The set of all values of `' a '` for which the quadratic equation `3x^2+2(a^2-3a+2)=0` possess roots of opposite sign, is
a.`(-oo,1)` b. `(-oo,0)` c. `(1,2)` d. `(3//2,2)`

Options

A. ` 1 lt a lt 2`
B. ` a in (2 , oo)`
C. `1 lt a lt 3 `
D. ` -1 lt a lt 0`

Answer

To find the set of all values of \( a \) for which the quadratic equation

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

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

### Step 1: Identify the Product of Roots Condition
For a quadratic equation \( Ax^2 + Bx + C = 0 \), the product of the roots \( \alpha \) and \( \beta \) is given by \( \alpha \beta = \frac{C}{A} \). The roots will be of opposite sign if their product is less than zero, i.e.,

\[ \alpha \beta < 0. \]

### Step 2: Determine \( A \), \( B \), and \( C \)
In our equation, we can identify:
- \( A = 3 \)
- \( B = 0 \) (since there is no \( x \) term)
- \( C = 2(a^2 - 3a + 2) \)

### Step 3: Set Up the Inequality
The condition for opposite sign roots translates to:

\[ \frac{C}{A} < 0 \implies \frac{2(a^2 - 3a + 2)}{3} < 0. \]

Since \( 3 > 0 \), we can simplify this to:

\[ 2(a^2 - 3a + 2) < 0 \implies a^2 - 3a + 2 < 0. \]

### Step 4: Factor the Quadratic
Now, we need to factor the quadratic expression \( a^2 - 3a + 2 \):

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

### Step 5: Solve the Inequality
We need to solve the inequality:

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

To find the intervals where this inequality holds, we identify the critical points, which are \( a = 1 \) and \( a = 2 \).

### Step 6: Test the Intervals
We will test the intervals determined by the critical points:
1. For \( a < 1 \) (e.g., \( a = 0 \)):
- \( (0 - 1)(0 - 2) = 1 \) (positive)

2. For \( 1 < a < 2 \) (e.g., \( a = 1.5 \)):
- \( (1.5 - 1)(1.5 - 2) = (0.5)(-0.5) = -0.25 \) (negative)

3. For \( a > 2 \) (e.g., \( a = 3 \)):
- \( (3 - 1)(3 - 2) = 2 \) (positive)

### Step 7: Conclusion
The inequality \( (a - 1)(a - 2) < 0 \) holds true for the interval \( (1, 2) \).

Thus, the set of all values of \( a \) for which the quadratic equation has roots of opposite sign is:

\[ \boxed{(1, 2)}. \]