Suppose $r$ and $s$ are the values of $x$ that satisfy the equation \[x^2 - 2mx + (m^2+2m+3) = 0\]for some real number $m$. find the minimum real value of $r^2+s^2$.

  -8

Step-by-step explanation:

For roots r and s, the quadratic can be factored ...

  f(x) = (x -r)(x -s) = x^2 -(r+s)x +rs

Then the value of r^2+s^2 can be determined from the coefficient of x (-(r+s)) and the constant (rs) by ...

  r^2 +s^2 = (-(r+s))^2 -2(rs) = (r^2 +2rs +s^2) -2rs = r^2 +s^2

Comparing this to your given equation, we have the coefficient of x as (-2m) and the constant term as (m^2+2m+3). Forming the expression ...

  (x-coefficient)^2 -2(constant term)

we get ...

  r^2 +s^2 = (-2m)^2 -2(m^2 +2m +3) = 2m^2 -4m -6

  r^2 +s^2 = 2(m -1)^2 -8

The minimum value of this quadratic expression is where m=1 and the squared term is zero. That minimum value is -8.

hi im bob : l

step-by-step explanation: