The equation `|x+1||x-1|=a^(2) - 2a - 3` can have real solutions for x, if a belongs to

Options

A. `(-oo, -1]uu[3, oo)`
B. `[1 - sqrt(5), 1 + sqrt(5)]`
C. `[1-sqrt(5), 1] uu [3, 1 + sqrt(5)]`
D. none of these

Explanation

To solve the equation \( |x+1||x-1| = a^2 - 2a - 3 \) for real solutions in terms of \( a \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ |x+1||x-1| = a^2 - 2a - 3 \]
We can express the left side as:
\[ |x^2 - 1| = a^2 - 2a - 3 \]

### Step 2: Analyze the right side
The right side can be factored:
\[ a^2 - 2a - 3 = (a - 3)(a + 1) \]
Thus, we rewrite the equation as:
\[ |x^2 - 1| = (a - 3)(a + 1) \]

### Step 3: Set up cases for the absolute value
The absolute value \( |x^2 - 1| \) leads to two cases:
1. \( x^2 - 1 = (a - 3)(a + 1) \)
2. \( x^2 - 1 = -(a - 3)(a + 1) \)

### Step 4: Solve Case 1
For the first case:
\[ x^2 - 1 = (a - 3)(a + 1) \]
This simplifies to:
\[ x^2 = (a - 3)(a + 1) + 1 \]
\[ x^2 = (a^2 - 2a - 2) \]

### Step 5: Determine conditions for real solutions
For \( x^2 \) to have real solutions, we need:
\[ a^2 - 2a - 2 \geq 0 \]
Factoring gives:
\[ (a - 1 - \sqrt{3})(a - 1 + \sqrt{3}) \geq 0 \]
The roots are \( 1 - \sqrt{3} \) and \( 1 + \sqrt{3} \). The intervals for \( a \) are:
\[ (-\infty, 1 - \sqrt{3}] \cup [1 + \sqrt{3}, \infty) \]

### Step 6: Solve Case 2
For the second case:
\[ x^2 - 1 = -(a - 3)(a + 1) \]
This simplifies to:
\[ x^2 = -(a^2 - 2a - 3) + 1 \]
\[ x^2 = -a^2 + 2a + 4 \]

### Step 7: Determine conditions for real solutions
For \( x^2 \) to have real solutions, we need:
\[ -a^2 + 2a + 4 \geq 0 \]
Factoring gives:
\[ -(a - 1 - \sqrt{5})(a - 1 + \sqrt{5}) \geq 0 \]
The roots are \( 1 - \sqrt{5} \) and \( 1 + \sqrt{5} \). The intervals for \( a \) are:
\[ [1 - \sqrt{5}, 1 + \sqrt{5}] \]

### Step 8: Find the intersection of intervals
To find the values of \( a \) for which both cases yield real solutions, we take the intersection of:
1. \( (-\infty, 1 - \sqrt{3}] \cup [1 + \sqrt{3}, \infty) \)
2. \( [1 - \sqrt{5}, 1 + \sqrt{5}] \)

### Step 9: Final solution
The intersection yields:
\[ [1 - \sqrt{5}, 1 - \sqrt{3}] \cup [1 + \sqrt{3}, 1 + \sqrt{5}] \]

### Conclusion
Thus, the values of \( a \) for which the equation has real solutions for \( x \) are:
\[ a \in [1 - \sqrt{5}, 1 - \sqrt{3}] \cup [1 + \sqrt{3}, 1 + \sqrt{5}] \]