Cannot understand this solution (max, min of a set)

Question

$$A = \{ |x|: x^2+x < 2\}$$

Solving the inequality we get $-2 < x < 1$. Now in the notes there is written that this corresponds to write $A$ in this way:

$$A = [0, 2)$$

But I don't get how and why. If $x = 1.9$, this doesn't verify the inequality.

I would say the set is instead $A = [0, 1)$.

If I'm wrong, can you correct me?

Answer 1

If you don't see it clearly you can write it as $B=\{ x:x^{2}+x<2\}$ and then $A=\{|x|:x\in B\}$

As you said, $B=(-2,1)$ and then $A=\{|x|:x\in(-2,1)\}=[0,2)$

Answer 2

$A$ means "the set of all absolute values of $x$ where $x^2 + x < 2$."

It doesn't mean that in order for $x \in A$, that $|x|$ itself must satisfy the inequality; after all, it doesn't say $|x|^2 + |x| < 2$. Rather, the condition $x^2 + x < 2$ is first satisfied, then take the absolute values of all such $x$ that do.