Find `k` if the equations `4x^(2)-11x+2k=0` and `x^(2)-3x-k=0` have a common root. Obtain the common root for this value of `k`.

Options

A. 0
B. `17//6`
C. either a or b
D. cannot be determined

Answer

To find the value of `k` such that the equations \( 4x^2 - 11x + 2k = 0 \) and \( x^2 - 3x - k = 0 \) have a common root, we can follow these steps:

### Step 1: Define the common root
Let \( \alpha \) be the common root of both equations.

### Step 2: Substitute \( \alpha \) into both equations
Since \( \alpha \) is a root of both equations, we can write:
1. From the first equation:
\[ 4\alpha^2 - 11\alpha + 2k = 0 \quad \text{(1)} \]
2. From the second equation:
\[ \alpha^2 - 3\alpha - k = 0 \quad \text{(2)} \]

### Step 3: Solve for \( k \) in terms of \( \alpha \)
From equation (2), we can express \( k \) in terms of \( \alpha \):
\[ k = \alpha^2 - 3\alpha \quad \text{(3)} \]

### Step 4: Substitute \( k \) from (3) into (1)
Now, substitute \( k \) from (3) into equation (1):
\[ 4\alpha^2 - 11\alpha + 2(\alpha^2 - 3\alpha) = 0 \]
This simplifies to:
\[ 4\alpha^2 - 11\alpha + 2\alpha^2 - 6\alpha = 0 \]
Combining like terms gives:
\[ 6\alpha^2 - 17\alpha = 0 \]

### Step 5: Factor the equation
Factoring out \( \alpha \):
\[ \alpha(6\alpha - 17) = 0 \]
This gives us two possible solutions:
1. \( \alpha = 0 \)
2. \( 6\alpha - 17 = 0 \) which simplifies to \( \alpha = \frac{17}{6} \)

### Step 6: Find corresponding values of \( k \)
Now we can find the values of \( k \) for both values of \( \alpha \):

1. For \( \alpha = 0 \):
\[ k = 0^2 - 3(0) = 0 \]

2. For \( \alpha = \frac{17}{6} \):
\[ k = \left(\frac{17}{6}\right)^2 - 3\left(\frac{17}{6}\right) \]
Calculating this:
\[ k = \frac{289}{36} - \frac{51}{6} = \frac{289}{36} - \frac{306}{36} = \frac{-17}{36} \]

### Conclusion
Thus, the values of \( k \) are \( 0 \) and \( -\frac{17}{36} \). The common roots for these values of \( k \) are \( 0 \) and \( \frac{17}{6} \) respectively.