+21 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.
+130458 r + s = 2m square both sides
r^2 + 2rs + s^2 = 4m^2
r^2 + s^2 = 4m^2 - 2rs
rs = m^2 - 6m + 11
2rs = 2m^2 -12m + 22
So
r^2 + s^2 = 4m^2 - 2rs
r^2 + s^2 = 4m^2 - (2m^2 - 12m + 22)
r^2 + s^2 = 2m^2 + 12m - 22
(r - s)^2 =
r^2 + s^2 - 2rs =
2m^2 + 12m - 22 - (2m^2 - 12m + 22) =
24m - 44
This is a linear expression.....there is no real number minimum
