If the equations `2x^(2)-7x+1=0` and `ax^(2)+bx+2=0` have a common root, then

Options

A. `a=2`, `b=-7`
B. `a=(-7)/(2)`, `b=1`
C. `a=4`, `b=-14`
D. `a=-4`, `b=1`

Explanation

To solve the problem, we need to find the values of \( a \) and \( b \) such that the equations \( 2x^2 - 7x + 1 = 0 \) and \( ax^2 + bx + 2 = 0 \) have a common root. Let's denote the common root by \( p \).

### Step-by-step Solution:

1. **Set up the equations with the common root**:
Since \( p \) is a common root, it must satisfy both equations. Therefore, we can write:
\[ 2p^2 - 7p + 1 = 0 \quad \text{(1)} \]
\[ ap^2 + bp + 2 = 0 \quad \text{(2)} \]

2. **Express \( p^2 \) from equation (1)**:
From equation (1), we can express \( p^2 \):
\[ 2p^2 = 7p - 1 \implies p^2 = \frac{7p - 1}{2} \quad \text{(3)} \]

3. **Substitute \( p^2 \) into equation (2)**:
Substitute equation (3) into equation (2):
\[ a\left(\frac{7p - 1}{2}\right) + bp + 2 = 0 \]
Multiply through by 2 to eliminate the fraction:
\[ a(7p - 1) + 2bp + 4 = 0 \]
Simplifying gives:
\[ (7a + 2b)p + (4 - a) = 0 \quad \text{(4)} \]

4. **Set coefficients to zero**:
Since equation (4) must hold for all \( p \), both coefficients must equal zero:
\[ 7a + 2b = 0 \quad \text{(5)} \]
\[ 4 - a = 0 \quad \text{(6)} \]

5. **Solve for \( a \) from equation (6)**:
From equation (6), we find:
\[ a = 4 \]

6. **Substitute \( a \) back into equation (5)**:
Substitute \( a = 4 \) into equation (5):
\[ 7(4) + 2b = 0 \]
\[ 28 + 2b = 0 \implies 2b = -28 \implies b = -14 \]

7. **Final values**:
Thus, we find:
\[ a = 4, \quad b = -14 \]

### Conclusion:
The values of \( a \) and \( b \) such that the equations have a common root are:
\[ \boxed{a = 4, \, b = -14} \]