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Grade 11Algebra

The smallest value of k, for which both the roots of the equation
x² – 8kx + 16 (k² – k + 1) = 0
are real, distinct and have values at least 4, is

Simran Bhatia

12 Years agoGrade 11

2 Answers

Aditi Chauhan

12 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.

SHAIK AASIF AHAMED

12 Years ago

Hello student,

for roots to be real and distinct its discriminant should be >0

hence (8k)^2-4(16)(k²-k+1)>0

we get k>1

Thanks and regards

Shaik Aasif

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