✔ 最佳答案
1. kx^2 + x + k > 0 for all real values of x
=> k > 0 and discriminant < 0
discriminant = [(-1)^2 - 4k(k)] < 0
=> 1 < 4k^2
=> k^2 > 1/4
=> k > 1/2 or k < -1/2 (rejected since k > 0)
Therefore k > 1/2.
2. Case 1: Suppose x >= 0. Then |x| = x and thus x|x| = x^2.
Hence, x^2 + 5x + 6 = 0
=> (x+2)(x+3) = 0
=> x = -2 or x = -3 (contradiction)
Case 2: Suppose x < 0. Then |x| = -x and thus x|x| = -x^2.
Hence, -x^2 + 5x + 6 = 0
=> (x-2)(x-3) = 0
=> x = 2 or x =3 (contradiction)
Therefore, the equation does not have any solution.
3. Solve | x - x^2 | = -4x
Case 1: x - x^2 = -4x
=> x^2 - 5x = 0
=> x(x - 5) = 0
=> x = 0 or x = 5 (rejected since -4x = |x - x^2| >= 0 or x <= 0).
Case 2: -(x - x^2) = -4x
=> x - x^2 = 4x
=> x^2 + 3x = 0
=> x(x + 3) = 0
=> x = 0 or x = -3
In conclusion, x = 0 or x = -3
2007-11-02 09:51:37 補充:
Correcton of case 2 in question 2:Case 2: Suppose x < 0. Then |x| = -x and thus x|x| = -x^2.Hence, -x^2 + 5x + 6 = 0=> (x+2)(x-3) = 0=> x = -2 or x =3 (contradiction)In conclusion, x = -2.
2007-11-02 09:52:54 補充:
Correcton of case 2 in question 2 again:Case 2: Suppose x < 0. Then |x| = -x and thus x|x| = -x^2.Hence, -x^2 5x 6 = 0=> (x-1)(x+6) = 0=> x = -1 or x =6 (contradiction)In conclusion, x = -1.