Find the set of possible values of k for each equation with the roots described.
1) x^2 - kx + k = 0 has two distinct real roots
2) 2x^2 + kx + k = 0 has no real roots
3) x^2 + (k+4) x+(4K+1) = 0 has real roots
4) (k-1) x^2 +(3k-1) x+ (k+5) = 0 has no real roots
1) x^2 - kx + k = 0 has two distinct real roots
This will occur when the discriminant is > 0
So
(-k)^2 - 4 (1)(k) > 0
k^2 - 4k > 0
k (k - 4) > 0
This will be true when k < 0 or when k > 4
2) 2x^2 + kx + k = 0 has no real roots
This will occur when the discriminant is < 0
So
k^2 - 4(2)(k) < 0
k^2 - 8k < 0
k(k - 8) < 0
This will be true when k lies in the interval (0, 8)
3) x^2 + (k+4) x+(4K+1) = 0 has real roots
This will occur when the discriminant = 0 (double real root) or the discriminant > 0 [two distinct real roots ]
So
(k + 4)^2 - 4 (1) (4k + 1) = 0
k^2 + 8k + 16 - 16k - 4 = 0
k^2 - 8k + 12 = 0
(k - 6) (k - 2) = 0
Setting both factors to 0 and solving for k produces k = 6 or k = 2
If the discriminant > 0...we arrive at
(k - 6) (k - 2) > 0
This will be true whenever k < 2 or when k > 6
4) (k-1) x^2 +(3k-1) x+ (k+5) = 0 has no real roots
This will be true when the discriminant < 0
So
(3k - 1)^2 - 4(k - 1)(k + 5) < 0
9k^2 - 6k + 1 - 4 (k^2 + 4k - 5) < 0
9k^2 - 6k + 1 - 4k^2 - 16k + 20 < 0
5k^2 - 22k + 21 < 0
(5k - 7)(k - 3) < 0
This will be true whenever k lies in the interval (7/5, 3)