To find the values of \(k\) for which the quadratic equation \(2x^2 + 3kx + k = 0\) has distinct real roots, we need to ensure that the discriminant is greater than zero.

The discriminant \(\Delta\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by \(b^2 - 4ac\).

Here, \(a = 2\), \(b = 3k\), and \(c = k\).

Thus, the discriminant is:

\(\Delta = (3k)^2 - 4 \times 2 \times k = 9k^2 - 8k\).

For distinct real roots, \(\Delta > 0\):

\(9k^2 - 8k > 0\).

Factorizing gives:

\(k(9k - 8) > 0\).

The critical points are \(k = 0\) and \(k = \frac{8}{9}\).

Using a sign chart or test intervals, we find that the inequality holds for:

\(k < 0\) or \(k > \frac{8}{9}\).

Thus, the set of values of \(k\) for which the equation has distinct real roots is \(k < 0, k > \frac{8}{9}\).