Algebrahard · Past Paper

For what value of k will the equation (k+1)x^2 - 2(k-1)x + 1 = 0 have real roots?

Ak ≥ 0
Bk ≤ 3
C0 ≤ k ≤ 3
DAll k

✓ Correct Answer: Bk ≤ 3

D ≥ 0 → 4(k-1)^2 - 4(k+1) ≥ 0 → k^2-2k+1-k-1 ≥ 0 → k^2-3k ≥ 0 → k(k-3) ≥ 0.

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