Question

$\sqrt{x+8} > x+2$ Решить неравенство помогите

Answer 1

$\sqrt{x+8}>x+2\; \; \Leftrightarrow \; \; \left \{ {{x+2\geq 0} \atop {x+8>x^2+4x+4}} \right.\; \; ili\; \; \left \{ {{x+2<0} \atop {x+8\geq 0}} \right.\\\\\left \{ {{x\geq -2} \atop {x^2+3x-4<0}} \right.\; \; ili\; \; \; \left \{ {{x<-2} \atop {x\geq -8}} \right.\\\\\star \; \; x^2+3x-4=0\; ,\; \; D=25\; ,\; \; x_{1}=-4\; ,\; \; x_2=1\; \; \star \\\\\left \{ {{x\geq -2} \atop {(x+4)(x-1)<0}} \right.\; \; ili\; \; \; -8\leq x<-2\\\\\left \{ {{x\geq -2} \atop {x\in (-4,1)}} \right.\; \; \; ili\; \; \; x\in [-8\, ;\, -2\, )\\\\x\in [-2\, ;\, 1)\quad ili\quad x\in [-8\, ;\, -2)\\\\Otvet:\; \; x\in [-8\,;\, -2)\cup [-2\, ;\, 1)=[-8\, ;\, 1)\; .$