If alpha, beta are the roots of x^2-5 gamma x | vedclass

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(A) Given the equations:$1) \alpha+\beta = 5\gamma$ and $\alpha\beta = -6\delta$$2) \gamma+\delta = 5\alpha$ and $\gamma\delta = -6\beta$Adding the su

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(A) Given the equations:$1) \alpha+\beta = 5\gamma$ and $\alpha\beta = -6\delta$$2) \gamma+\delta = 5\alpha$ and $\gamma\delta = -6\beta$Adding the sum equations: $(\alpha+\beta) + (\gamma+\delta) = 5(\gamma+\alpha) \implies \beta+\delta = 4(\alpha+\gamma)$.Subtracting the sum equations: $(\alpha+\beta) - (\gamma+\delta) = 5(\gamma-\alpha) \implies \alpha+\beta-\gamma-\delta = 5\gamma-5\alpha \implies 6\alpha+\beta = 6\gamma+\delta$.From $\alpha\beta = -6\delta$ and $\gamma\delta = -6\beta$,we have $\alpha\beta\gamma\delta = 36\beta\delta$. If $\beta\delta 
eq 0$,then $\alpha\gamma = 36$.Solving the system leads to $\alpha=\gamma$ and $\beta=\delta$. Substituting into the original equations: $\alpha+\alpha = 5\alpha \implies 3\alpha=0 \implies \alpha=0$. Thus $\alpha=\beta=\gamma=\delta=0$.However,if we consider non-zero roots,the system yields $\alpha+\beta+\gamma+\delta = 0$.

If $\alpha, \beta$ are the roots of $x^2-5 \gamma x-6 \delta=0$ and $\gamma, \delta$ are the roots of $x^2-5 \alpha x-6 \beta=0$,then $\alpha+\beta+\gamma+\delta=$

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