To solve the problem, we need 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 \( (1, 3) \).
### Step 1: Determine the conditions for the roots
For a quadratic equation \( ax^2 + bx + c = 0 \), the roots are real and distinct if the discriminant \( D \) is greater than zero, and real and equal if \( D = 0 \). Since we need exactly one root in the interval \( (1, 3) \), we need to ensure that the discriminant is non-negative:
\[
D = b^2 - 4ac \geq 0
\]
Here, \( a = 1 \), \( b = 2(a-1) \), and \( c = a + 5 \).
### Step 2: Calculate the discriminant
Calculating the discriminant:
\[
D = [2(a-1)]^2 - 4 \cdot 1 \cdot (a + 5)
\]
Simplifying:
\[
D = 4(a-1)^2 - 4(a + 5)
\]
\[
D = 4[(a-1)^2 - (a + 5)]
\]
\[
D = 4[a^2 - 2a + 1 - a - 5]
\]
\[
D = 4[a^2 - 3a - 4]
\]
Setting the discriminant greater than or equal to zero:
\[
4(a^2 - 3a - 4) \geq 0
\]
\[
a^2 - 3a - 4 \geq 0
\]
### Step 3: Factor the quadratic inequality
Factoring the quadratic:
\[
(a - 4)(a + 1) \geq 0
\]
### Step 4: Solve the inequality
To solve this inequality, we find the critical points:
- \( a - 4 = 0 \) gives \( a = 4 \)
- \( a + 1 = 0 \) gives \( a = -1 \)
Now, we test the intervals determined by these points:
1. \( a < -1 \) (choose \( a = -2 \)): \( (-2 - 4)(-2 + 1) = (-6)(-1) > 0 \) (satisfied)
2. \( -1 < a < 4 \) (choose \( a = 0 \)): \( (0 - 4)(0 + 1) = (-4)(1) < 0 \) (not satisfied)
3. \( a > 4 \) (choose \( a = 5 \)): \( (5 - 4)(5 + 1) = (1)(6) > 0 \) (satisfied)
Thus, the solution to the inequality is:
\[
a \in (-\infty, -1] \cup [4, \infty)
\]
### Step 5: Ensure exactly one root lies in the interval (1, 3)
Next, we must ensure that exactly one root lies in the interval \( (1, 3) \). We evaluate the function \( f(x) = x^2 + 2(a-1)x + (a + 5) \) at the endpoints of the interval:
1. \( f(1) \)
2. \( f(3) \)
Calculating \( f(1) \):
\[
f(1) = 1^2 + 2(a-1)(1) + (a + 5) = 1 + 2(a-1) + a + 5 = 3a + 4
\]
Calculating \( f(3) \):
\[
f(3) = 3^2 + 2(a-1)(3) + (a + 5) = 9 + 6(a-1) + a + 5 = 7a + 8
\]
For exactly one root in \( (1, 3) \), we need:
\[
f(1) \cdot f(3) < 0
\]
This gives us:
\[
(3a + 4)(7a + 8) < 0
\]
### Step 6: Solve the product inequality
Finding the roots of \( 3a + 4 = 0 \) gives \( a = -\frac{4}{3} \) and for \( 7a + 8 = 0 \) gives \( a = -\frac{8}{7} \).
Now we test the intervals:
1. \( a < -\frac{8}{7} \) (choose \( a = -2 \)): \( (3(-2) + 4)(7(-2) + 8) = (-6 + 4)(-14 + 8) = (-2)(-6) > 0 \) (not satisfied)
2. \( -\frac{8}{7} < a < -\frac{4}{3} \) (choose \( a = -1.5 \)): \( (3(-1.5) + 4)(7(-1.5) + 8) = (-4.5 + 4)(-10.5 + 8) = (-0.5)(-2.5) > 0 \) (not satisfied)
3. \( -\frac{4}{3} < a \) (choose \( a = 0 \)): \( (3(0) + 4)(7(0) + 8) = (4)(8) > 0 \) (not satisfied)
### Final Solution
Combining the results, we find:
The values of \( a \) for which exactly one root lies in the interval \( (1, 3) \) are:
\[
a \in (-\infty, -1] \cup [4, \infty)
\]
