If `a, b in R` and `ax^2 + bx +6 = 0,a!= 0` does not have two distinct real roots, then :

Options

A. (a) minimum possible value of `3a+b` is `-2`
B. (b) minimum possible value of `3a+b` is 2
C. (c) minimum possible value of `6a+b` is `-1`
D. (d) minimum possible value of `6a+b` is 1

Explanation

To solve the problem, we need to analyze the quadratic equation given by \( ax^2 + bx + 6 = 0 \) under the condition that it does not have two distinct real roots.

### Step-by-Step Solution:

1. **Understanding the Condition for Roots**:
A quadratic equation \( ax^2 + bx + c = 0 \) has two distinct real roots if its discriminant \( D \) is greater than zero. The discriminant is given by:
\[ D = b^2 - 4ac \]
For our equation, \( c = 6 \), so we have:
\[ D = b^2 - 4a \cdot 6 = b^2 - 24a \]

2. **Setting Up the Condition**:
Since the equation does not have two distinct real roots, the discriminant must be less than or equal to zero:
\[ b^2 - 24a \leq 0 \]
This implies:
\[ b^2 \leq 24a \]

3. **Rearranging the Inequality**:
We can rearrange this inequality to express \( a \) in terms of \( b \):
\[ a \geq \frac{b^2}{24} \]

4. **Finding the Minimum Value of \( 3a + b \)**:
We want to find the minimum value of \( 3a + b \). Substituting \( a \) from the inequality into the expression:
\[ 3a + b \geq 3\left(\frac{b^2}{24}\right) + b \]
Simplifying this gives:
\[ 3a + b \geq \frac{b^2}{8} + b \]

5. **Combining Terms**:
To combine the terms, we can express \( b \) as \( \frac{8b}{8} \):
\[ 3a + b \geq \frac{b^2 + 8b}{8} \]

6. **Completing the Square**:
Now we complete the square for the expression \( b^2 + 8b \):
\[ b^2 + 8b = (b + 4)^2 - 16 \]
Therefore:
\[ 3a + b \geq \frac{(b + 4)^2 - 16}{8} \]

7. **Finding the Minimum Value**:
The minimum value of \( (b + 4)^2 \) is \( 0 \) (which occurs when \( b = -4 \)). Thus:
\[ 3a + b \geq \frac{0 - 16}{8} = -2 \]

8. **Conclusion**:
The minimum possible value of \( 3a + b \) is \( -2 \).

### Final Answer:
The minimum possible value of \( 3a + b \) is \( -2 \).