Question

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

• A. $(-4,1)$
• B. $(-4,-3]$ $\cup$ $[-2,1)$
• C. $(-4,-3)$ $\cup$ $(-2,1)$
• D. $[-4,-3]$ $\cup$ $[-2,1]$

Answer 1

Answer: (B)

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

$x^2+5x+6\ge 0\Rightarrow (x+2)(x+3)\ge 0\Rightarrow x\in (-\infty ,-3]\cup [-2,\infty )$

and $x^2+3x-4<0\Rightarrow (x+4)(x-1)<0\Rightarrow x\in (-4,1)$

Hence set of solution satisfying both is $(-4,-3]\cup [-2,1)$