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.
i thought 4 was the answer but it is incorrect...
+9481 \(x = {-2m \pm \sqrt{(2m)^2-4(m^2+2m+3)} \over 2} = {-2m \pm \sqrt{4m^2-4m^2-8m-12} \over 2}= -m \pm \sqrt{-2m-3} \)
\(r = -m + \sqrt{-2m-3} \\~\\ r^2=(-m + \sqrt{-2m-3})(-m + \sqrt{-2m-3}) \\~\\ r^2=m^2-2m\sqrt{-2m-3}+(-2m-3)\)
and
\(s = -m-\sqrt{-2m-3} \\~\\ s^2=(-m-\sqrt{-2m-3})(-m-\sqrt{-2m-3}) \\~\\ s^2=m^2+2m\sqrt{-2m-3}+(-2m-3)\)
\(r^2+s^2=m^2-2m\sqrt{-2m-3}+(-2m-3)+m^2+2m\sqrt{-2m-3}+(-2m-3) \\~\\ r^2+s^2=2m^2+2(-2m-3) \\~\\ r^2+s^2=2m^2-4m-6\)
The minimum value of r^2 + s^2 occurs when m = 4/( 2 * 2 ) = 1
When m = 1 , r2 + s2 = 2 - 4 - 6 = -8
+130071 Here's my take on this....
The sum of the roots is r + s = 2m .....so....
(r + s)^2 =
r^2 + 2rs + s^2 = 4m^2 (1)
The product of the roots is r * s = m^2 + 2m + 3 (2)
Manipulating (1), we have that
r^2 + s^2 = 4m^2 - 2rs
Sub (2) into this
r^2 + s^2 =
4m^2 - 2(m^2 + 2m + 3) =
4m^2 - 2m^2 - 4m - 6
2m^2 - 4m - 6 (3)
(3) is minimized when m =
4/ (2*2) = 4 / 4 = 1
And the minimum value is 2(1)^2 - 4(1) - 6 = 2 - 4 - 6 = -8
So......
The minimum of r^2 + s^2 = [ min ] 2m^2 - 4m - 6 = - 8
