Let f ( x ) = x 2 − bx + c , b is a odd positive integer, f ( x ) = 0 have two prime numbers as roots and b + c = 35. Then the global minimum value of f(x) is

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Let α , β be roots of x 2 − b x + c = 0 . Then α + β = b . One of the roots is '2' (since α , β are integers). Therefore, f ( 2 ) = 0 ⇒ 2 b − c = 4 and b + c = 35 . Thus, b = 13 and c = 22 . The minimum value is f ( 2 13 ​ ) = − 4 81 ​ .

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