Let f (x) = the square of AX + BX + C (a > 0) and f (1) = A / 2 (1)) prove that the function has two zeros

Let f (x) = the square of AX + BX + C (a > 0) and f (1) = A / 2 (1)) prove that the function has two zeros

Let f (1) = A / 2 = > A / 2 = a + B + C = > - B = A / 2 + CB & sup2; - 4ac = A & sup2; + 4 + C & sup2; + ac-4ac = A & sup2; + 4 + C & sup2; - 3aca & sup2; - 4 + 9A & sup2; / 4-9a & sup2; = - 2A & sup2; + (3a / 2-C) & sup2; depend on whether the title is wrong