Sum of the squares of all integral values of a for which the inequality `x^(2) + ax + a^(2) + 6a lt 0` is satisfied for all `x in ( 1,2)` must be equal to

Options

A. 90
B. 89
C. 88
D. 91

Answer

To solve the problem, we need to find the sum of the squares of all integral values of \( a \) for which the inequality \( x^2 + ax + a^2 + 6a < 0 \) is satisfied for all \( x \) in the interval \( (1, 2) \).

### Step-by-Step Solution:

1. **Set up the inequality for \( x = 1 \) and \( x = 2 \)**:
- For \( x = 1 \):
\[ 1^2 + a(1) + a^2 + 6a < 0 \implies 1 + a + a^2 + 6a < 0 \implies a^2 + 7a + 1 < 0 \]
- For \( x = 2 \):
\[ 2^2 + a(2) + a^2 + 6a < 0 \implies 4 + 2a + a^2 + 6a < 0 \implies a^2 + 8a + 4 < 0 \]

2. **Find the roots of the quadratic inequalities**:
- For \( a^2 + 7a + 1 < 0 \):
- The roots can be found using the quadratic formula:
\[ a = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-7 \pm \sqrt{49 - 4}}{2} = \frac{-7 \pm \sqrt{45}}{2} = \frac{-7 \pm 3\sqrt{5}}{2} \]
- Approximate values:
\[ a_1 \approx \frac{-7 + 6.7}{2} \approx -0.15, \quad a_2 \approx \frac{-7 - 6.7}{2} \approx -6.85 \]
- Thus, the interval for this inequality is \( (-6.85, -0.15) \).

- For \( a^2 + 8a + 4 < 0 \):
- The roots are:
\[ a = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{-8 \pm \sqrt{64 - 16}}{2} = \frac{-8 \pm \sqrt{48}}{2} = \frac{-8 \pm 4\sqrt{3}}{2} = -4 \pm 2\sqrt{3} \]
- Approximate values:
\[ a_1 \approx -4 + 3.46 \approx -0.54, \quad a_2 \approx -4 - 3.46 \approx -7.46 \]
- Thus, the interval for this inequality is \( (-7.46, -0.54) \).

3. **Find the intersection of the intervals**:
- The intervals are \( (-6.85, -0.15) \) and \( (-7.46, -0.54) \).
- The intersection is \( (-6.85, -0.54) \).

4. **Identify integral values of \( a \)**:
- The integral values of \( a \) in the interval \( (-6.85, -0.54) \) are \( -6, -5, -4, -3, -2, -1 \).

5. **Calculate the sum of the squares of these integral values**:
\[ (-6)^2 + (-5)^2 + (-4)^2 + (-3)^2 + (-2)^2 + (-1)^2 = 36 + 25 + 16 + 9 + 4 + 1 = 91 \]

### Final Answer:
The sum of the squares of all integral values of \( a \) is \( \boxed{91} \).