If 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 determine the values of 'a' for which the graph of the function \( y = 16x^2 + 8(a+5)x - 7a - 5 \) is strictly above the x-axis, we need to analyze the quadratic function.

### Step-by-Step Solution:

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

2. **Condition for the quadratic to be above the x-axis**:
A quadratic function is strictly above the x-axis if it has no real roots. This occurs when the discriminant \( D \) is less than 0. The discriminant for a quadratic equation \( ax^2 + bx + c \) is given by:
\[ D = b^2 - 4ac \]

3. **Calculate the discriminant**:
Substitute the values of \( a \), \( b \), and \( c \) into the discriminant formula:
\[ D = [8(a + 5)]^2 - 4(16)(-7a - 5) \]

Expanding this:
\[ D = 64(a + 5)^2 + 64(7a + 5) \]

Simplifying further:
\[ D = 64[(a + 5)^2 + (7a + 5)] \]
\[ D = 64[a^2 + 10a + 25 + 7a + 5] \]
\[ D = 64[a^2 + 17a + 30] \]

4. **Set the discriminant less than 0**:
For the quadratic to be strictly above the x-axis, we need:
\[ 64(a^2 + 17a + 30) < 0 \]
Since \( 64 \) is a positive constant, we can simplify this to:
\[ a^2 + 17a + 30 < 0 \]

5. **Find the roots of the quadratic**:
We can find the roots of the quadratic equation \( a^2 + 17a + 30 = 0 \) using the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( b = 17 \) and \( c = 30 \):
\[ a = \frac{-17 \pm \sqrt{17^2 - 4 \cdot 1 \cdot 30}}{2 \cdot 1} \]
\[ a = \frac{-17 \pm \sqrt{289 - 120}}{2} \]
\[ a = \frac{-17 \pm \sqrt{169}}{2} \]
\[ a = \frac{-17 \pm 13}{2} \]
This gives us the roots:
\[ a_1 = \frac{-4}{2} = -2 \quad \text{and} \quad a_2 = \frac{-30}{2} = -15 \]

6. **Determine the intervals**:
The quadratic \( a^2 + 17a + 30 \) opens upwards (since the coefficient of \( a^2 \) is positive). It will be negative between its roots:
\[ -15 < a < -2 \]

### Final Answer:
Thus, the values of \( a \) for which the graph of the function is strictly above the x-axis are:
\[ -15 < a < -2 \]