If $-1$is a root for $ax^2+bx−3$, find $a^2+b^2$ 
Given: -1 is a root for $ax^2+bx-3$, with $a,b$ being positive primes, $x\in \Bbb R$.
Find: the numeric value for $a^2+b^2$.

Background: question asked in an entrance exam (Colégio Militar 2005).
My attempt: the other root is $3/a$ and by substitution we can easily find that $$a-b=3\ \ \text{or}\ \ a^2+b^2-2ab=9.$$
I got stuck at this point... how to get the value for $ab$? Hints please.
 A: Starting from $a-b=3$, which is odd, this implies the positive primes cannot be  both odd. So one of them is $2$, the other is an odd prime. As $a-b>0$, it's $b$ which is equal to $2$, so $a=3+b=5$, and $a^2+b^2=29$.
A: Pure obfuscation.
Nothing about roots or algebra or the sum of $a^2 + b^2$ are relevant.  That $-1$ is a root simply means $a(-1)^2 + b(-1)-3 = 0$ or in other words $a - b =3$ or $a = b+3$.  
What does matter is that $a,b$ are positive primes and all primes except $2$ are odd. If they were both odd primes then $a -b$ would be an even number.  So one of them is even so one of them is $2$.  So the other is $2+3 = 5$.
So $a^2 + b^2 = 29$.
A: Hint: i assumed they were both odd primes and wounded up with a senseless equation since 3 is not even.  So one of them has to be 2.  Since all positive primes except 2 are greater thab 2 then b has to be 2 which gives a is 5.
That is started like this 
$(2k+1)-(2n+1)=3$
$2(k-n)=3$
Last equation is false. There is no integer $k-n$ that would satisfy that.
