Answer
$x\in(-\sqrt{2},\sqrt{2})$
Work Step by Step
The square root function is defined for nonnegative real numbers, and is an increasing function.
This means that if a and b are nonnegative,
$a \lt b\Rightarrow\sqrt{a} \lt \sqrt{b}$
An alternate definition of the absolute value is $|x|=\sqrt{x^{2}}$.
Both sides of the inequality sign are nonnegative, so we may take the square root of both sides, with the direction of inequality being unchanged:
$\sqrt{x^{2}} \lt \sqrt{2}$
$|x| \lt \sqrt{2}$
by property 6 from the table 'Absolute Values and Intervals",
$-\sqrt{2} \lt x \lt \sqrt{2}$
In interval form, we can write this result as
$x\in(-\sqrt{2},\sqrt{2})$
