+) Case 1: a = 0
=> f(x) = 0×x²+c = c
=> for all values of x, f(x) always = c (does not satisfy the requirement)
+) Case 2: a≠0
=> f(x) = ax²+c
=> for every non-zero a, f(x) has only one solution x
Ans: a≠0, c ∈ R
P/s: c can be any value (in case you don't know the symbols above)
Ok done. Thank to me >:333