If x2+5x+6≥0 then x lies in the interval

x2+5x+6≥0 (x+3)(x+2)≥0 x≥−2(or)x≤−3

x2+5x+6≥0 (x+3)(x+2)≥0 x≥−2(or)x≤−3

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If $x^{2} + 5 x + 6 \geq 0$ then $x$ lies in the interval

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$(- \infty, - 3) \cup (- 2, \infty)$

$(- 3, - 2)$

$[- 3, - 2]$

$(- \infty, - 3] \cup [- 2, \infty)$

answer is B.

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$x^{2} + 5 x + 6 \geq 0 (x + 3) (x + 2) \geq 0 x \geq - 2 (o r) x \leq - 3$

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