Problem 1: Solve the inequality (x ^ 2 - 1)/(x ^ 2 - 4) + (x - 2)/(x + 2) >= 0

To solve the inequality:

\[
\frac{x^2 - 1}{x^2 - 4} + \frac{x - 2}{x + 2} \geq 0,
\]

we'll follow these steps:

### Step 1: Simplify the expression
First, factor both the numerator and denominator of the rational terms.

1. \(x^2 - 1\) can be factored as \((x - 1)(x + 1)\).
2. \(x^2 - 4\) can be factored as \((x - 2)(x + 2)\).

Thus, we can rewrite the inequality as:

\[
\frac{(x - 1)(x + 1)}{(x - 2)(x + 2)} + \frac{x - 2}{x + 2} \geq 0.
\]

### Step 2: Combine the terms
We need a common denominator to combine the two fractions:

The denominator is already \( (x + 2) \) for both fractions, but the first fraction has an additional \( (x - 2) \) factor in its denominator. So, we'll multiply the second fraction by \( \frac{x - 2}{x - 2} \) to align the denominators.

This gives:

\[
\frac{(x - 1)(x + 1) + (x - 2)^2}{(x - 2)(x + 2)} \geq 0.
\]

### Step 3: Expand and simplify the numerator
Expand the terms in the numerator:

- \( (x - 1)(x + 1) = x^2 - 1 \).
- \( (x - 2)^2 = x^2 - 4x + 4 \).

Now, adding these:

\[
x^2 - 1 + (x^2 - 4x + 4) = 2x^2 - 4x + 3.
\]

Thus, the inequality becomes:

\[
\frac{2x^2 - 4x + 3}{(x - 2)(x + 2)} \geq 0.
\]

### Step 4: Solve the inequality
We need to find where this rational expression is greater than or equal to zero. First, solve for the zeros of the numerator and the undefined points from the denominator.

- The numerator \( 2x^2 - 4x + 3 \) is a quadratic equation. Solve it using the quadratic formula:
  \[
  x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(3)}}{2(2)} = \frac{4 \pm \sqrt{16 - 24}}{4} = \frac{4 \pm \sqrt{-8}}{4}.
  \]

  Since the discriminant is negative, there are no real solutions for the numerator.

- The denominator is zero when \( x = 2 \) or \( x = -2 \). These are the points of discontinuity.

### Step 5: Sign analysis
We will perform a sign analysis to check the behavior of the expression in the different intervals created by the critical points \( x = -2 \) and \( x = 2 \).

Consider the intervals:

1. \( (-\infty, -2) \),
2. \( (-2, 2) \),
3. \( (2, \infty) \).

We now check the sign of the expression in each interval. Since the quadratic in the numerator has no real roots and opens upwards (leading coefficient is positive), it is always positive. So, the sign of the whole expression depends on the denominator:

- For \( x \in (-\infty, -2) \), both \( x - 2 \) and \( x + 2 \) are negative, so the product is positive.
- For \( x \in (-2, 2) \), \( x - 2 \) is negative and \( x + 2 \) is positive, so the product is negative.
- For \( x \in (2, \infty) \), both \( x - 2 \) and \( x + 2 \) are positive, so the product is positive.

### Step 6: Conclusion
The inequality is satisfied when the expression is greater than or equal to zero. Therefore, the solution is:

\[
(-\infty, -2) \cup (2, \infty).
\]

The points \( x = -2 \) and \( x = 2 \) are excluded because they make the denominator zero, causing the expression to be undefined.

---

Would you like further clarification on any step or additional details? Here are five related questions for further exploration:

1. How would the inequality change if the quadratic had real roots?
2. How do you determine where a rational expression is undefined?
3. Can you solve inequalities involving absolute values similarly?
4. How does the sign analysis change if the quadratic opens downwards?
5. How can this method apply to inequalities with more complex numerators?

**Tip**: Always factor and simplify expressions before solving inequalities; this helps in determining critical points more easily!

Solve the inequality (x ^ 2 - 1)/(x ^ 2 - 4) + (x - 2)/(x + 2) >= 0

Step-by-step solution of the inequality (x^2 - 1)/(x^2 - 4) + (x - 2)/(x + 2) >= 0, involving quadratic equations, rational expressions, and sign analysis. Suitable for high school students in Grades 10-12.

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