To find the values of the parameter \( a \) for which the roots of the quadratic equation
\[
x^2 + 2(a-1)x + (a+5) = 0
\]
lie in the interval \( (1, 3) \), we will follow these steps:
### Step 1: Identify the Quadratic Equation
The given quadratic equation is:
\[
f(x) = x^2 + 2(a-1)x + (a+5)
\]
### Step 2: Determine the Conditions for Roots
For both roots \( \alpha \) and \( \beta \) to lie in the interval \( (1, 3) \), we need to satisfy the following conditions:
1. The discriminant \( D \) must be non-negative (ensuring real roots).
2. The vertex of the parabola must lie within the interval \( (1, 3) \).
3. The function values at the endpoints \( f(1) \) and \( f(3) \) must be positive.
### Step 3: Calculate the Discriminant
The discriminant \( D \) of the quadratic equation is given by:
\[
D = b^2 - 4ac
\]
Here, \( a = 1 \), \( b = 2(a-1) \), and \( c = a + 5 \).
Calculating \( D \):
\[
D = [2(a-1)]^2 - 4(1)(a+5)
\]
\[
D = 4(a-1)^2 - 4(a+5)
\]
\[
D = 4[(a-1)^2 - (a+5)]
\]
\[
D = 4[a^2 - 2a + 1 - a - 5]
\]
\[
D = 4[a^2 - 3a - 4]
\]
For the roots to be real, we need:
\[
D \geq 0 \implies a^2 - 3a - 4 \geq 0
\]
### Step 4: Solve the Inequality
Factoring the quadratic:
\[
a^2 - 3a - 4 = (a - 4)(a + 1)
\]
The critical points are \( a = 4 \) and \( a = -1 \). The intervals to test are \( (-\infty, -1) \), \( (-1, 4) \), and \( (4, \infty) \).
Testing intervals:
- For \( a < -1 \), choose \( a = -2 \): \( (-2-4)(-2+1) = (-6)(-1) > 0 \)
- For \( -1 < a < 4 \), choose \( a = 0 \): \( (0-4)(0+1) = (-4)(1) < 0 \)
- For \( a > 4 \), choose \( a = 5 \): \( (5-4)(5+1) = (1)(6) > 0 \)
Thus, the solution for the discriminant is:
\[
a \in (-\infty, -1] \cup [4, \infty)
\]
### Step 5: Vertex Condition
The vertex \( x_v \) of the parabola is given by:
\[
x_v = -\frac{b}{2a} = -\frac{2(a-1)}{2} = 1 - a
\]
We need:
\[
1 < 1 - a < 3
\]
This gives two inequalities:
1. \( 1 - a > 1 \) implies \( a < 0 \)
2. \( 1 - a < 3 \) implies \( a > -2 \)
Thus, combining these gives:
\[
-2 < a < 0
\]
### Step 6: Function Values at Endpoints
Now we check \( f(1) > 0 \) and \( f(3) > 0 \):
Calculating \( f(1) \):
\[
f(1) = 1^2 + 2(a-1)(1) + (a+5) = 1 + 2(a-1) + (a+5)
\]
\[
= 1 + 2a - 2 + a + 5 = 3a + 4
\]
Setting \( f(1) > 0 \):
\[
3a + 4 > 0 \implies a > -\frac{4}{3}
\]
Calculating \( f(3) \):
\[
f(3) = 3^2 + 2(a-1)(3) + (a+5) = 9 + 6(a-1) + (a+5)
\]
\[
= 9 + 6a - 6 + a + 5 = 7a + 8
\]
Setting \( f(3) > 0 \):
\[
7a + 8 > 0 \implies a > -\frac{8}{7}
\]
### Step 7: Combine All Conditions
Now we combine all conditions:
1. From the discriminant: \( a \in (-\infty, -1] \cup [4, \infty) \)
2. From vertex condition: \( -2 < a < 0 \)
3. From function values: \( a > -\frac{4}{3} \) and \( a > -\frac{8}{7} \)
The final solution is:
\[
a \in \left(-\frac{4}{3}, 0\right) \cap (-\infty, -1] = \left(-\frac{4}{3}, -1\right]
\]
### Final Answer
Thus, the values of \( a \) for which both roots lie in the interval \( (1, 3) \) are:
\[
a \in \left(-\frac{4}{3}, -1\right]
\]
