If $\left( {\alpha ,\beta } \right),\left( {\alpha ,\gamma } \right),\left( {\alpha ,\delta } \right)$ are the roots of equations$a{x^2} + 2bx + c = 0,\,2b{x^2} + cx + a = 0,\,$ $c{x^2} + ax + 2b = 0$ respectively, where, $a,b,c \in + ve$ real no, then $\alpha + {\alpha ^2}$.

Since alpha is a root of all three equations, we have a(alpha)^2 + 2b(alpha) + c = 0, 2b(alpha)^2 + c(alpha) + a = 0, and c(alpha)^2 + a(alpha) + 2b = 0. Adding these three equations yields (a + 2b + c)(alpha)^2 + (2b + c + a)(alpha) + (c + a + 2b) = 0, which factors to (a + 2b + c)(alpha^2 + alpha + 1) = 0. Since a, b, and c are positive real numbers, their sum cannot be zero, meaning alpha^2 + alpha + 1 = 0, and thus alpha + alpha^2 = -1.

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