Let f (x) = ax ^ 2 + BX + C, if 6A + 2B + C = 0, f (1) * f (3) > 0, prove that the equation f (x) = 0 must have two unequal real roots, and 3 < X1 + x2 < 5

We can only prove the latter question: F (1) f (3) = (a + B + C) (9a + 3B + C) > 0, because 6A + 2B + C = 0, so C = - 6a-2b is brought into f (1) f (3) = (- 5a-b) (3a + b) > 0, and both sides are divided by a ^ 2, (- 5-b / a) (3 + B / a) > 0 at the same time

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