The set of solutions satisfying both x^2+5x+6 | vedclass

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(B) Step $1$: Solve $x^2+5x+6 \geq 0$. Factorizing,we get $(x+2)(x+3) \geq 0$. The roots are $x = -3$ and $x = -2$. Testing intervals,the solution is

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

Vedclass · Answer

(B) Step $1$: Solve $x^2+5x+6 \geq 0$. Factorizing,we get $(x+2)(x+3) \geq 0$. The roots are $x = -3$ and $x = -2$. Testing intervals,the solution is $x \in (-\infty, -3] \cup [-2, \infty)$. Step $2$: Solve $x^2+3x-4 Factorizing,we get $(x+4)(x-1) The roots are $x = -4$ and $x = 1$. Testing intervals,the solution is $x \in (-4, 1)$. Step $3$: Find the intersection of the two sets. Intersection of $(-\infty, -3] \cup [-2, \infty)$ and $(-4, 1)$ is $(-4, -3] \cup [-2, 1)$. Thus,the correct option is $B$.

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

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