If the sum of two roots of x^3+p x^2+q x-5=0 | vedclass

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(C) Let the roots be $\alpha, \beta, \gamma$. Given $\alpha + \beta = \gamma$. From Vieta's formulas,$\alpha + \beta + \gamma = -p$. Substituting $\al

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(C) Let the roots be $\alpha, \beta, \gamma$. Given $\alpha + \beta = \gamma$. From Vieta's formulas,$\alpha + \beta + \gamma = -p$. Substituting $\alpha + \beta = \gamma$,we get $2\gamma = -p$,so $\gamma = -\frac{p}{2}$. Since $\gamma$ is a root,it satisfies the equation: $(-\frac{p}{2})^3 + p(-\frac{p}{2})^2 + q(-\frac{p}{2}) - 5 = 0$. $-\frac{p^3}{8} + \frac{p^3}{4} - \frac{pq}{2} - 5 = 0$. Multiplying by $8$,we get $-p^3 + 2p^3 - 4pq - 40 = 0$. $p^3 - 4pq = 40$. $p(p^2 - 4q) = 40$.

If the sum of two roots of $x^3+p x^2+q x-5=0$ is equal to its third root,then $p(p^2-4q)=$

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