Question

Решить неравенство √(х+8)≥х+2

Answer 1

$\sqrt{x+8} \geq x+2\\\\1)\left \{ {{x+2<0} \atop {x+8\geq 0}} \right.\\\\\\\left \{ {{x<-2} \atop {x\geq -8}} \right.\Rightarrow x\in[-8;-2)\\\\\\2)\left \{ {{x+2\geq0 } \atop {x+8\geq(x+2)^{2}}} \right.\\\\\\\left \{ {{x\geq-2} \atop {x+8\geq x^{2}+4x+4}} \right.\\\\\\\left \{ {{x\geq-2 } \atop {x^{2}+3x-4\leq0}} \right.\\\\\\\left \{ {{x\geq -2} \atop {x\in[-4;1]}} \right. \Rightarrow x\in [- 2;1]\\\\Otvet:\boxe{[-8;1]}$