If `ax^(2) + bx + 10 = 0` does not have two distinct real roots, then the least value of 5a + b, is

Options

A. -3
B. -2
C. 3
D. none of these

Explanation

To solve the problem, we need to find the least value of \( 5a + b \) given that the quadratic equation \( ax^2 + bx + 10 = 0 \) does not have two distinct real roots.

### Step-by-Step Solution:

1. **Understanding the Condition for Roots**:
A quadratic equation \( ax^2 + bx + c = 0 \) does not have two distinct real roots if its discriminant \( D \) is less than or equal to zero. The discriminant is given by:
\[ D = b^2 - 4ac \]
In our case, \( c = 10 \), so:
\[ D = b^2 - 4a \cdot 10 = b^2 - 40a \]
Therefore, the condition becomes:
\[ b^2 - 40a \leq 0 \]
Rearranging gives:
\[ b^2 \leq 40a \]

2. **Expressing \( a \) in terms of \( b \)**:
From the inequality \( b^2 \leq 40a \), we can express \( a \) as:
\[ a \geq \frac{b^2}{40} \]

3. **Substituting \( a \) into \( 5a + b \)**:
We want to minimize the expression \( 5a + b \). Substituting the minimum value of \( a \):
\[ 5a + b \geq 5\left(\frac{b^2}{40}\right) + b = \frac{5b^2}{40} + b = \frac{b^2}{8} + b \]

4. **Finding the Minimum Value of \( \frac{b^2}{8} + b \)**:
Let \( y = \frac{b^2}{8} + b \). To find the minimum value, we can complete the square:
\[ y = \frac{1}{8}(b^2 + 8b) \]
Completing the square inside the parentheses:
\[ b^2 + 8b = (b + 4)^2 - 16 \]
Thus:
\[ y = \frac{1}{8}((b + 4)^2 - 16) = \frac{(b + 4)^2}{8} - 2 \]

5. **Finding the Minimum Value**:
The expression \( \frac{(b + 4)^2}{8} \) is always non-negative and achieves its minimum value of 0 when \( b + 4 = 0 \) (i.e., \( b = -4 \)). Therefore, the minimum value of \( y \) occurs at:
\[ y_{\text{min}} = 0 - 2 = -2 \]

### Conclusion:
The least value of \( 5a + b \) is \( -2 \).