The graph of the function `y=16x^2+8(a+5)x-7a-5` is strictly above the x axis, then 'a' must satisfy the inequality

Options

A. `-15 lt a lt - 2`
B. `-2 lt a lt -1`
C. `5 lt a lt 7`
D. none of these

Answer

To solve the problem, we need to determine the values of \( a \) for which the quadratic function \( y = 16x^2 + 8(a+5)x - (7a + 5) \) is strictly above the x-axis. This means that the quadratic must not have any real roots, which occurs when the discriminant is less than zero.

### Step-by-Step Solution:

1. **Identify the coefficients**:
The quadratic can be expressed in the standard form \( ax^2 + bx + c \), where:
- \( a = 16 \)
- \( b = 8(a + 5) \)
- \( c = - (7a + 5) \)

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 of \( a \), \( b \), and \( c \):
\[ D = [8(a + 5)]^2 - 4 \cdot 16 \cdot (-(7a + 5)) \]

3. **Simplify the discriminant**:
\[ D = 64(a + 5)^2 + 64(7a + 5) \]
Factor out 64:
\[ D = 64 \left( (a + 5)^2 + (7a + 5) \right) \]

4. **Expand the expression inside the parentheses**:
\[ (a + 5)^2 = a^2 + 10a + 25 \]
So,
\[ D = 64 \left( a^2 + 10a + 25 + 7a + 5 \right) \]
Combine like terms:
\[ D = 64 \left( a^2 + 17a + 30 \right) \]

5. **Set the discriminant less than zero**:
For the quadratic to be strictly above the x-axis, we need:
\[ a^2 + 17a + 30 < 0 \]

6. **Factor the quadratic**:
We can factor \( a^2 + 17a + 30 \):
\[ a^2 + 17a + 30 = (a + 15)(a + 2) \]

7. **Determine the intervals**:
We need to find when:
\[ (a + 15)(a + 2) < 0 \]
The roots of the equation are \( a = -15 \) and \( a = -2 \).

8. **Test the intervals**:
The critical points divide the number line into intervals:
- \( (-\infty, -15) \)
- \( (-15, -2) \)
- \( (-2, \infty) \)

We test a point from each interval:
- For \( a = -16 \) (in \( (-\infty, -15) \)): \( (-)(-) > 0 \)
- For \( a = -10 \) (in \( (-15, -2) \)): \( (+)(-) < 0 \)
- For \( a = -1 \) (in \( (-2, \infty) \)): \( (+)(+) > 0 \)

Thus, the quadratic is negative (and hence the function is above the x-axis) in the interval:
\[ -15 < a < -2 \]

### Final Answer:
The values of \( a \) must satisfy the inequality:
\[ -15 < a < -2 \]