For what value of

We know that, in a quadratic equation, ax² + bx + c = 0.

For real and equal roots, b² – 4ac = 0

Given Equation: 2x² – (2k + 1)x + k = 0

So, we have, a = 2, b = -(2k + 1), and c = k

Therefore,

{-(2k + 1)}² – 4 × 2 × k = 0

4k² + 1 + 4k – 8k = 0

4k² – 4k + 1 = 0

This is a formula for (2k – 1)², so,

(2k – 1)² = 0

2k – 1 = 0

k = 1/2

Hence, for k = 1/2 the given quadratic equation has real and equal roots.

OR

Given: f(x) = 2kx² – x + k

If x = 1, is a root of the equation, it will satisfy the equation. Therefore,

f(1) = 0

2k(1)² – 1 + k = 0

2k – 1 + k = 0

3k – 1 = 0

k = 1/3