Solve the inequality $|x^2-2|<1$

Question

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.

Answer 1

Hints:

$$|x^2-2|<1\iff -1<x^2-2<1\iff 1<x^2<3\ldots$$

Answer 2

$|x^2-2|<1$

$-1<x^2-2<1$

$1<x^2$ or $x^2<3$

$1<x$, $-1>x$, or $-\sqrt{3}<x<\sqrt{3}$

Therefore, $-\sqrt{3}<x<-1$ and $1<x<\sqrt{3}$

Answer 3

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