Answer:
[tex]x=-1, S=\{x\:\in\mathbb{R}:x=-1\}[/tex]
Step-by-step explanation:
For the sake of clarity, assuming you meant:
[tex]|x^{2}+2x+3|\le2[/tex]
1) Absolute Value or Modulus functions has one property that assures us that:
[tex]If |a|<b \therefore \: b<a<-b\\|x^{2}+2x+3|\le2 \Rightarrow |x^{2}+2x+3|= x^{2}+2x+3\geqslant -2 \\\:And\:x^{2}+2x+3\leqslant 2[/tex]
2) Solving for x
[tex]I) x^{2}+2x+3\geqslant -2\Rightarrow x^{2}+2x+5\geqslant 0 \Rightarrow x\geqslant \frac{-2\pm \sqrt{2^{2}-4*1*5}}{2} x'\geqslant \frac{-2+4i }{2} \:and\:x''\geqslant \frac{-2-4i }{2}[/tex]
Not defined in Real Set of Numbers
Evaluating [tex]x^{2}+2x+3\leqslant 2[/tex]:
[tex]x^{2}+2x+3\leqslant 2\\x^{2}+2x+1\leqslant 0\\(x+1)(x+1)\leqslant 0 \Rightarrow S=\{-1\}[/tex]
3) So, since for the first case the Discriminant Δ <0, then the solutions presented for [tex]x^{2}+2x+3\geqslant -2\: \in \mathbb{C}[/tex]. The only solution in the Real Set for the inequality [tex]|x^{2}+2x+3|\le2[/tex] is [tex]S=\{-1\}[/tex], i.e. x=-1.
