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(C) The given quadratic equation is $x^2 + px + \frac{3p}{4} = 0$.By the properties of roots,we have $\alpha + \beta = -p$ and $\alpha \beta = \frac{3

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$\{-2, 5\}$

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(C) The given quadratic equation is $x^2 + px + \frac{3p}{4} = 0$.By the properties of roots,we have $\alpha + \beta = -p$ and $\alpha \beta = \frac{3p}{4}$.Given that $|\alpha - \beta| = \sqrt{10}$,squaring both sides gives $(\alpha - \beta)^2 = 10$.Using the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta$,we substitute the values:$(-p)^2 - 4 \left( \frac{3p}{4} \right) = 10$.This simplifies to $p^2 - 3p = 10$,or $p^2 - 3p - 10 = 0$.Factoring the quadratic equation,we get $(p - 5)(p + 2) = 0$.Thus,$p = 5$ or $p = -2$.Therefore,$p \in \{-2, 5\}$.

If $\alpha$ and $\beta$ are roots of the equation $x^2 + px + \frac{3p}{4} = 0$ such that $|\alpha - \beta| = \sqrt{10}$,then $p$ belongs to the set

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