Find the positive value of k for which x2+kx+64=0 and x2-8x+k=0 will have real roots.
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f(x) = x² + kx +64 = 0 and g(x) = x²-8x + k = 0
For real roots : b²-4ac ≥ 0
1) k²- 4. 64 ≥ 0 ⇒ (k-16)(k+16) ≥ 0 ⇒ k ∈ (- ∞ , -16] U [16 , ∞)
2) 64 - 4k ≥ 0 ⇒ 16 - k ≥ 0 ⇒ k ∈ [16 , ∞)
Intersection gives k ∈ [16, ∞)
So least positive of k is 16.
For real roots : b²-4ac ≥ 0
1) k²- 4. 64 ≥ 0 ⇒ (k-16)(k+16) ≥ 0 ⇒ k ∈ (- ∞ , -16] U [16 , ∞)
2) 64 - 4k ≥ 0 ⇒ 16 - k ≥ 0 ⇒ k ∈ [16 , ∞)
Intersection gives k ∈ [16, ∞)
So least positive of k is 16.
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