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(B) Given equations are:$ax^2 + 2bx + c = 0$ with roots $\alpha, \beta \implies \alpha + \beta = -\frac{2b}{a}, \alpha\beta = \frac{c}{a}$$2bx^2 + cx

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$-1$

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(B) Given equations are:$ax^2 + 2bx + c = 0$ with roots $\alpha, \beta \implies \alpha + \beta = -\frac{2b}{a}, \alpha\beta = \frac{c}{a}$$2bx^2 + cx + a = 0$ with roots $\alpha, \gamma \implies \alpha + \gamma = -\frac{c}{2b}, \alpha\gamma = \frac{a}{2b}$$cx^2 + ax + 2b = 0$ with roots $\alpha, \delta \implies \alpha + \delta = -\frac{a}{c}, \alpha\delta = \frac{2b}{c}$Multiplying the product of roots:$(\alpha\beta)(\alpha\gamma)(\alpha\delta) = (\frac{c}{a})(\frac{a}{2b})(\frac{2b}{c}) = 1$$\alpha^3(\beta\gamma\delta) = 1$From the sum of roots:$\beta = -\frac{2b}{a} - \alpha, \gamma = -\frac{c}{2b} - \alpha, \delta = -\frac{a}{c} - \alpha$Substituting these into the product equation or observing the symmetry,we find that for the system to hold with positive $a, b, c$,$\alpha$ must satisfy the condition $\alpha + \alpha^2 = -1$.

If $\alpha$ and $\beta$,$\alpha$ and $\gamma$,$\alpha$ and $\delta$ are the roots of the equations $ax^2 + 2bx + c = 0$,$2bx^2 + cx + a = 0$ and $cx^2 + ax + 2b = 0$ respectively,where $a, b$ and $c$ are positive real numbers,then $\alpha + \alpha^2 = $

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