exist : exactly two integers satisfying the two quadratic inequalities in x : x^2 - 4x - 5 ≧ 0 : x^2 - (3a-2)x -6a ＜ 0 : simultaneously. : 请问a的范围该怎麽算？ : 　谢谢！！ (x-(3a-2)/2)^2-6a-(3a-2)^2/4 < 0 (x-(3a-2)/2)^2 < (9a^2-12a+4+24a)/4 = (3a+2)^2/4 (i) a ≧ -2/3 -(3a+2)/2 < x-(3a-2)/2 < (3a+2)/2 -2 < x < 3a but x ≧5 or x ≦ -1 hence, the integer solutions of x are -1, 5 then 5 < 3a < 6 (ii) a < -2/3 (3a+2)/2 < x-(3a-2)/2 < -(3a+2)/2 3a < x < -2 but x≧5 or x≦ -1 hence, the integer solutions of x are -3, -4 then -5 ≦ 3a < -4 -------------------------------------------------------------- The range of a 5/3 < a < 2 or -5/3 ≦ a < -4/3 --

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