If $\displaystyle \alpha, : \beta$ are roots of equation $\displaystyle ax^{2} + bx + c = 0$ which are real and opposite in sign, then roots of the equation $\displaystyle \alpha \left ( x - \beta \right )^{2} + \beta \left ( x - \alpha \right )^{2} = 0$ are

Expanding the given equation alpha(x - beta) squared + beta(x - alpha) squared = 0 results in alpha(x squared - 2*beta*x + beta squared) + beta(x squared - 2*alpha*x + alpha squared) = 0. Grouping the terms gives (alpha + beta)x squared - 4*alpha*beta*x + alpha*beta(alpha + beta) = 0. The discriminant is 16*alpha squared*beta squared - 4*alpha*beta(alpha + beta) squared, which factors to 4*alpha*beta(4*alpha*beta - (alpha + beta) squared). Because alpha and beta have opposite signs, the product alpha*beta is negative, making the entire discriminant positive and ensuring both roots are real. Furthermore, since the sum of the roots is 4*alpha*beta divided by (alpha + beta) and the product of the roots is alpha*beta, both of these values are negative, meaning both individual roots must be positive.

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