The condition that x^3 - p x^2 + q x - r = 0 | vedclass

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Question

(A) Given the cubic equation $x^3 - p x^2 + q x - r = 0$ ... $(i)$.Let the roots be $\alpha, \beta, \gamma$.From Vieta's formulas:$\alpha + \beta + \g

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$r = pq$

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(A) Given the cubic equation $x^3 - p x^2 + q x - r = 0$ ... $(i)$.Let the roots be $\alpha, \beta, \gamma$.From Vieta's formulas:$\alpha + \beta + \gamma = p$ ... $(ii)$$\alpha \beta + \beta \gamma + \gamma \alpha = q$ ... $(iii)$$\alpha \beta \gamma = r$ ... $(iv)$Given that two roots are equal in magnitude but opposite in sign,let $\alpha = -\beta$.Substituting $\alpha = -\beta$ into $(ii)$:$-\beta + \beta + \gamma = p \implies \gamma = p$.Substituting $\gamma = p$ into $(iv)$:$\alpha \beta (p) = r \implies -\beta^2 p = r \implies \beta^2 = -\frac{r}{p}$.Substituting $\alpha = -\beta$ and $\gamma = p$ into $(iii)$:$-\beta^2 + \beta p - \beta p = q \implies -\beta^2 = q$.Equating the two expressions for $\beta^2$:$-\frac{r}{p} = -q \implies r = pq$.

The condition that $x^3 - p x^2 + q x - r = 0$ may have two of its roots equal to each other but of opposite sign is

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