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If the equation $x^{2} + 4 x + k = 0$ has real and distinct roots, then

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$k > 4$

$k < 4$

$k \leq 0$

$k \geq 0$

answer is B.

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Detailed Solution

$b^{2} - 4 ac > 0 \Rightarrow 4^{2} - 4 \times 1 \times k > 0 \Rightarrow 16 - 4 k > 0$

$16 > 4 k \Rightarrow 4 k < 16 \Rightarrow k < 4$

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