Solve : `(x^2 +x -4)/(x) lt 1 and x^(2) lt 64`

Answer

To solve the inequalities \(\frac{x^2 + x - 4}{x} < 1\) and \(x^2 < 64\), we will follow these steps:

### Step 1: Solve the first inequality \(\frac{x^2 + x - 4}{x} < 1\)

1. Start by rewriting the inequality:
\[ \frac{x^2 + x - 4}{x} - 1 < 0 \]
This simplifies to:
\[ \frac{x^2 + x - 4 - x}{x} < 0 \]
Which further simplifies to:
\[ \frac{x^2 - 4}{x} < 0 \]

2. Factor the numerator:
\[ \frac{(x - 2)(x + 2)}{x} < 0 \]

### Step 2: Determine the critical points

The critical points occur when the numerator or denominator is zero:
- \(x - 2 = 0 \Rightarrow x = 2\)
- \(x + 2 = 0 \Rightarrow x = -2\)
- \(x = 0\) (from the denominator)

Thus, the critical points are \(x = -2, 0, 2\).

### Step 3: Test intervals around the critical points

We will test the sign of the expression in the intervals:
1. \( (-\infty, -2) \)
2. \( (-2, 0) \)
3. \( (0, 2) \)
4. \( (2, \infty) \)

- For \(x < -2\) (e.g., \(x = -3\)):
\[ \frac{(-3 - 2)(-3 + 2)}{-3} = \frac{(-5)(-1)}{-3} = \frac{5}{-3} < 0 \quad \text{(True)} \]

- For \(-2 < x < 0\) (e.g., \(x = -1\)):
\[ \frac{(-1 - 2)(-1 + 2)}{-1} = \frac{(-3)(1)}{-1} = \frac{-3}{-1} > 0 \quad \text{(False)} \]

- For \(0 < x < 2\) (e.g., \(x = 1\)):
\[ \frac{(1 - 2)(1 + 2)}{1} = \frac{(-1)(3)}{1} = -3 < 0 \quad \text{(True)} \]

- For \(x > 2\) (e.g., \(x = 3\)):
\[ \frac{(3 - 2)(3 + 2)}{3} = \frac{(1)(5)}{3} > 0 \quad \text{(False)} \]

### Step 4: Combine the intervals

From the tests, the solution for the first inequality is:
\[ x \in (-\infty, -2) \cup (0, 2) \]

### Step 5: Solve the second inequality \(x^2 < 64\)

1. Rewrite the inequality:
\[ x^2 - 64 < 0 \]
Factor it:
\[ (x - 8)(x + 8) < 0 \]

### Step 6: Determine the critical points

The critical points are:
- \(x - 8 = 0 \Rightarrow x = 8\)
- \(x + 8 = 0 \Rightarrow x = -8\)

### Step 7: Test intervals around the critical points

We will test the sign of the expression in the intervals:
1. \( (-\infty, -8) \)
2. \( (-8, 8) \)
3. \( (8, \infty) \)

- For \(x < -8\) (e.g., \(x = -9\)):
\[ (-9 - 8)(-9 + 8) = (-17)(-1) > 0 \quad \text{(False)} \]

- For \(-8 < x < 8\) (e.g., \(x = 0\)):
\[ (0 - 8)(0 + 8) = (-8)(8) < 0 \quad \text{(True)} \]

- For \(x > 8\) (e.g., \(x = 9\)):
\[ (9 - 8)(9 + 8) = (1)(17) > 0 \quad \text{(False)} \]

### Step 8: Combine the intervals

From the tests, the solution for the second inequality is:
\[ x \in (-8, 8) \]

### Step 9: Find the intersection of the two solutions

We need to find the intersection of:
1. \(x \in (-\infty, -2) \cup (0, 2)\)
2. \(x \in (-8, 8)\)

The intersection is:
1. From \((-8, -2)\)
2. From \((0, 2)\)

Thus, the final solution is:
\[ x \in (-8, -2) \cup (0, 2) \]