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...

Guest Feb 24, 2018

#2**+1 **

\(x = {-2m \pm \sqrt{(2m)^2-4(m^2+2m+3)} \over 2} = {-2m \pm \sqrt{4m^2-4m^2-8m-12} \over 2}= -m \pm \sqrt{-2m-3} \)

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

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 , r^{2} + s^{2} = 2 - 4 - 6 = -8

hectictar
Feb 24, 2018

#3**+2 **

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

CPhill
Feb 24, 2018