If one root of the quadratic equation x^2 + p | vedclass

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Question

(A) Let the roots of the equation $x^2 + px + q = 0$ be $\alpha$ and $\alpha^2$. From the relation between roots and coefficients: Sum of roots: $\alp

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the roots of the equation $x^2 + px + q = 0$ be $\alpha$ and $\alpha^2$. From the relation between roots and coefficients: Sum of roots: $\alpha + \alpha^2 = -p$ Product of roots: $\alpha \cdot \alpha^2 = \alpha^3 = q$ So,$\alpha = q^{1/3}$. Substituting $\alpha$ into the sum equation: $q^{1/3} + (q^{1/3})^2 = -p$ $q^{1/3} + q^{2/3} = -p$ Cubing both sides: $(q^{1/3} + q^{2/3})^3 = (-p)^3$ Using $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$: $q + q^2 + 3(q^{1/3})(q^{2/3})(q^{1/3} + q^{2/3}) = -p^3$ $q + q^2 + 3q(q^{1/3} + q^{2/3}) = -p^3$ Since $q^{1/3} + q^{2/3} = -p$: $q + q^2 + 3q(-p) = -p^3$ $p^3 + q^2 - 3pq + q = 0$ $p^3 + q(q - 3p) + q = 0$ $p^3 - q(3p - q) + q = 0$.

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

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