The value of k for which the equation `3x^(2) + 2x (k^(2) + 1) + k^(2) - 3k + 2 = 0` has roots of opposite signs, lies in the interval

Options

A. `(-oo, 0)`
B. `(-oo, -1)`
C. `(1, 2)`
D. `(3//2, 2)`

Answer

To find the value of \( k \) for which the equation

\[ 3x^2 + 2x(k^2 + 1) + (k^2 - 3k + 2) = 0 \]

has roots of opposite signs, we can follow these steps:

### Step 1: Understand the condition for opposite signs
For the roots of a quadratic equation \( ax^2 + bx + c = 0 \) to be of opposite signs, the product of the roots must be negative. The product of the roots can be given by the formula:

\[ \text{Product of roots} = \frac{c}{a} \]

### Step 2: Identify coefficients
In our equation, we have:
- \( a = 3 \)
- \( b = 2(k^2 + 1) \)
- \( c = k^2 - 3k + 2 \)

### Step 3: Set up the inequality
Since we want the product of the roots to be less than zero, we set up the inequality:

\[ \frac{k^2 - 3k + 2}{3} < 0 \]

This simplifies to:

\[ k^2 - 3k + 2 < 0 \]

### Step 4: Factor the quadratic expression
Next, we factor the quadratic expression \( k^2 - 3k + 2 \):

\[ k^2 - 3k + 2 = (k - 1)(k - 2) \]

### Step 5: Solve the inequality
Now we need to determine when the product \( (k - 1)(k - 2) < 0 \). This occurs between the roots of the equation, which are \( k = 1 \) and \( k = 2 \).

### Step 6: Analyze the intervals
To find the intervals where the product is negative, we can test the intervals created by the roots:
- For \( k < 1 \): Choose \( k = 0 \) → \( (0 - 1)(0 - 2) = 2 > 0 \)
- For \( 1 < k < 2 \): Choose \( k = 1.5 \) → \( (1.5 - 1)(1.5 - 2) = (0.5)(-0.5) = -0.25 < 0 \)
- For \( k > 2 \): Choose \( k = 3 \) → \( (3 - 1)(3 - 2) = 2 > 0 \)

Thus, the inequality \( (k - 1)(k - 2) < 0 \) holds true in the interval:

\[ (1, 2) \]

### Conclusion
The value of \( k \) for which the equation has roots of opposite signs lies in the interval \( (1, 2) \).