Suppose r and s are the values of x that satisfy the equation x^2 - 2mx + (m^2 - 6m + 11) = 0 for some real number m. Find the minimum real value of (r - s)^2.
r + s = 2m
(r + s)^2 = 4m^2 = r^2 + s^2 + 2rs → r^2 + s^2 = 4m^2 - 2rs
And
rs = (m^2 - 6m + 11)
So
(r - s)^2 = (r^2 + s^2) - 2rs = (4m^2 - 2rs) - 2rs = 4m^2 - 4rs = 4m^2 - 4(m^2 -6m + 11) =
24m - 44
24m - 44 has no min value
+456 
