Find the product of roots of quadratic equation if |x1-x2|=1 and x^2=2x-m.

We'll re-write the constraint x^2=2x-m moving all terms to
the left side.

x^2-2x+m =
0

Since we know that m represents the value of the product
of the roots, we'll write the Viete's 2nd relation:

x1*x2 =
m

From the module definition we'll write the given
constraint:

-1<(x1-x2)<1

If
we are square raising, we'll
get:

(x1-x2)^2=1

x1^2-2x1*x2+x2^2=1

x1^2+x2^2=1+2x1*x2 
(1)

We know that if we are substituting the roots into the
equation, they are verifying it. So, we'll introduce the roots, one by one, into the
equation.

x1^2-2x1+m=0
(2)

x2^2-2x2+m=0 (3)

Now,
we'll add (2) and
(3):

x1^2-2x1+m+x2^2-2x2+m=0

We'll
combine like
terms:

(x1^2+x2^2)-2(x1+x2)+2m=0

From
the Viete's relations, we'll have:

x1+x2= -
(-2)/1

x1+x2= 2

So, x1^2+x2^2=
2(x1+x2)-2m, where x1+x2= 2

x1^2+x2^2= 2*2
-2m

x1^2+x2^2= 4-2m  (4)

But
from relation (1), x1^2+x2^2=1+2x1*x2, where
x1*x2=m 

From (1) and
(4):

 1+2m=
4-2m

We'll move the terms in m to the left side and the
numbers alone to the right
side:

4m=3

The
value of the product of the roots is m = 3/4.