Đáp án:

p+q = 2 +3 = 5 Ans.

Giải thích các bước giải:

Given : p and q are prime in x^2 -px +q =0
and it has distinct positive roots
As we know that
for distinct roots
D > 0
or, (-p)^2 -4q >0
or, p^2 - 4q >0
or, p^2 > 4q
or, p > √(4q)
or, p > 2√q
But p and q are primes and here we got p is greater than the twice of the sqrt of q.
Let a and b be the roots of the eqn.,
Also we can right the above equation as
(x-a)(x-b) where a and b are distinct positive roots
so, x^2 -ax -bx +ab =0
or, x^2 -(a+b)x +(ab) = 0
comparing this equation to the given eqn.
we get (a+b) = p
(ab) = q
but since q is a prime no. so it can be written as (1×itself ) according to the definition of prime
so, either a or b is equal to 1.
hence ab = q
or, 1×b = q
so, b = q
Now , as (a +b) = p
so, 1+b = p
or, 1+q = p
or, p - q = 1
And we have only one pair of such primer nos.
whose difference is 1 and those are 2 & 3
so, q= 2 and p = 3.
{Also we can see that , p > 2√q
or. 3 > 2×√2 = 2×1.414 = 2.828}
Hence roots are distinct too.
Now solving , p+q = 2 +3 = 5 Ans.