Find the values of the parameter a for which the roots of the quadratic equation `x^(2)+2(a-1)x+a+5=0` are
such that one root is greater than 3 and the other root is smaller than 1.

Answer

To find the values of the parameter \( a \) for which the roots of the quadratic equation \( x^2 + 2(a-1)x + (a+5) = 0 \) satisfy the conditions that one root is greater than 3 and the other root is smaller than 1, we will follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation is:
\[ x^2 + 2(a-1)x + (a+5) = 0 \]
Here, we can identify:
- \( A = 1 \)
- \( B = 2(a-1) \)
- \( C = a + 5 \)

### Step 2: Use the condition for the roots
Let the roots be \( \alpha \) and \( \beta \). We want:
- \( \alpha > 3 \)
- \( \beta < 1 \)

### Step 3: Apply Vieta's formulas
From Vieta's formulas, we know:
- \( \alpha + \beta = -\frac{B}{A} = -2(a-1) \)
- \( \alpha \beta = \frac{C}{A} = a + 5 \)

### Step 4: Set up inequalities from the conditions
From the conditions on the roots, we can derive:
1. Since \( \alpha > 3 \) and \( \beta < 1 \):
\[ \alpha + \beta > 3 + \beta \quad \text{(since } \beta < 1\text{)} \]
Thus,
\[ -2(a-1) > 3 + \beta \]
Since \( \beta < 1 \), we can use \( \beta = 1 - \epsilon \) for some small \( \epsilon > 0 \):
\[ -2(a-1) > 3 + (1 - \epsilon) \implies -2(a-1) > 4 - \epsilon \]

2. Also, since \( \alpha > 3 \):
\[ \alpha + \beta < 3 + 1 = 4 \]
Thus,
\[ -2(a-1) < 4 \implies -2a + 2 < 4 \implies -2a < 2 \implies a > -1 \]

### Step 5: Analyze the product of the roots
The product of the roots gives:
\[ \alpha \beta < 3 \cdot 1 = 3 \]
Thus,
\[ a + 5 < 3 \implies a < -2 \]

### Step 6: Combine the inequalities
We have derived the following inequalities:
1. \( a > -1 \)
2. \( a < -2 \)

However, these inequalities cannot be satisfied simultaneously. We need to check the conditions again.

### Step 7: Re-evaluate the conditions
To satisfy both conditions, we need to ensure:
- The discriminant \( D \) must be greater than zero for the roots to be real:
\[ D = B^2 - 4AC = (2(a-1))^2 - 4(1)(a+5) > 0 \]
This simplifies to:
\[ 4(a^2 - 2a + 1) - 4(a + 5) > 0 \]
\[ 4a^2 - 8a + 4 - 4a - 20 > 0 \implies 4a^2 - 12a - 16 > 0 \]
Dividing by 4:
\[ a^2 - 3a - 4 > 0 \]
Factoring gives:
\[ (a - 4)(a + 1) > 0 \]
The solution to this inequality is:
\[ a < -1 \quad \text{or} \quad a > 4 \]

### Step 8: Final conditions
Now we combine:
- From the product condition: \( a < -2 \)
- From the discriminant condition: \( a < -1 \) or \( a > 4 \)

Thus, the only valid range that satisfies all conditions is:
\[ a < -2 \]

### Conclusion
The values of the parameter \( a \) for which the roots of the quadratic equation \( x^2 + 2(a-1)x + (a+5) = 0 \) satisfy the given conditions are:
\[ \boxed{(-\infty, -2)} \]