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If the roots of $x^{2} + 2 x + k = 0$ are real and distinct, then:

Competition practiceAlgebra

Ak < 1
Bk > 1
C $k = 1$
Dk ≥ 1

Step-by-Step Solution

Real and distinct roots need

D > 0

, so

2^{2} - 4 (1) (k) = 4 - 4 k > 0

, which gives

k < 1

.

Answer: A) k < 1

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