1條al問題,, quadratic??

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.

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