The graph of the function `y=16x^(2)+8(a+2)x-3a-2` is strictly above the x - axis, then number of integral velues of 'a' is

Answer

To solve the problem, we need to determine the number of integral values of \( a \) such that the quadratic function \( y = 16x^2 + 8(a + 2)x - (3a + 2) \) is strictly above the x-axis. This means that the discriminant of the quadratic must be less than zero.

### Step-by-Step Solution:

1. **Identify the coefficients**:
The given quadratic function is in the form \( y = ax^2 + bx + c \).
- Here, \( a = 16 \)
- \( b = 8(a + 2) \)
- \( c = -(3a + 2) \)

2. **Calculate the discriminant**:
The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c \) is given by:
\[ D = b^2 - 4ac \]
Substituting the values:
\[ D = [8(a + 2)]^2 - 4 \cdot 16 \cdot (-(3a + 2)) \]

3. **Simplify the discriminant**:
\[ D = 64(a + 2)^2 + 64(3a + 2) \]
Factor out 64:
\[ D = 64\left[(a + 2)^2 + (3a + 2)\right] \]

4. **Expand and combine like terms**:
\[ D = 64\left[a^2 + 4a + 4 + 3a + 2\right] \]
\[ D = 64\left[a^2 + 7a + 6\right] \]

5. **Set the discriminant less than zero**:
For the quadratic to be strictly above the x-axis, we need:
\[ D < 0 \implies a^2 + 7a + 6 < 0 \]

6. **Factor the quadratic**:
\[ a^2 + 7a + 6 = (a + 6)(a + 1) \]

7. **Determine the intervals**:
The quadratic \( (a + 6)(a + 1) < 0 \) is negative between its roots:
- The roots are \( a = -6 \) and \( a = -1 \).
- The inequality holds for \( -6 < a < -1 \).

8. **Identify integral values**:
The integral values of \( a \) in the interval \( (-6, -1) \) are:
- \( -5, -4, -3, -2 \)

9. **Count the integral values**:
There are 4 integral values: \( -5, -4, -3, -2 \).

### Final Answer:
The number of integral values of \( a \) is **4**.