-1}} \right." alt="\left \{ {{x^2+4x<1} \atop {x^2+4x>-1}} \right." align="absmiddle" class="latex-formula">
0}} \right." alt="\left \{ {{x^2+4x-1<0} \atop {x^2+4x+1>0}} \right." align="absmiddle" class="latex-formula">
решим оба квадратных уравнения:
1)x^2+4x-1<0;</p>
D=16+4*1=20=4*5;
x1=(-4+2√5)/2=-2+√5;
x1=(-4-2√5)/2=-2-√5;
+ - +
___-2-√5____-2+√5____
-2-√5
2)x^2+4x+1>0;
D=16-4*1=12=4*3;
x1=(-4+2√3)/2=-2+√3;
x1=(-4-2√3)/2=-2-√3;
+ - +
___-2-√3____-2+√3____
x<-2-√3;</strong>
x>-2+√3;
x ∈ (-2-√5;-2-√3) ∪ (-2+√3;-2+√5);