If 2 and 6 are the roots of the equation `ax^2 + bx + 1 = 0`, then the quadratic equation, whose roots are `1/(2a + b)` and `1/(6a + b)`, is:

Options

A. `x^2 + 10 x + 16 = 0`
B. `2x + 11 x + 12 = 0`
C. `4x^2 + 14 x + 12 = 0`
D. `x^2 + 8 x + 12 = 0`

Answer

To solve the problem step by step, we start with the given information:

1. **Identify the roots of the original quadratic equation**:
The roots of the equation \( ax^2 + bx + 1 = 0 \) are given as \( 2 \) and \( 6 \).

2. **Use Vieta's formulas**:
According to Vieta's formulas, for a quadratic equation \( ax^2 + bx + c = 0 \):
- The sum of the roots \( r_1 + r_2 = -\frac{b}{a} \)
- The product of the roots \( r_1 \cdot r_2 = \frac{c}{a} \)

Here, the roots are \( 2 \) and \( 6 \):
- Sum of the roots: \( 2 + 6 = 8 \)
- Product of the roots: \( 2 \cdot 6 = 12 \)

Therefore, we can write:
\[ -\frac{b}{a} = 8 \quad \text{(1)} \]
\[ \frac{1}{a} = 12 \quad \text{(2)} \]

3. **Solve for \( a \) and \( b \)**:
From equation (2), we can find \( a \):
\[ a = \frac{1}{12} \]
Substituting \( a \) into equation (1):
\[ -\frac{b}{\frac{1}{12}} = 8 \implies -12b = 8 \implies b = -\frac{2}{3} \]

4. **Find the new roots**:
We need to find the roots \( \frac{1}{2a + b} \) and \( \frac{1}{6a + b} \).
- First, calculate \( 2a + b \):
\[ 2a + b = 2 \cdot \frac{1}{12} - \frac{2}{3} = \frac{2}{12} - \frac{8}{12} = -\frac{6}{12} = -\frac{1}{2} \]
- Now, calculate \( 6a + b \):
\[ 6a + b = 6 \cdot \frac{1}{12} - \frac{2}{3} = \frac{6}{12} - \frac{8}{12} = -\frac{2}{12} = -\frac{1}{6} \]

5. **Calculate the new roots**:
- The first root is:
\[ \frac{1}{2a + b} = \frac{1}{-\frac{1}{2}} = -2 \]
- The second root is:
\[ \frac{1}{6a + b} = \frac{1}{-\frac{1}{6}} = -6 \]

6. **Form the new quadratic equation**:
The roots of the new quadratic equation are \( -2 \) and \( -6 \). Using Vieta's formulas again:
- Sum of the roots: \( -2 + (-6) = -8 \)
- Product of the roots: \( (-2)(-6) = 12 \)

Therefore, the new quadratic equation can be formed as:
\[ x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \]
This gives us:
\[ x^2 + 8x + 12 = 0 \]

### Final Answer:
The quadratic equation whose roots are \( \frac{1}{2a + b} \) and \( \frac{1}{6a + b} \) is:
\[ \boxed{x^2 + 8x + 12 = 0} \]