SOLUTION: Show that the roots of the equation (x-p)(x-q)=2 are real and distinct for all real values of p and q.

        (x-p)(x-q) = 2

 x² - qx - px + pq = 2

  x² - (q+p)x + pq = 2

x² - (q+p)x + pq-2 = 0

We must show that the discriminant is always positive.

Discriminant = b² - 4ac = [-(q+p)]² - 4(1)(pq-2)

We simplify this discriminant

(q+p)² - 4(pq-2)

(q+p)² - 4pq + 8

q² + 2pq + p² - 4pq + 8

q² - 2pq + p² + 8

(q - p)² + 8

This will always be positive because it is greater 
than or equal to 8.

Therefore the roots are real and distinct.

Edwin

1.  The quadratic function f(x) = (x-p)*(x-q)  is zero  at  x = p  (and x= q),  and goes 

    to infinity as  x ---> infinity  or  x ---> - infinity.

    Hence, there is a value of x, where the function f(x) takes the intermediate value of 2:  f(x) = 2.


    So, the equation f(x) really has the roots.


2.  The roots can not be x= p  or  x= q,  since at these values the finction is zero (and, therefore, can not be 2).