(QUICK)F4..today AMATH test...DIFFICULT for me

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.

1/
If kx^2 + x + k > 0 for all real values of x,

then Δ < 0
Δ = (1)^2 - 4 k (k) < 0
4 k^2 >1
k^2 > 1/4
k < -1/2 or k> 1/2


2/
Solve x|x| + 5x + 6 = 0

Case1: for x> or = 1,

x* x + 5x + 6 = 0
x^2 + 5x + 6 = 0
(x+2) (x+3) = 0
x = -2 or x = -3
so, no solution

Case 2: x<0
x * (-x) + 5x + 6 = 0
-x^2 + 5x + 6 = 0
x^2 - 5x - 6 = 0
(x+1)(x-6) = 0
x = -1 or x = 6 (rejected, since xis not <0 here)

combine the 2 cases, x = -1

3/
| x - x^2 | = -4x

{ Square both sides }

(x - x^2 )^2 = 16 x^2
x^4 - 2x^3 + x^2 = 16x^2
x^4 - 2x^3 - 15 x^2 = 0
x^2 (x^2 - 2x - 15 ) = 0
x^2 (x-5 ) (x+3) = 0
so, x =0 or x=5 or x=-3