If `x^(2) > 9` then x belongs to -

Options

A. (-3,3)
B. (0,3)
C. `(3,infty)`
D. `(-infty,-3)cup(3,infty)`

Answer

To solve the inequality \( x^2 > 9 \), we can follow these steps:

### Step 1: Rewrite the inequality
We start with the inequality:
\[ x^2 > 9 \]

### Step 2: Move all terms to one side
We can rewrite this as:
\[ x^2 - 9 > 0 \]

### Step 3: Factor the expression
Next, we can factor the left-hand side:
\[ (x - 3)(x + 3) > 0 \]

### Step 4: Find the critical points
The critical points occur when the expression equals zero:
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]
\[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \]
Thus, the critical points are \( x = -3 \) and \( x = 3 \).

### Step 5: Determine the intervals
The critical points divide the number line into three intervals:
1. \( (-\infty, -3) \)
2. \( (-3, 3) \)
3. \( (3, \infty) \)

### Step 6: Test each interval
We will test a point from each interval to see where the inequality holds true.

- **Interval 1: \( (-\infty, -3) \)**
- Test point: \( x = -4 \)
- Calculation: \( (-4 - 3)(-4 + 3) = (-7)(-1) = 7 > 0 \) (True)

- **Interval 2: \( (-3, 3) \)**
- Test point: \( x = 0 \)
- Calculation: \( (0 - 3)(0 + 3) = (-3)(3) = -9 < 0 \) (False)

- **Interval 3: \( (3, \infty) \)**
- Test point: \( x = 4 \)
- Calculation: \( (4 - 3)(4 + 3) = (1)(7) = 7 > 0 \) (True)

### Step 7: Combine the results
The inequality \( (x - 3)(x + 3) > 0 \) holds true for the intervals:
- \( (-\infty, -3) \)
- \( (3, \infty) \)

### Final Answer
Thus, the solution to the inequality \( x^2 > 9 \) is:
\[ x \in (-\infty, -3) \cup (3, \infty) \]