We are given the inequality:

$\left|x + \frac{2}{x}\right| < 3$

We need to determine the values of $x$ for which this inequality holds true.

Step 1: Consider the two cases for absolute value.

The absolute value inequality can be split into two separate inequalities:

1. $x + \frac{2}{x} < 3$
2. $x + \frac{2}{x} > -3$

Step 2: Solve the inequality $x + \frac{2}{x} < 3$.

Rearranging the inequality:

$x + \frac{2}{x} - 3 < 0 \quad \text{or} \quad x - 3 + \frac{2}{x} < 0$

Multiply through by $x$ to clear the denominator (keeping in mind that $x \neq 0$):

$x^2 - 3x + 2 < 0$

Factor this quadratic expression:

$(x - 1)(x - 2) < 0$

This inequality holds when $1 < x < 2$. Thus, for this part of the solution, $x \in (1, 2)$.

Step 3: Solve the inequality $x + \frac{2}{x} > -3$.

Rearranging the inequality:

$x + \frac{2}{x} + 3 > 0 \quad \text{or} \quad x + 3 + \frac{2}{x} > 0$

Multiply through by $x$ (again, $x \neq 0$):

$x^2 + 3x + 2 > 0$

Factor the quadratic expression:

$(x + 1)(x + 2) > 0$

This inequality holds when $x \in (-\infty, -2) \cup (-1, 0)$.

Step 4: Combine the two inequalities.

From Step 2, we know that $1 < x < 2$, and from Step 3, $x \in (-\infty, -2) \cup (-1, 0)$.

Step 5: Conclusion.

The solution set is:

$x \in (-\infty, -2) \cup (-1, 1) \cup (2, \infty)$

This corresponds to option B.

Let me know if you need further clarification or more details.

5 Related Questions:

1. How do you solve inequalities involving absolute values?
2. How do you factor quadratic expressions to solve inequalities?
3. What is the significance of testing intervals in inequality solutions?
4. Can you graph the function $f(x) = x + \frac{2}{x}$ to visualize the solution?
5. How do different transformations affect the solutions of inequalities?

Tip: When solving absolute value inequalities, always split the inequality into two cases—one for the positive value and one for the negative value.