Question

Find the positive vakue of K for which tje equation X²+KX+64=0 and X²-8X+K=0 will have real roots.

Answer 1

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Answer 2

Solution:-

Let D₁ and D₂ be the discriminant of the given equations and these will have equal roots only

if D₁, D₂ ≥ 0

Or, if D₁ = (k² -4 × 64) ≥ 0 and D₂ = (64 - 4k) ≥ 0

Or, if k² ≥ 256 and 4k ≤ 64

Or, if k ≥ 16 and k ≤ 16

Or, k = 16

Hence the positive value of k is 16
Answer.