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.

### Step 1: Find the roots of the first equation
We start with the quadratic equation:
\[ 2x^2 - 7x + 1 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we identify \( a = 2 \), \( b = -7 \), and \( c = 1 \).

Calculating the discriminant:
\[ D = b^2 - 4ac = (-7)^2 - 4 \cdot 2 \cdot 1 = 49 - 8 = 41 \]

Now, substituting into the quadratic formula:
\[ x = \frac{7 \pm \sqrt{41}}{4} \]
Thus, the roots are:
\[ x_1 = \frac{7 + \sqrt{41}}{4}, \quad x_2 = \frac{7 - \sqrt{41}}{4} \]

### Step 2: Assume a common root
Let’s assume that the common root is \( \alpha = \frac{7 + \sqrt{41}}{4} \) (it can also be \( \frac{7 - \sqrt{41}}{4} \), but it will lead to the same results).

### Step 3: Substitute the common root into the second equation
Now we substitute \( \alpha \) into the second equation:
\[ a\left(\frac{7 + \sqrt{41}}{4}\right)^2 + b\left(\frac{7 + \sqrt{41}}{4}\right) + 2 = 0 \]

### Step 4: Simplify the equation
Calculating \( \left(\frac{7 + \sqrt{41}}{4}\right)^2 \):
\[ \left(\frac{7 + \sqrt{41}}{4}\right)^2 = \frac{(7 + \sqrt{41})^2}{16} = \frac{49 + 14\sqrt{41} + 41}{16} = \frac{90 + 14\sqrt{41}}{16} \]

Now substituting back:
\[ a\left(\frac{90 + 14\sqrt{41}}{16}\right) + b\left(\frac{7 + \sqrt{41}}{4}\right) + 2 = 0 \]
Multiplying through by 16 to eliminate the denominator:
\[ a(90 + 14\sqrt{41}) + 4b(7 + \sqrt{41}) + 32 = 0 \]

### Step 5: Collect terms
This gives us:
\[ (90a + 28b + 32) + (14a + 4b)\sqrt{41} = 0 \]
For this equation to hold, both coefficients must equal zero:
1. \( 90a + 28b + 32 = 0 \)
2. \( 14a + 4b = 0 \)

### Step 6: Solve the system of equations
From the second equation:
\[ 4b = -14a \implies b = -\frac{14}{4}a = -\frac{7}{2}a \]

Substituting \( b \) into the first equation:
\[ 90a + 28\left(-\frac{7}{2}a\right) + 32 = 0 \]
\[ 90a - 98a + 32 = 0 \implies -8a + 32 = 0 \implies 8a = 32 \implies a = 4 \]

Now substituting \( a = 4 \) back to find \( b \):
\[ b = -\frac{7}{2}(4) = -14 \]

### Final Answer
Thus, the values of \( a \) and \( b \) are:
\[ a = 4, \quad b = -14 \]