Solve `sqrt(x-2)(x^2-4x-5)=0.`

Explanation

To solve the equation \(\sqrt{x-2}(x^2-4x-5)=0\), we will follow these steps:

### Step 1: Identify the conditions for the square root
The expression \(\sqrt{x-2}\) is defined only when \(x-2 \geq 0\). This implies:
\[ x \geq 2 \]

### Step 2: Set the equation to zero
The equation can be expressed as a product of two factors:
\[ \sqrt{x-2} \cdot (x^2 - 4x - 5) = 0 \]
For this product to equal zero, either factor must be zero. Therefore, we have two cases to consider:
1. \(\sqrt{x-2} = 0\)
2. \(x^2 - 4x - 5 = 0\)

### Step 3: Solve the first case
For the first case:
\[ \sqrt{x-2} = 0 \]
Squaring both sides gives:
\[ x - 2 = 0 \implies x = 2 \]

### Step 4: Solve the second case
Now, we solve the quadratic equation:
\[ x^2 - 4x - 5 = 0 \]
We can factor this quadratic:
\[ (x - 5)(x + 1) = 0 \]
Setting each factor to zero gives us:
\[ x - 5 = 0 \implies x = 5 \]
\[ x + 1 = 0 \implies x = -1 \]

### Step 5: Check the validity of the solutions
We have found three potential solutions: \(x = 2\), \(x = 5\), and \(x = -1\). However, we must check which of these satisfy the condition \(x \geq 2\):
- \(x = 2\) is valid.
- \(x = 5\) is valid.
- \(x = -1\) is not valid since it does not satisfy \(x \geq 2\).

### Final Solution
The valid solutions to the equation \(\sqrt{x-2}(x^2-4x-5)=0\) are:
\[ x = 2 \quad \text{and} \quad x = 5 \]