If x, y, and z are real numbers such that (x-3)^2+(y-4)^2 +(z-5)^2&=0, then x+y+z=

1997 AMC 12 Problem 3

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x,

y,

and

z

are real numbers such that

\begin{aligned} (x-3)^2+(y-4)^2\\ {}+(z-5)^2&=0, \end{aligned}

then

x+y+z=

-12

0

8

12

50

Each squared real quantity is nonnegative

A sum of nonnegative terms is zero only when every term is zero

All three squares are nonnegative, so each must be

0.

Hence

x=3,

y=4,

z=5,

and

x+y+z=12.

The correct answer is D.