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...
+9488 x=−2m±√(2m)2−4(m2+2m+3)2=−2m±√4m2−4m2−8m−122=−m±√−2m−3
r=−m+√−2m−3 r2=(−m+√−2m−3)(−m+√−2m−3) r2=m2−2m√−2m−3+(−2m−3)
and
s=−m−√−2m−3 s2=(−m−√−2m−3)(−m−√−2m−3) s2=m2+2m√−2m−3+(−2m−3)
r2+s2=m2−2m√−2m−3+(−2m−3)+m2+2m√−2m−3+(−2m−3) r2+s2=2m2+2(−2m−3) r2+s2=2m2−4m−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
+130477 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
