If $\alpha$ and $\beta$ are the roots of the equation $ax^2+bx+c=0$ $(a\neq 0,$ a, b, c being different), then the sum of the roots of the equation $a^{2}x^{2}+(b^{2}-2ac)x+b^{2}-4ac=0$ is

Using the sum and product of roots formulas for ax^2 + bx + c = 0, we establish that alpha + beta = -b/a and alpha beta = c/a. For the new equation a^2x^2 + (b^2 - 2ac)x + (b^2 - 4ac) = 0, the sum of its roots is found by dividing the linear coefficient by the negative of the leading coefficient, yielding -(b^2 - 2ac)/a^2. Substituting c/a for the product alpha beta simplifies the expression to (2a^2(alpha beta) - b^2)/a^2 = 2 alpha beta - (b/a)^2. Rewriting this in terms of the original sum and product gives 2 alpha beta - (alpha + beta)^2, which equals -(alpha^2 + beta^2), a strictly negative value for real distinct roots.

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