If $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2+(p-3)x-2p=3$ $(p\in R)$, then the minimum value of $(\alpha^2+\beta^2+\alpha\beta)$, is

Rearranging the equation gives x^2 + (p-3)x - 2p - 3 = 0, meaning the sum of the roots is -(p-3) and the product is -2p - 3. The target expression alpha^2 + beta^2 + alpha*beta can be rewritten as (alpha + beta)^2 - alpha*beta. Substituting the Vieta's formulas yields (3-p)^2 - (-2p-3), which expands to p^2 - 6p + 9 + 2p + 3. Simplifying this gives p^2 - 4p + 12, which can be completed to a square as (p-2)^2 + 8. Since the square of a real number is always non-negative, the minimum possible value for this expression is 8. The correct result is 8.

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