Find the vlaue of k for which the roots of the following equation are real and equal `k^(2)x^(2)-2(2k-1) x + 4=0`

Answer

To find the value of \( k \) for which the roots of the equation \( k^2 x^2 - 2(2k - 1)x + 4 = 0 \) are real and equal, we will use the condition that the discriminant of the quadratic equation must be zero.

### Step-by-Step Solution:

1. **Identify coefficients**:
The given quadratic equation is in the form \( ax^2 + bx + c = 0 \). Here, we can identify:
- \( a = k^2 \)
- \( b = -2(2k - 1) = -4k + 2 \)
- \( c = 4 \)

2. **Set up the discriminant condition**:
For the roots to be real and equal, the discriminant \( D \) must be zero. The discriminant is given by:
\[ D = b^2 - 4ac \]
Setting \( D = 0 \):
\[ (-4k + 2)^2 - 4(k^2)(4) = 0 \]

3. **Expand the discriminant**:
Calculate \( (-4k + 2)^2 \):
\[ (-4k + 2)^2 = 16k^2 - 16k + 4 \]
Now substitute this back into the discriminant equation:
\[ 16k^2 - 16k + 4 - 16k^2 = 0 \]

4. **Simplify the equation**:
The \( 16k^2 \) terms cancel out:
\[ -16k + 4 = 0 \]

5. **Solve for \( k \)**:
Rearranging the equation gives:
\[ -16k = -4 \quad \Rightarrow \quad k = \frac{4}{16} = \frac{1}{4} \]

### Conclusion:
The value of \( k \) for which the roots of the equation are real and equal is:
\[ \boxed{\frac{1}{4}} \]