Question

If p and q are distinct primes and x^2-PX+q=0 has distinct positive roots ,then p+q equals

Answer 1

see above image......

Answer 2

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.