If the sum of two of the roots of x^3+p x^2-q | vedclass

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(A) Let the roots of the equation $x^3+p x^2-q x+r=0$ be $\alpha, \beta, \gamma$. From the relation between roots and coefficients: $\alpha+\beta+\gam

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$-r$

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(A) Let the roots of the equation $x^3+p x^2-q x+r=0$ be $\alpha, \beta, \gamma$. From the relation between roots and coefficients: $\alpha+\beta+\gamma = -p$ $\alpha \beta + \beta \gamma + \gamma \alpha = -q$ $\alpha \beta \gamma = -r$ Given that the sum of two roots is zero,let $\alpha + \beta = 0$,which implies $\beta = -\alpha$. Substituting $\alpha + \beta = 0$ into the first relation: $0 + \gamma = -p \implies \gamma = -p$. Substituting $\gamma = -p$ into the third relation: $\alpha \beta (-p) = -r \implies \alpha \beta p = r$. Now,substitute $\beta = -\alpha$ into the second relation: $\alpha(-\alpha) + \gamma(\alpha + \beta) = -q$ $-\alpha^2 + \gamma(0) = -q -\alpha^2 = -q \alpha^2 = q$. Since $\alpha \beta = \alpha(-\alpha) = -\alpha^2 = -q$,we have $\alpha \beta = -q$. From $\alpha \beta p = r$,we get $(-q)p = r$,which means $-pq = r$,or $pq = -r$.

If the sum of two of the roots of $x^3+p x^2-q x+r=0$ is zero,then $pq$ is equal to

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