Find the values of a, for which the quadratic expression `ax^2 + (a - 2) x - 2` is negative for exactly two integral values of `x`.

Answer

To find the values of \( a \) for which the quadratic expression \( ax^2 + (a - 2)x - 2 \) is negative for exactly two integral values of \( x \), we will follow these steps:

### Step 1: Define the quadratic function
Let \( f(x) = ax^2 + (a - 2)x - 2 \).

### Step 2: Evaluate the function at specific integer points
We need to evaluate the function at \( x = 0, 1, 2, -1, -2 \) to determine when it is negative.

- \( f(0) = -2 \) (This is already negative)
- \( f(1) = a(1)^2 + (a - 2)(1) - 2 = a + a - 2 - 2 = 2a - 4 \)
- \( f(2) = a(2)^2 + (a - 2)(2) - 2 = 4a + 2a - 4 - 2 = 6a - 6 \)
- \( f(-1) = a(-1)^2 + (a - 2)(-1) - 2 = a - (a - 2) - 2 = 2 - 2 = 0 \)
- \( f(-2) = a(-2)^2 + (a - 2)(-2) - 2 = 4a - 2a + 4 - 2 = 2a + 2 \)

### Step 3: Set conditions for negativity
For \( f(x) \) to be negative for exactly two integral values of \( x \), we can consider the following cases:

1. **Case 1**: \( f(1) < 0 \) and \( f(2) \geq 0 \)
- From \( f(1) < 0 \):
\[ 2a - 4 < 0 \implies a < 2 \]
- From \( f(2) \geq 0 \):
\[ 6a - 6 \geq 0 \implies a \geq 1 \]

2. **Case 2**: \( f(-1) > 0 \) and \( f(-2) \geq 0 \)
- Since \( f(-1) = 0 \), it does not contribute to negativity.
- From \( f(-2) \geq 0 \):
\[ 2a + 2 \geq 0 \implies a \geq -1 \]

### Step 4: Combine the conditions
From Case 1, we have:
- \( 1 \leq a < 2 \)

From Case 2, we have:
- \( a \geq -1 \) (which is already satisfied by \( a \geq 1 \))

### Final Result
Thus, the values of \( a \) for which the quadratic expression \( ax^2 + (a - 2)x - 2 \) is negative for exactly two integral values of \( x \) are:
\[ \boxed{[1, 2)} \]