If one of the zeros of the quadratic polynomial f(x) = `4x^2 - 8kx - 9` is equal in magnitude but opposite in sign of the other, find the value of k.

Answer

To find the value of \( k \) such that one of the zeros of the quadratic polynomial \( f(x) = 4x^2 - 8kx - 9 \) is equal in magnitude but opposite in sign to the other, we can follow these steps:

### Step 1: Understand the relationship between the zeros
Let the two zeros be \( \alpha \) and \( \beta \). According to the problem, we have:
\[ \beta = -\alpha \]

### Step 2: Use Vieta's formulas
From Vieta's formulas, we know that for a quadratic polynomial of the form \( ax^2 + bx + c \):
- The sum of the zeros \( \alpha + \beta = -\frac{b}{a} \)
- The product of the zeros \( \alpha \beta = \frac{c}{a} \)

For our polynomial \( f(x) = 4x^2 - 8kx - 9 \):
- Here, \( a = 4 \), \( b = -8k \), and \( c = -9 \).

### Step 3: Calculate the sum of the zeros
Using the sum of the zeros:
\[ \alpha + \beta = -\frac{-8k}{4} = \frac{8k}{4} = 2k \]
Since \( \beta = -\alpha \), we can write:
\[ \alpha + (-\alpha) = 0 \]
Thus, we have:
\[ 2k = 0 \implies k = 0 \]

### Step 4: Verify the product of the zeros
Now, we can check the product of the zeros:
\[ \alpha \beta = \frac{-9}{4} \]
Substituting \( \beta = -\alpha \):
\[ \alpha(-\alpha) = -\alpha^2 = \frac{-9}{4} \]
This implies:
\[ \alpha^2 = \frac{9}{4} \implies \alpha = \pm \frac{3}{2} \]
Thus, the zeros are \( \frac{3}{2} \) and \( -\frac{3}{2} \), confirming that they are equal in magnitude but opposite in sign.

### Conclusion
The value of \( k \) is:
\[ \boxed{0} \]