If `ax^(2)+bx+6=0` does not have two distinct real roots, `a in R` and `b in R`, then the least value of `3a+b` is

Options

A. 4
B. `-1`
C. 1
D. `-2`

Answer

To solve the problem, we need to determine the least value of \(3a + b\) under the condition that the quadratic equation \(ax^2 + bx + 6 = 0\) does not have two distinct real roots. This condition is satisfied when the discriminant of the quadratic equation is less than or equal to zero.

### Step-by-step Solution:

1. **Identify the Discriminant**:
The discriminant \(D\) of the quadratic equation \(ax^2 + bx + c = 0\) 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. **Set the Discriminant Condition**:
Since the equation does not have two distinct real roots, we require:
\[ D \leq 0 \implies b^2 - 24a \leq 0 \]
Rearranging this gives:
\[ b^2 \leq 24a \implies a \geq \frac{b^2}{24} \]

3. **Express \(3a + b\)**:
We need to find the least value of \(3a + b\). Substituting \(a\) from the inequality:
\[ 3a + b \geq 3\left(\frac{b^2}{24}\right) + b = \frac{b^2}{8} + b \]

4. **Rearranging the Expression**:
We can rewrite the expression:
\[ 3a + b \geq \frac{b^2}{8} + b = \frac{b^2 + 8b}{8} \]

5. **Finding the Minimum Value**:
To find the minimum value of \(\frac{b^2 + 8b}{8}\), we can complete the square:
\[ b^2 + 8b = (b + 4)^2 - 16 \]
Therefore,
\[ \frac{b^2 + 8b}{8} = \frac{(b + 4)^2 - 16}{8} = \frac{(b + 4)^2}{8} - 2 \]

6. **Determine the Minimum**:
The term \(\frac{(b + 4)^2}{8}\) is always non-negative and achieves its minimum value of 0 when \(b + 4 = 0\) (i.e., \(b = -4\)). Thus:
\[ \text{Minimum of } \frac{(b + 4)^2}{8} = 0 \]
Therefore,
\[ \text{Minimum of } 3a + b = 0 - 2 = -2 \]

### Conclusion:
The least value of \(3a + b\) is \(-2\).