Question

The smallest value of k for which both the roots of the equation x^2-8kkx+16(k^2-k+1)=0 are real, distinct and have values at least 4 is

Answer 1

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.

Answer 2

THE SMALLEST VALUE OF K IS 2

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