Let $f(x) = x^2 - bx + c$, b is an odd positive integer. If f(x)=0 has two prime numbers as roots and b + c = 35, then the global minimum value of f(x) is

Given: $f(x) = x^2 - bx + c$
where $b$ is an odd positive integer and $b + c = 35$
Let $\alpha, \beta$ be two prime roots of $f(x) = 0$
By Vieta's formulae:
$\alpha + \beta = b$
$\alpha \beta = c$
Now, $b + c = 35$
$\Rightarrow (\alpha + \beta) + \alpha\beta = 35$
Since $b$ is odd and $\alpha + \beta = b$ is odd
$\Rightarrow$ one of $\alpha, \beta$ must be even prime
$\therefore \alpha = 2$ (only even prime)
$\Rightarrow 2 + \beta + 2\beta = 35$
$\Rightarrow 3\beta = 33$
$\Rightarrow \beta = 11$
$\therefore b = \alpha + \beta = 2 + 11 = 13$
$\therefore c = \alpha\beta = 2 \times 11 = 22$
Verification: $b + c = 13 + 22 = 35$ ✓
So $f(x) = x^2 - 13x + 22$
Global minimum occurs at $x = \dfrac{b}{2} = \dfrac{13}{2}$
Minimum value $= f\left(\dfrac{13}{2}\right) = \left(\dfrac{13}{2}\right)^2 - 13\left(\dfrac{13}{2}\right) + 22$
$= \dfrac{169}{4} - \dfrac{169}{2} + 22$
$= \dfrac{169 - 338 + 88}{4}$
$= \dfrac{-81}{4}$
$\therefore \boxed{\text{Global Minimum} = -\dfrac{81}{4}}$