Everything Prime

Let the roots of required quadratic equation be $\alpha , \beta$ ( Where $\alpha , \beta$ are prime numbers.) So, the equation can be written as

$(x-\alpha)(x-\beta)$ $=x^{2}-(\alpha+\beta)+\alpha\beta$

Since the sum of coefficients of the equation is prime:

$1-\alpha-\beta+\alpha\beta$ is prime

$(\alpha-1)(\beta-1)$ is prime

The product of two numbers can be prime if and only if one of the numbers is $1$

So, $\alpha-1=1$

$\alpha=2$

Now,

$(\beta-1)$ is prime

also $\beta$ is prime

So, we're talking about two consecutive prime numbers and fortunately we have just one pair of consecutive prime numbers, i.e., $(2 , 3)$

$\beta=3$

Therefore, sum of roots of the equation $=\alpha+\beta = 2+3 = 5$

Let the quadratic polynomial be denoted by f (x) . Now sum of coefficients =f (1)=a prime m. Let primes p,q be such that f (p)=f (q)=0. Thus p-1 divides m and so does q-1. Since m is a prime, one of p & q must be equal to 2. We note that m must be 2. So, the other root must be 3. So, sum of roots is 5.