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 \(3a + b\) given that the quadratic equation \(ax^2 + bx + 6 = 0\) does not have two distinct real roots. This condition implies that the discriminant of the quadratic must be less than or equal to zero.

### Step-by-step Solution:

1. **Identify the Discriminant Condition**:
The discriminant \(D\) of the quadratic equation \(ax^2 + bx + 6 = 0\) is given by:
\[ D = b^2 - 4ac \]
Here, \(c = 6\). Therefore, the discriminant becomes:
\[ D = b^2 - 24a \]
For the quadratic to not have two distinct real roots, we require:
\[ D \leq 0 \implies b^2 - 24a \leq 0 \implies b^2 \leq 24a \]

2. **Express \(a\) in terms of \(b\)**:
Rearranging the inequality gives:
\[ a \geq \frac{b^2}{24} \]

3. **Substitute \(a\) into the expression \(3a + b\)**:
We want to minimize:
\[ 3a + b \]
Substituting \(a\) from the previous step:
\[ 3a + b \geq 3\left(\frac{b^2}{24}\right) + b = \frac{b^2}{8} + b \]

4. **Rearranging the expression**:
We can rewrite the expression as:
\[ 3a + b \geq \frac{b^2}{8} + b \]
To find the minimum value, we can complete the square for the quadratic in \(b\):
\[ \frac{b^2}{8} + b = \frac{1}{8}(b^2 + 8b) = \frac{1}{8}((b + 4)^2 - 16) = \frac{(b + 4)^2}{8} - 2 \]

5. **Finding the minimum value**:
The term \((b + 4)^2\) is always non-negative, and its minimum value is \(0\) (when \(b = -4\)). Therefore, the minimum value of the entire expression is:
\[ \frac{(b + 4)^2}{8} - 2 \geq -2 \]
Thus, the least value of \(3a + b\) is:
\[ -2 \]

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