Given quadratic function, f (x) = ax & # 178; + BX + C (a ≠ 0), this paper proves that the equation f (x) = 1 / 2 [f (0) + F (1)] has two unequal real roots, and one root is in the interval (0,1)

Given quadratic function, f (x) = ax & # 178; + BX + C (a ≠ 0), this paper proves that the equation f (x) = 1 / 2 [f (0) + F (1)] has two unequal real roots, and one root is in the interval (0,1)

It is proved that: F (0) + F (1) = C + (a + B + C) = a + B + 2C; f (x) = 0.5 [f (0) + F (1)] that is ax ^ 2 + BX + C = 0.5A + 0.5B + C; → ax ^ 2 + bx-0.5 (a + b) = 0; its discriminant △ = B ^ 2 - 4A × [- 0.5 (a + b)] = B ^ 2 + (2a ^ 2 + 2Ab) = (a + b) ^ 2 + A ^ 2 because f (x) is a quadratic function, a ≠ 0; then a ^ 2 > 0; then