To find the values of the parameter \( a \) for which the roots of the quadratic equation \( x^2 + 2(a-1)x + (a+5) = 0 \) have exactly one root in the interval \( (1, 3) \), we can follow these steps:
### Step 1: Define the quadratic function
We start with the quadratic function:
\[
f(x) = x^2 + 2(a-1)x + (a+5)
\]
### Step 2: Evaluate the function at the endpoints of the interval
We need to evaluate \( f(1) \) and \( f(3) \):
1. **Calculate \( f(1) \)**:
\[
f(1) = 1^2 + 2(a-1)(1) + (a+5) = 1 + 2(a-1) + (a+5) = 1 + 2a - 2 + a + 5 = 3a + 4
\]
2. **Calculate \( f(3) \)**:
\[
f(3) = 3^2 + 2(a-1)(3) + (a+5) = 9 + 6(a-1) + (a+5) = 9 + 6a - 6 + a + 5 = 7a + 8
\]
### Step 3: Set up the condition for exactly one root in the interval
For exactly one root to lie in the interval \( (1, 3) \), we need:
\[
f(1) \cdot f(3) < 0
\]
This means one of the values must be positive and the other must be negative.
### Step 4: Solve the inequality \( (3a + 4)(7a + 8) < 0 \)
To find the critical points, we set each factor to zero:
1. \( 3a + 4 = 0 \) gives \( a = -\frac{4}{3} \)
2. \( 7a + 8 = 0 \) gives \( a = -\frac{8}{7} \)
Now, we can test the intervals determined by these critical points:
- Interval \( (-\infty, -\frac{8}{7}) \)
- Interval \( (-\frac{8}{7}, -\frac{4}{3}) \)
- Interval \( (-\frac{4}{3}, \infty) \)
Testing a value from each interval:
- For \( a = -3 \) (in \( (-\infty, -\frac{8}{7}) \)): \( (3(-3) + 4)(7(-3) + 8) = (-9 + 4)(-21 + 8) = (-5)(-13) > 0 \)
- For \( a = -1 \) (in \( (-\frac{8}{7}, -\frac{4}{3}) \)): \( (3(-1) + 4)(7(-1) + 8) = (-3 + 4)(-7 + 8) = (1)(1) > 0 \)
- For \( a = 0 \) (in \( (-\frac{4}{3}, \infty) \)): \( (3(0) + 4)(7(0) + 8) = (0 + 4)(0 + 8) = (4)(8) > 0 \)
Thus, the product is negative in the interval:
\[
(-\frac{8}{7}, -\frac{4}{3})
\]
### Step 5: Ensure the roots are real
For the roots to be real, the discriminant \( D \) must be greater than zero:
\[
D = b^2 - 4ac = (2(a-1))^2 - 4(1)(a+5) > 0
\]
Calculating the discriminant:
\[
D = 4(a-1)^2 - 4(a+5) = 4[(a-1)^2 - (a+5)] = 4[a^2 - 2a + 1 - a - 5] = 4[a^2 - 3a - 4]
\]
Setting the discriminant greater than zero:
\[
a^2 - 3a - 4 > 0
\]
Factoring gives:
\[
(a - 4)(a + 1) > 0
\]
The critical points are \( a = -1 \) and \( a = 4 \). Testing intervals:
- For \( a < -1 \): \( (-)(-) > 0 \)
- For \( -1 < a < 4 \): \( (-)(+) < 0 \)
- For \( a > 4 \): \( (+)(+) > 0 \)
Thus, \( D > 0 \) for \( a < -1 \) or \( a > 4 \).
### Step 6: Find the intersection of conditions
We have two conditions:
1. \( a \in (-\frac{8}{7}, -\frac{4}{3}) \)
2. \( a < -1 \) or \( a > 4 \)
The intersection of these conditions gives:
\[
a \in (-\frac{8}{7}, -1)
\]
### Final Answer
The values of the parameter \( a \) for which the roots of the quadratic equation have exactly one root in the interval \( (1, 3) \) are:
\[
a \in \left(-\frac{8}{7}, -1\right)
\]
