Quadratic Equation and Inequalities

We have x2−8kx+16(k2−k+1) = 0
D > 0 ⇒ k > 1 (1)
−

b

2a

> 4 ⇒

8k

2

> 4
⇒ k > 1 (2)
Now, f (4) ≥ 0 ⇒ 16 - 32k +16(k2 - k +1) ≥ 0. Hence,
k2 − 3k + 2 ≥ 0
k ≤ 1 ∪ k ≥ 2 (3)
Using Eqs. (1), (2) and (3), we get kmin 2.