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

\[ x^2 + 2(a-1)x + (a+5) = 0 \]

are such that one root is greater than 3 and the other root is smaller than 1, we can follow these steps:

### Step 1: Determine the conditions for the roots
Given that one root is greater than 3 and the other root is smaller than 1, we can denote the roots as \( r_1 \) and \( r_2 \) such that \( r_1 > 3 \) and \( r_2 < 1 \).

### Step 2: Use the quadratic formula
The roots of the quadratic equation \( ax^2 + bx + c = 0 \) can be found using the quadratic formula:

\[ r_{1,2} = \frac{-b \pm \sqrt{D}}{2a} \]

where \( D = b^2 - 4ac \) is the discriminant. In our case, \( a = 1 \), \( b = 2(a-1) \), and \( c = a + 5 \).

### Step 3: Ensure the roots are real and unequal
For the roots to be real and unequal, the discriminant must be greater than zero:

\[ D = [2(a-1)]^2 - 4(1)(a+5) > 0 \]

Calculating the discriminant:

\[ D = 4(a-1)^2 - 4(a+5) \]
\[ = 4[(a-1)^2 - (a+5)] \]
\[ = 4[a^2 - 2a + 1 - a - 5] \]
\[ = 4[a^2 - 3a - 4] \]

Thus, we need:

\[ a^2 - 3a - 4 > 0 \]

### Step 4: Factor the quadratic inequality
Factoring \( a^2 - 3a - 4 \):

\[ (a - 4)(a + 1) > 0 \]

### Step 5: Solve the inequality
To solve the inequality \( (a - 4)(a + 1) > 0 \), we find the critical points \( a = 4 \) and \( a = -1 \). We test the intervals:

1. \( (-\infty, -1) \): Choose \( a = -2 \) → \( (-2 - 4)(-2 + 1) = (-6)(-1) > 0 \) (True)
2. \( (-1, 4) \): Choose \( a = 0 \) → \( (0 - 4)(0 + 1) = (-4)(1) < 0 \) (False)
3. \( (4, \infty) \): Choose \( a = 5 \) → \( (5 - 4)(5 + 1) = (1)(6) > 0 \) (True)

Thus, the solution to the inequality is:

\[ a < -1 \quad \text{or} \quad a > 4 \]

### Step 6: Apply the conditions from the roots
Next, we also need to ensure that one root is greater than 3 and the other is less than 1. We can use the fact that the sum and product of the roots can help us here.

The sum of the roots \( r_1 + r_2 = -\frac{b}{a} = -(2(a-1)) = -2a + 2 \).

The product of the roots \( r_1 r_2 = \frac{c}{a} = a + 5 \).

### Step 7: Set up inequalities based on the roots
From \( r_1 > 3 \) and \( r_2 < 1 \):

1. \( r_1 + r_2 < 3 + 1 = 4 \) implies \( -2a + 2 < 4 \) → \( -2a < 2 \) → \( a > -1 \).
2. \( r_1 r_2 < 3 \cdot 1 = 3 \) implies \( a + 5 < 3 \) → \( a < -2 \).

### Step 8: Combine the conditions
We have two conditions from the roots:

1. \( a < -2 \)
2. \( a > -1 \)

The valid range for \( a \) is:

\[ a < -2 \]

### Final Result
Thus, the values of the parameter \( a \) for which the roots of the quadratic equation satisfy the given conditions are:

\[ \boxed{(-\infty, -2)} \]