Which of the following is the empty set

Options

A. {x : x is a real number and `x^(2) - 1 = 0`}
B. {x : x is a real number and `x^(2) + 1 = 0`}
C. {x : x is a real number and `x^(2) - 9 = 0`}
D. {x : x is a real number and `x^(2) = x + 2`}

Explanation

To determine which of the given options represents the empty set, we will analyze each option step by step.

### Step 1: Analyze Option 1
**Option 1:** \( x \) such that \( x \) is a real number and \( x^2 - 1 = 0 \)

1. Solve the equation:
\[ x^2 - 1 = 0 \]
This can be factored as:
\[ (x + 1)(x - 1) = 0 \]
Therefore, the solutions are:
\[ x = 1 \quad \text{and} \quad x = -1 \]
2. Since both solutions are real numbers, this set is not empty.

### Step 2: Analyze Option 2
**Option 2:** \( x \) such that \( x^2 + 1 = 0 \)

1. Solve the equation:
\[ x^2 + 1 = 0 \]
Rearranging gives:
\[ x^2 = -1 \]
Taking the square root results in:
\[ x = \pm i \]
where \( i \) is the imaginary unit.
2. Since the solutions are complex numbers, this set is empty.

### Step 3: Analyze Option 3
**Option 3:** \( x \) such that \( x^2 - 9 = 0 \)

1. Solve the equation:
\[ x^2 - 9 = 0 \]
This can be factored as:
\[ (x - 3)(x + 3) = 0 \]
Therefore, the solutions are:
\[ x = 3 \quad \text{and} \quad x = -3 \]
2. Since both solutions are real numbers, this set is not empty.

### Step 4: Analyze Option 4
**Option 4:** \( x \) such that \( x^2 = x + 2 \)

1. Rearranging gives:
\[ x^2 - x - 2 = 0 \]
We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -1, c = -2 \):
\[ x = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \]
Simplifying gives:
\[ x = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm \sqrt{9}}{2} = \frac{1 \pm 3}{2} \]
Therefore, the solutions are:
\[ x = 2 \quad \text{and} \quad x = -1 \]
2. Since both solutions are real numbers, this set is not empty.

### Conclusion
After analyzing all options, we find that **Option 2** is the only one that represents the empty set.