Step-by-step Solution:

To determine whether the roots of the equation \(2x^2 - 4p^2x + 4p^4 - 2r = 0\) are imaginary, we can proceed with the following steps: 1. Given Equations: \[ \alpha + \beta = -p \quad \text{(1)} \] \[ \alpha\beta = \frac{p^2}{2} \quad \text{(2)} \] \[ \alpha^4 + \beta^4 = r \quad \text{(3)} \] 2. Find \(\alpha^2 + \beta^2\): Square equation (1): \[ (\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta = p^2 \] Using equation (2): \[ \alpha^2 + \beta^2 = p^2 - 2 \cdot \frac{p^2}{2} = p^2 - p^2 = 0 \] 3. Find \(\alpha^4 + \beta^4\): Square \(\alpha^2 + \beta^2\): \[ (\alpha^2 + \beta^2)^2 = \alpha^4 + \beta^4 + 2\alpha^2\beta^2 = 0 \] Using equation (2): \[ \alpha^4 + \beta^4 = -2\alpha^2\beta^2 = -2\left(\frac{p^2}{2}\right)^2 = -2 \cdot \frac{p^4}{4} = -\frac{p^4}{2} \] Therefore: \[ r = -\frac{p^4}{2} \quad \text{(4)} \] 4. Calculate the Discriminant: The discriminant \(D\) of the quadratic equation \(2x^2 - 4p^2x + 4p^4 - 2r = 0\) is: \[ D = (-4p^2)^2 - 4 \cdot 2 \cdot (4p^4 - 2r) \] Simplify: \[ D = 16p^4 - 32p^4 + 16r \] Substitute \(r\) from equation (4): \[ D = 16p^4 - 32p^4 + 16\left(-\frac{p^4}{2}\right) = 16p^4 - 32p^4 - 8p^4 = -24p^4 \] 5. Determine the Nature of the Roots: Since the discriminant \(D = -24p^4\) is negative, the roots of the equation are imaginary. Therefore, the roots of the equation \(2x^2 - 4p^2x + 4p^4 - 2r = 0\) are: \[ \boxed{\text{imaginary}} \]