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(A) Given the quadratic equation $x^2 + px + p^3 = 0$,the sum and product of the roots are $\alpha + \beta = -p$ and $\alpha \beta = p^3$.Since $(\alp

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$4, -2$

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(A) Given the quadratic equation $x^2 + px + p^3 = 0$,the sum and product of the roots are $\alpha + \beta = -p$ and $\alpha \beta = p^3$.Since $(\alpha, \beta)$ lies on the parabola $y^2 = x$,we have $\beta^2 = \alpha$.Substituting $\alpha = \beta^2$ into the product of roots: $\beta^2 \cdot \beta = p^3$ $\Rightarrow \beta^3 = p^3$ $\Rightarrow \beta = p$.Now,substitute $\beta = p$ and $\alpha = p^2$ into the sum of roots equation: $p^2 + p = -p$.This simplifies to $p^2 + 2p = 0$,which gives $p(p + 2) = 0$.Since $p 
eq 0$,we must 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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