If one root of the equation x^2 + px + q = 0 | vedclass

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(D) Let the roots of the given equation $x^2 + px + q = 0$ be $\alpha$ and $\alpha^2$.From the relationship between roots and coefficients:Product of

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$p^3 + q^2 + q(1 - 3p) = 0$

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(D) Let the roots of the given equation $x^2 + px + q = 0$ be $\alpha$ and $\alpha^2$.From the relationship between roots and coefficients:Product of roots: $\alpha \cdot \alpha^2 = \alpha^3 = q$Sum of roots: $\alpha + \alpha^2 = -p$Now,cube both sides of the sum equation: $(\alpha + \alpha^2)^3 = (-p)^3$Using the identity $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$:$\alpha^3 + (\alpha^2)^3 + 3\alpha \cdot \alpha^2(\alpha + \alpha^2) = -p^3$Substitute $\alpha^3 = q$ and $\alpha + \alpha^2 = -p$ into the equation:$q + q^2 + 3q(-p) = -p^3$Rearranging the terms,we get:$p^3 + q^2 + q - 3pq = 0$$p^3 + q^2 + q(1 - 3p) = 0$

If one root of the equation $x^2 + px + q = 0$ is the square of the other,then

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