Aditi Chauhan
Last Activity: 10 Years ago
Sol. The given equation is
x2 – 8kx + 16(k2 – k + 1) = 0
∵ Both the roots are real and distinct
∴ D > 0 ⇒ (8k)2 – 4 x 16 (k2 – k + 1) > 0
⇒ k > 1 . . . . . . . . . . . . . . . . . . (i)
∵ Both the roots are greater than or equal to 4
∴ α + β > 8 and f (4) ≥ 0
⇒ k > . . . . . . . . . . . . . . . . . . . . (ii)
And 16 – 32k + 16(k2 – k + 1) ≥ 0
⇒ k2 – 3k + 2 ≥ 0 ⇒ (k – 1) (k – 2) ≥ 0
⇒ k ∈ (- ∞, 1)] ∪ [ 2, ∞) . . . . . . . . . . . . . . . . . (iii)
Combining (i), (ii) and (iii), we get k ≥ 2 or the smallest value of k = 2.
