If p and q are distinct primes and x^2 - px + q = 0?

Hello,

Let x₁ and x₂ be the roots of the equation. Then, we have:

(x - x₁)(x - x₂) = x² - (x₁ + x₂)x + x₁.x₂ = x² - px + q

By mere comparison, we can deduce:

{ x₁ + x₂ = p

{ x₁.x₂ = q

Thus:

p + q = x₁ + x₂ + x₁.x₂

    = (x₁ + 1)(x₂ + 1) - 1

Regards,

Dragon.Jade :-)

it is got positive integral roots so roots are 1 and q as q is prime

equation is (x-1)(x-q)

= x^2 - (q+1) + q = 0

so p = q + 1

so q and q +1 are primes so q = 2 and q +1 =3 so p = 3, q = 2 and p+q = 5

Let r and s be the distinct positive integral roots of x^2 - px + q = 0.

Since (x - r)(x - s) = x^2 - (r + s)x + rs, r + s = p and rs = q.

Since q is prime and r & s are distinct positive integers,

r = 1 and s = q OR r = q and s = 1.

In either case, p = r + s = 1 + q. So p and q are primes, and p is 1 more than q.

The only possibility is p = 3 and q = 2.

The answer is p + q = 3 + 2 = 5.

Lord bless you today!

If the roots are x1 and x2, we have

x1+x2=p and

x1x2=q

But q is prime, so x1=1 and x2=q (both must be positive and integral)

Therefore 1+q=p

But the only way this can be true is p=2, q=3

So p+q=5.

Ans 5

P To Q