Browse Questions

# The set of solutions satisfying both $x^2+5x+6 \geq 0$ and $x^2+3x-4 < 0$ is:

$(a)\;(-4,1) \quad (b)\;(-4,-3]\;\cup\;[-2,1) \quad (c)\;(-4,-3) \; \cup \; (-2,1) \quad (d)\;[-4,-3] \cup [-2,1]$

$(b)\;(-4,-3]\;\cup\;[-2,1)$