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(A) Given the quadratic equation $x^2 + px + p^3 = 0$ with roots $\alpha$ and $\beta$.From the properties of roots,we have $\alpha + \beta = -p$ and $

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$4, -2$

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(A) Given the quadratic equation $x^2 + px + p^3 = 0$ with roots $\alpha$ and $\beta$.From the properties of roots,we have $\alpha + \beta = -p$ and $\alpha \beta = p^3$.Since the point $(\alpha, \beta)$ lies on the parabola $y^2 = x$,we have $\beta^2 = \alpha$.Substituting $\alpha = \beta^2$ into the product of roots equation: $\beta^2 \cdot \beta = p^3 \implies \beta^3 = p^3 \implies \beta = p$.Now,substitute $\beta = p$ and $\alpha = p^2$ into the sum of roots equation $\alpha + \beta = -p$:$p^2 + p = -p \implies p^2 + 2p = 0$.Factoring gives $p(p + 2) = 0$. Since $p 
eq 0$,we have $p = -2$.Substituting $p = -2$ back into the roots: $\beta = p = -2$ and $\alpha = p^2 = (-2)^2 = 4$.Thus,the roots are $4$ and $-2$.

Let $\alpha, \beta$ be the roots of the quadratic equation $x^2 + px + p^3 = 0$ $(p \neq 0)$. If $(\alpha, \beta)$ is a point on the parabola $y^2 = x$,then the roots of the quadratic equation are:

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