There is only one real value of a for which the quadratic equation `ax^2+(a+3)x+a-3=0` has two positive integral solutions. The product of these two solutions is

Options

A. 0.09
B. 0.08
C. 0.06
D. 12

Explanation

To find the product of the two positive integral solutions of the quadratic equation \( ax^2 + (a+3)x + (a-3) = 0 \), we will follow these steps:

### Step 1: Identify the coefficients
The quadratic equation can be expressed in the standard form \( Ax^2 + Bx + C = 0 \), where:
- \( A = a \)
- \( B = a + 3 \)
- \( C = a - 3 \)

### Step 2: Use the relationships for roots
For a quadratic equation \( Ax^2 + Bx + C = 0 \):
- The sum of the roots \( (x_1 + x_2) \) is given by \( -\frac{B}{A} = -\frac{a + 3}{a} \)
- The product of the roots \( (x_1 \cdot x_2) \) is given by \( \frac{C}{A} = \frac{a - 3}{a} \)

### Step 3: Set conditions for positive integral solutions
Since we want two positive integral solutions, we need:
1. The sum of the roots \( -\frac{a + 3}{a} \) to be negative (which implies \( a < 0 \)).
2. The product of the roots \( \frac{a - 3}{a} \) to be positive (which implies \( a - 3 > 0 \) or \( a > 3 \)).

However, since \( a < 0 \) and \( a > 3 \) cannot be satisfied simultaneously, we need to analyze the conditions further.

### Step 4: Express \( a \) in terms of integers
For the sum of the roots to be an integer, \( -\frac{a + 3}{a} \) must be an integer. This can be rewritten as:
\[ 1 + \frac{3}{a} \text{ must be an integer.} \]
This implies \( \frac{3}{a} \) must be an integer, which means \( a \) must be a divisor of 3. The divisors of 3 are \( \pm 1, \pm 3 \).

### Step 5: Evaluate possible values of \( a \)
Given \( a < 0 \), the possible values of \( a \) are:
- \( a = -1 \)
- \( a = -3 \)

### Step 6: Check each value of \( a \)
1. **For \( a = -1 \)**:
- The equation becomes \( -x^2 + 2x + 2 = 0 \).
- The roots are \( x = 1 \) and \( x = 2 \) (both positive integers).
- The product of the roots is \( 1 \cdot 2 = 2 \).

2. **For \( a = -3 \)**:
- The equation becomes \( -3x^2 + 0x + 0 = 0 \).
- The roots are not positive integers.

Thus, the only valid value is \( a = -1 \).

### Step 7: Calculate the product of the roots
The product of the roots for \( a = -1 \) is:
\[ x_1 \cdot x_2 = 1 \cdot 2 = 2. \]

### Conclusion
The product of the two positive integral solutions is \( 2 \).