Solve the inequality $|x^2-2|<1$ I'm trying to solve inequalities that include absolute values. The way we solved these questions in class is different than what I am used to. The teacher would take the positive and negative possibilities of the inequality and treat them separately. I am trying to solve this as a union of intervals.
$$|x^2-2|<1$$
Normally I would just write this as:
$$-\sqrt{3} < x < -1 \text{ and } 1 < x < \sqrt{3}$$
However this is not correct I don't believe.

$$|x^2-2|<1\iff-1< x^2-2 <1$$
$$-1< x^2-2 <1\iff1<x^2<3$$
$$1<x^2<3\iff x^2-1>0\space \&\space x^2-3<0$$
$$x^2-1>0\space \&\space x^2-3<0 \iff x\in((-\infty,-1)\cup(1,\infty))\cap(-\sqrt 3,\sqrt 3)$$
$$ x\in((-\infty,-1)\cup(1,\infty))\cap(-\sqrt 3,\sqrt 3)\iff x\in(-\sqrt3,-1)\cup(1,\sqrt3)$$
