Prime Zeroes!

Let the prime-number roots of $x^2 - ax + 2b$ be $p$ and $q$ . Then, by Vieta's formula , we have $pq = 2b$ . Since $p$ and $q$ are primes and $2$ is also a prime, it means that $p=2$ and $q=b$ or vice versa. By Vieta's formula again, $p+q = 2+b = a$ , $\implies a-b = \boxed{2}$ .

$Product~ of~ the~ roots~=2b.~(^*_*)\\ Our~roots~are~both~primes.\\ But~only~ even~ prime~ is~2.\\ \therefore~one~root~is~2,~\implies~other ~root~is~b.\\ -(-a)=sum~of~roots=2+b .~(^*_*)\\ \implies~a - b=2.\\ Both~(^*_*) ~are~due~to~Vieta's~Formula.$