Question : If $x$,$y$ and $z$ are real numbers such that $(x-3)^2+(y-4)^2+(z-5)^2=0$, then $(x+y+z)$ is equal to:
Option 1: –12
Option 2: 0
Option 3: 8
Option 4: 12
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Correct Answer: 12
Solution :
Given: $x,y$ and $z$ are real numbers such that $(x-3)^2+(y-4)^2+(z-5)^2=0$
We know,
The addition of three perfect squares can be zero only when the individual numbers are zero.
So, $(x-3)^2=0, (y-4)^2=0$ and $(z-5)^2=0$
⇒ $x = 3, y = 4$ and $z = 5$.
Then, $(x+y+z)$ = 3 + 4 + 5 = 12
Hence, the correct answer is 12.
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