Find the value of `k` such that `3x^(2)+2kx+x-k-5` has the sum of zeros as half of their product.

Answer

To find the value of \( k \) such that the polynomial \( 3x^2 + 2kx + x - k - 5 \) has the sum of its zeros equal to half of their product, we will follow these steps:

### Step 1: Rewrite the polynomial in standard form
The given polynomial is:
\[ 3x^2 + 2kx + x - k - 5 \]
We can combine the terms involving \( x \):
\[ 3x^2 + (2k + 1)x - (k + 5) \]
Thus, we have:
\[ A = 3, \quad B = 2k + 1, \quad C = -k - 5 \]

### Step 2: Use the formulas for the sum and product of the roots
For a quadratic polynomial \( Ax^2 + Bx + C \):
- The sum of the roots \( \alpha + \beta \) is given by:
\[ -\frac{B}{A} = -\frac{2k + 1}{3} \]
- The product of the roots \( \alpha \beta \) is given by:
\[ \frac{C}{A} = \frac{-k - 5}{3} \]

### Step 3: Set up the equation based on the problem statement
According to the problem, the sum of the roots is half of the product of the roots:
\[ -\frac{2k + 1}{3} = \frac{1}{2} \cdot \frac{-k - 5}{3} \]

### Step 4: Eliminate the denominators
To eliminate the denominators, we can multiply both sides by \( 6 \) (the least common multiple of 3 and 2):
\[ 6 \left(-\frac{2k + 1}{3}\right) = 6 \left(\frac{1}{2} \cdot \frac{-k - 5}{3}\right) \]
This simplifies to:
\[ -2(2k + 1) = -k - 5 \]

### Step 5: Expand and simplify the equation
Expanding the left side:
\[ -4k - 2 = -k - 5 \]

### Step 6: Rearrange the equation to solve for \( k \)
Bringing all terms involving \( k \) to one side and constant terms to the other side:
\[ -4k + k = -5 + 2 \]
This simplifies to:
\[ -3k = -3 \]

### Step 7: Solve for \( k \)
Dividing both sides by -3:
\[ k = 1 \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{1} \]