To solve the problem, we need to find the number of integral values of \( m \) for which exactly one root of the quadratic equation
\[
x^2 + 3mx + (m^2 - 4) = 0
\]
lies in the interval \((-6, 2)\).
### Step 1: Determine the conditions for the roots
For a quadratic equation \( ax^2 + bx + c = 0 \), the roots are given by the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In our case, \( a = 1 \), \( b = 3m \), and \( c = m^2 - 4 \).
The discriminant \( D \) must be non-negative for the roots to be real:
\[
D = b^2 - 4ac = (3m)^2 - 4(1)(m^2 - 4) = 9m^2 - 4m^2 + 16 = 5m^2 + 16
\]
Since \( 5m^2 + 16 > 0 \) for all \( m \), the roots are always real.
### Step 2: Find the roots
The roots of the equation are:
\[
x = \frac{-3m \pm \sqrt{5m^2 + 16}}{2}
\]
### Step 3: Analyze the roots
Let the roots be \( r_1 \) and \( r_2 \). For exactly one root to lie in the interval \((-6, 2)\), one root must be in the interval while the other must be outside it.
### Step 4: Evaluate the roots at the endpoints
We need to evaluate the function at the endpoints:
1. At \( x = -6 \):
\[
f(-6) = (-6)^2 + 3m(-6) + (m^2 - 4) = 36 - 18m + m^2 - 4 = m^2 - 18m + 32
\]
2. At \( x = 2 \):
\[
f(2) = (2)^2 + 3m(2) + (m^2 - 4) = 4 + 6m + m^2 - 4 = m^2 + 6m
\]
### Step 5: Set up the conditions
For exactly one root to be in \((-6, 2)\), we need:
\[
f(-6) \cdot f(2) < 0
\]
This means one of the values must be positive and the other must be negative.
### Step 6: Solve the inequalities
1. \( f(-6) = m^2 - 18m + 32 < 0 \)
2. \( f(2) = m^2 + 6m > 0 \)
#### Solving \( m^2 - 18m + 32 < 0 \):
The roots of the quadratic equation \( m^2 - 18m + 32 = 0 \) can be found using the quadratic formula:
\[
m = \frac{18 \pm \sqrt{(-18)^2 - 4 \cdot 1 \cdot 32}}{2 \cdot 1} = \frac{18 \pm \sqrt{324 - 128}}{2} = \frac{18 \pm \sqrt{196}}{2} = \frac{18 \pm 14}{2}
\]
Calculating the roots:
\[
m = \frac{32}{2} = 16 \quad \text{and} \quad m = \frac{4}{2} = 2
\]
The inequality \( m^2 - 18m + 32 < 0 \) holds for:
\[
2 < m < 16
\]
#### Solving \( m^2 + 6m > 0 \):
Factoring gives:
\[
m(m + 6) > 0
\]
The roots are \( m = 0 \) and \( m = -6 \). The inequality holds for:
\[
m < -6 \quad \text{or} \quad m > 0
\]
### Step 7: Combine the intervals
We need to find the intersection of the intervals:
1. From \( f(-6) < 0 \): \( (2, 16) \)
2. From \( f(2) > 0 \): \( (-\infty, -6) \cup (0, \infty) \)
The only overlapping interval is:
\[
(2, 16)
\]
### Step 8: Count the integral values
The integral values of \( m \) in the interval \( (2, 16) \) are:
\[
3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15
\]
This gives us a total of \( 13 \) integral values.
### Step 9: Calculate \( k \) and find \( \left\lfloor \frac{k}{5} \right\rfloor \)
Let \( k = 13 \). Then:
\[
\frac{k}{5} = \frac{13}{5} = 2.6
\]
Thus, the greatest integer function gives:
\[
\left\lfloor \frac{13}{5} \right\rfloor = 2
\]
### Final Answer
\[
\boxed{2}
\]
