If one root of the equation 2x^2 - 5px + 2p^2 = 0 exceeds the other by 4, then what is the value of p?

Explanation

Let roots be \alpha and \beta. Sum of roots \alpha + \beta = \frac{5p}{2} and product \alpha\beta = p^2. Given \alpha - \beta = 4. Using (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta, we get 16 = (\frac{5p}{2})^2 - 4(p^2) = \frac{25p^2}{4} - 4p^2 = \frac{9p^2}{4}. Thus, p^2 = \frac{64}{9}, meaning p = \pm \frac{8}{3}.

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