If the sum of two of the roots of {x^3} + p{x | vedclass

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(B) Let the roots of the cubic equation ${x^3} + p{x^2} + qx + r = 0$ be $\alpha, \beta, \gamma$.Given that the sum of two roots is zero,let $\alpha +

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$r$

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(B) Let the roots of the cubic equation ${x^3} + p{x^2} + qx + r = 0$ be $\alpha, \beta, \gamma$.Given that the sum of two roots is zero,let $\alpha + \beta = 0$.From the relation between roots and coefficients,the sum of the roots is $\alpha + \beta + \gamma = -p$.Substituting $\alpha + \beta = 0$,we get $0 + \gamma = -p$,so $\gamma = -p$.Since $\gamma$ is a root of the equation,it must satisfy the equation: $(-p)^3 + p(-p)^2 + q(-p) + r = 0$.$-p^3 + p^3 - pq + r = 0$.$-pq + r = 0$.Therefore,$pq = r$.

If the sum of two of the roots of ${x^3} + p{x^2} + qx + r = 0$ is zero,then $pq =$

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