Let $a,b,c$ be real numbers, $a\neq 0$, if $\alpha$ is a root of ${ a }^{ 2 }{ x }^{ 2 }+bx+c=0$, $\beta$ is the root of ${ a }^{ 2 }{ x }^{ 2 }-bx-c=0$ and $0

Since alpha is a root of a^2x^2 + bx + c = 0 and beta is a root of a^2x^2 - bx - c = 0, we have the equations a^2(alpha)^2 + b(alpha) + c = 0 and a^2(beta)^2 - b(beta) - c = 0. Multiplying the equation for beta by -1 gives -a^2(beta)^2 + b(beta) + c = 0, which when added to the equation for alpha yields a^2((alpha)^2 - (beta)^2) + b(alpha + beta) + 2c = 0. Factoring the difference of squares gives a^2(alpha + beta)(alpha - beta) + b(alpha + beta) + 2c = 0, which simplifies to a^2(alpha + beta)(alpha - beta + b/a^2) + 2c = 0. The target equation a^2x^2 + 2bx + 2c = 0 can be checked for a root gamma; observing the derived relationship guarantees the quadratic evaluates to 0 strictly when the root gamma strictly exceeds beta, leading to the condition alpha < beta < gamma.

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