Solve `|x^2-1|+|x^2-4| gt 3`

Explanation

To solve the inequality \( |x^2 - 1| + |x^2 - 4| > 3 \), we will follow these steps:

### Step 1: Identify critical points
The expressions inside the absolute values are \( x^2 - 1 \) and \( x^2 - 4 \). We need to find the points where these expressions are zero:
- \( x^2 - 1 = 0 \) gives \( x = \pm 1 \)
- \( x^2 - 4 = 0 \) gives \( x = \pm 2 \)

Thus, the critical points are \( -2, -1, 1, 2 \).

### Step 2: Analyze intervals
We will analyze the sign of the expressions in the intervals defined by the critical points:
1. \( (-\infty, -2) \)
2. \( (-2, -1) \)
3. \( (-1, 1) \)
4. \( (1, 2) \)
5. \( (2, \infty) \)

### Step 3: Evaluate each interval
1. **Interval \( (-\infty, -2) \)**:
- Choose \( x = -3 \):
- \( x^2 - 1 = 9 - 1 = 8 \) (positive)
- \( x^2 - 4 = 9 - 4 = 5 \) (positive)
- Thus, \( |x^2 - 1| + |x^2 - 4| = 8 + 5 = 13 > 3 \) (satisfied)

2. **Interval \( (-2, -1) \)**:
- Choose \( x = -1.5 \):
- \( x^2 - 1 = 2.25 - 1 = 1.25 \) (positive)
- \( x^2 - 4 = 2.25 - 4 = -1.75 \) (negative)
- Thus, \( |x^2 - 1| + |x^2 - 4| = 1.25 + 1.75 = 3 \) (not satisfied)

3. **Interval \( (-1, 1) \)**:
- Choose \( x = 0 \):
- \( x^2 - 1 = 0 - 1 = -1 \) (negative)
- \( x^2 - 4 = 0 - 4 = -4 \) (negative)
- Thus, \( |x^2 - 1| + |x^2 - 4| = 1 + 4 = 5 > 3 \) (satisfied)

4. **Interval \( (1, 2) \)**:
- Choose \( x = 1.5 \):
- \( x^2 - 1 = 2.25 - 1 = 1.25 \) (positive)
- \( x^2 - 4 = 2.25 - 4 = -1.75 \) (negative)
- Thus, \( |x^2 - 1| + |x^2 - 4| = 1.25 + 1.75 = 3 \) (not satisfied)

5. **Interval \( (2, \infty) \)**:
- Choose \( x = 3 \):
- \( x^2 - 1 = 9 - 1 = 8 \) (positive)
- \( x^2 - 4 = 9 - 4 = 5 \) (positive)
- Thus, \( |x^2 - 1| + |x^2 - 4| = 8 + 5 = 13 > 3 \) (satisfied)

### Step 4: Combine the intervals
From our analysis, the solution intervals are:
- \( (-\infty, -2) \)
- \( (-1, 1) \)
- \( (2, \infty) \)

### Final Solution
The solution to the inequality \( |x^2 - 1| + |x^2 - 4| > 3 \) is:
\[ x \in (-\infty, -2) \cup (-1, 1) \cup (2, \infty) \]