The set of values of a for which the inequality, `x^2 + ax + a^2 + 6a < 0` is satisfied for all `x belongs (1, 2)` lies in the interval:

Options

A. (1,2)
B. `[1,2]`
C. `[-7,4]`
D. none of these

Answer

To solve the inequality \( x^2 + ax + a^2 + 6a < 0 \) for all \( x \) in the interval \( (1, 2) \), we will follow these steps:

### Step 1: Define the function
Let \( f(x) = x^2 + ax + a^2 + 6a \).

### Step 2: Evaluate the function at the endpoints of the interval
We need to ensure that \( f(x) < 0 \) for all \( x \) in \( (1, 2) \). Therefore, we will evaluate \( f(1) \) and \( f(2) \).

#### Evaluate \( f(1) \):
\[ f(1) = 1^2 + a(1) + a^2 + 6a = 1 + a + a^2 + 6a = a^2 + 7a + 1 \]
We require:
\[ a^2 + 7a + 1 < 0 \tag{1} \]

#### Evaluate \( f(2) \):
\[ f(2) = 2^2 + a(2) + a^2 + 6a = 4 + 2a + a^2 + 6a = a^2 + 8a + 4 \]
We require:
\[ a^2 + 8a + 4 < 0 \tag{2} \]

### Step 3: Solve the inequalities
Now we will solve both inequalities (1) and (2).

#### Solve inequality (1):
The quadratic inequality \( a^2 + 7a + 1 < 0 \) can be solved by finding its roots using the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-7 \pm \sqrt{49 - 4}}{2} = \frac{-7 \pm \sqrt{45}}{2} = \frac{-7 \pm 3\sqrt{5}}{2} \]
The roots are:
\[ a_1 = \frac{-7 - 3\sqrt{5}}{2}, \quad a_2 = \frac{-7 + 3\sqrt{5}}{2} \]
The quadratic opens upwards (since the coefficient of \( a^2 \) is positive), so \( a^2 + 7a + 1 < 0 \) is satisfied between the roots:
\[ \frac{-7 - 3\sqrt{5}}{2} < a < \frac{-7 + 3\sqrt{5}}{2} \]

#### Solve inequality (2):
Similarly, for \( a^2 + 8a + 4 < 0 \):
\[ a = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{-8 \pm \sqrt{64 - 16}}{2} = \frac{-8 \pm \sqrt{48}}{2} = \frac{-8 \pm 4\sqrt{3}}{2} = -4 \pm 2\sqrt{3} \]
The roots are:
\[ b_1 = -4 - 2\sqrt{3}, \quad b_2 = -4 + 2\sqrt{3} \]
The quadratic opens upwards, so \( a^2 + 8a + 4 < 0 \) is satisfied between the roots:
\[ -4 - 2\sqrt{3} < a < -4 + 2\sqrt{3} \]

### Step 4: Find the intersection of the intervals
We need to find the intersection of the two intervals:
1. \( \left( \frac{-7 - 3\sqrt{5}}{2}, \frac{-7 + 3\sqrt{5}}{2} \right) \)
2. \( (-4 - 2\sqrt{3}, -4 + 2\sqrt{3}) \)

### Step 5: Conclusion
The set of values of \( a \) for which the inequality \( x^2 + ax + a^2 + 6a < 0 \) is satisfied for all \( x \in (1, 2) \) lies in the intersection of these two intervals.