For the equation `|x^(2)-2x-3|=b`, which of the following statements is true?

Options

A. For `b lt0`, there are no solutions
B. For b = 0, there are three solutions
C. For `0ltblt4`, there are two solutions
D. For b = 4, there are four solutions

Answer

To solve the equation \( |x^2 - 2x - 3| = b \), we will analyze the expression inside the absolute value and determine the number of solutions for different values of \( b \).

### Step 1: Solve the quadratic equation
First, we need to find the roots of the quadratic equation \( x^2 - 2x - 3 = 0 \).

Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -2, c = -3 \).

Calculating the discriminant:
\[ D = (-2)^2 - 4 \cdot 1 \cdot (-3) = 4 + 12 = 16 \]
Since \( D > 0 \), there are two distinct real roots.

Calculating the roots:
\[ x = \frac{2 \pm \sqrt{16}}{2 \cdot 1} = \frac{2 \pm 4}{2} \]
This gives:
\[ x_1 = \frac{6}{2} = 3, \quad x_2 = \frac{-2}{2} = -1 \]

### Step 2: Analyze the quadratic function
The quadratic function \( f(x) = x^2 - 2x - 3 \) opens upwards (since the coefficient of \( x^2 \) is positive). The vertex of the parabola can be found using:
\[ x_v = -\frac{b}{2a} = -\frac{-2}{2 \cdot 1} = 1 \]
Calculating \( f(1) \):
\[ f(1) = 1^2 - 2 \cdot 1 - 3 = 1 - 2 - 3 = -4 \]
The minimum value of the function is \( -4 \) at \( x = 1 \).

### Step 3: Determine the number of solutions for different values of \( b \)
1. **When \( b < 0 \)**:
The equation \( |x^2 - 2x - 3| = b \) has no solutions because the left side is always non-negative.

2. **When \( b = 0 \)**:
The equation \( |x^2 - 2x - 3| = 0 \) implies \( x^2 - 2x - 3 = 0 \). We found two roots: \( x = -1 \) and \( x = 3 \). Thus, there are **2 solutions**.

3. **When \( 0 < b < 4 \)**:
The graph of \( y = x^2 - 2x - 3 \) intersects the line \( y = b \) at four points since the maximum value of the quadratic function is \( 4 \) (at \( x = 1 \)). Thus, there are **4 solutions**.

4. **When \( b = 4 \)**:
The equation \( |x^2 - 2x - 3| = 4 \) results in two cases:
- \( x^2 - 2x - 3 = 4 \) leading to \( x^2 - 2x - 7 = 0 \) (2 solutions).
- \( x^2 - 2x - 3 = -4 \) leading to \( x^2 - 2x + 1 = 0 \) (1 solution).
Thus, there are **3 solutions**.

5. **When \( b > 4 \)**:
The equation will have no solutions since the maximum value of \( |x^2 - 2x - 3| \) is \( 4 \).

### Conclusion
- For \( b < 0 \): No solutions.
- For \( b = 0 \): 2 solutions.
- For \( 0 < b < 4 \): 4 solutions.
- For \( b = 4 \): 3 solutions.
- For \( b > 4 \): No solutions.

### Final Answer
The correct statement is: For \( b < 0 \), there are no solutions.