The quadratic function y = f (x) satisfies f (3 + x) = (3-x), and f (x) = 0 has two real roots x1, x2?
According to the condition f (3 + x) = f (3-x), the symmetry axis of quadratic function [let y = ax ^ 2 + BX + C] be: x = - B / 2A = 3, then - B / a = 6. The two roots of ax ^ 2 + BX + C = 0 are x1, x2. From the relationship between root and coefficient, we can get: X1 + x2 = - B / a = 6
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- 1. If the quadratic function y = f (x) satisfies f (3 + x) = f (3-x), and f (x) = 0 has two real roots X1 and X2, then X1 + x2 = () A. 0b. 3C. 6D. Not sure
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