Find the product of all real values of r for which 1/(2x) = (r - x)/5 has exactly one real solution.
1/ (2x) = (r - x) / 5
5 = 2x ( r - x)
5 = 2rx- 2x^2
2x^2 - 2rx + 5
If this has one real solution then the discriminant = 0 .....so.....
(-2r)^2 - 4 (2) ( 5) = 0
4r^2 - 40 = 0
4r^2= 40
r^2 = 10
r = sqrt (10) or r = -sqrt (10)
Their product = -10
