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

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$\{-2, 5\}$

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(C) The given quadratic equation is $x^2 + px + \frac{3p}{4} = 0.$From the properties of roots,we have $\alpha + \beta = -p$ and $\alpha \beta = \frac{3p}{4}.$We are given that $|\alpha - \beta| = \sqrt{10}.$Squaring both sides,we get $(\alpha - \beta)^2 = 10.$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.$$p^2 - 3p = 10.$$p^2 - 3p - 10 = 0.$Factoring the quadratic equation: $(p - 5)(p + 2) = 0.$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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