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Given the real number a > b > C and a + B + C = 0, the two different real number roots of the equation AX ^ 2 + BX + C = 0 are x1, X2 (1) proof - 1 / 2C and a + B + C = 0, and the two different real number roots of the equation AX ^ 2 + BX + C = 0 are x1, X2 (1) proof - 1 / 2
(1) Thus, (1) can not be proved
It should be - 1 / 2C and a + B + C = 0, there must be a > 0, C0, that is a > - 2b, divide both sides by - 2A (less than 0), get - 1 / 2B, divide both sides by a, get: B / A
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