Question

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

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

Answer 1

The correct option is B

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

$x^2+5x+6\ge 0\Rightarrow x\le -3orx\ge -2$
$x^2+3x-4<0\Rightarrow -4<x<1$
$\Rightarrow xϵ(-4,-3]\cup [-2,1)$