Problem: If x,yx, yx,y, and zzz are real numbers such that
(x−3)2+(y−4)2+(z−5)2=0(x-3)^{2}+(y-4)^{2}+(z-5)^{2}=0 (x−3)2+(y−4)2+(z−5)2=0
then x+y+z=x+y+z=x+y+z=
Answer Choices:
A. −12-12−12
B. 000
C. 888
D. 121212
E. 505050
Solution:
Since each summand is nonnegative, the sum is zero only when each term is zero. Hence the only solution is x=3,y=4x=3, y=4x=3,y=4, and z=5z=5z=5, so the desired sum is 121212.