F.4 A.maths Discriminant3

Determine the value of k for which each equation has 2 different complex solutions(D<0)
1. x2 – 4kx + (2k + 1)2 = 0
D = (-4k)2 – 4(2k + 1)2
= 16k2 – 4(4k2 + 4k + 1)
= 16k2 – 16k2 – 16k – 4
= - 16k – 4 < 0
16k + 4 > 0
16k > - 4
k > -1/4
2. kx2 – 2(k + 1)x + (k – 1) = 0 (k <> 0)
D = [-2(k + 1)]2 – 4k(k – 1)
= 4k2 + 8k + 4 – 4k2 + 4k
= 12k + 4 < 0
12k < - 4
k < -1/3
Discriminate the solutions of the following quadratic equations.
1. x2 – 3x + k = 0
D = (-3)2 – 4k = 9 – 4k
If 9 – 4k >= 0 or k <= 9/4, real roots
If 9 – 4k = 0 or k = 9/4, double roots
If 9 – 4k < 0 or k > 9/4, complex roots
2. 2x2 – 4x – 3m = 0
D = (-4)2 – 4(2)(-3m) = 16 + 24m
If 16 + 24m >= 0 or m >= -2/3, real roots
If 16 + 24m = 0 or m = -2/3, double roots
If 16 + 24m < 0 or m < -2/3, complex roots