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