The value of `a` for which the quadratic equation
`3x^(2) + 2a^(2) + 1x + a^(2) - 3a + 2 = 0`
Possesses roots of opposite signs lies in

Options

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

Answer

To find the value of \( a \) for which the quadratic equation

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

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

### Step 1: Identify the roots' product condition
For a quadratic equation \( ax^2 + bx + c = 0 \), the product of the roots \( \alpha \) and \( \beta \) is given by

\[ \alpha \beta = \frac{c}{a} \]

In our case, \( a = 3 \), \( b = 2a^2 + 1 \), and \( c = a^2 - 3a + 2 \). Therefore, we have:

\[ \alpha \beta = \frac{a^2 - 3a + 2}{3} \]

### Step 2: Condition for opposite signs
For the roots to have opposite signs, their product must be negative:

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

This simplifies to:

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

### Step 3: Factor the quadratic expression
Next, we factor the quadratic expression:

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

### Step 4: Solve the inequality
Now we need to solve the inequality:

\[ (a - 1)(a - 2) < 0 \]

### Step 5: Determine the intervals
To find the intervals where the product is negative, we can test the intervals determined by the roots \( a = 1 \) and \( a = 2 \):

- For \( a < 1 \): Choose \( a = 0 \) → \( (0 - 1)(0 - 2) = 2 > 0 \) (positive)
- For \( 1 < a < 2 \): Choose \( a = 1.5 \) → \( (1.5 - 1)(1.5 - 2) = (0.5)(-0.5) = -0.25 < 0 \) (negative)
- For \( a > 2 \): Choose \( a = 3 \) → \( (3 - 1)(3 - 2) = 2 > 0 \) (positive)

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

\[ 1 < a < 2 \]

### Conclusion
The values of \( a \) for which the quadratic equation possesses roots of opposite signs lie in the interval \( (1, 2) \).

---