Question

The smallest value of k, for which both the roots of the equation

x^2 - 8kx + 16(k^2 – k + 1) = 0

are real and have values at least 4, is

(A) 1

(B) 2

(C) 0

(D) 4

Solution

The given equation is

x^2 - 8kx + 16(k2 - k + 1) = 0

∵ Both the roots are real and distinct

∴ D > 0 ⇒ (8k)^2 - 4 x 16 (k^2 - k + 1) > 0

k > 1............ (i)

∵ Both the roots are greater than or equal to 4

∴a +β >8 and f (4) ≥ 0

⇒k >.............................. (ii)

And 16 - 32k + 16(k2 - k + 1) ≥ 0

⇒ k^2 - 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.