Suppose r and s are the values of c 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.

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 $r^2+s^2$.

Use the quadratic formula:

$ax^2 + bx + c = 0 \implies x = \frac{1}{2}\left(-b \pm \sqrt{b^2 - 4ac}\right)$

(here, $a = 1$, $b = -2m$ and $c = m^2 + 2m + 3$)

to find that

$r = m + \sqrt{-(2m + 3)}$

and

$s = m - \sqrt{-(2m + 3)}$.

Then $r^2 + s^2 = 2\left(m^2 - 2m +3\right)$.

This is a concave-up parabola with vertex at $m = 1$.

Now $\left(r^2 + s^2\right)\big|_{m = 1} = 2\left(1 - 2 + 3\right) = 4$.