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I have been solving some questions on algebra until I came across this question:

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

find the value of $$x + y + z$$.

So first, I expanded the equation. Then, I factorised it again leaving x, y and z out which left me with this equation:

$$x(x - 6) + y(y - 8) + z(z - 10) = -50$$

I thought this was the correct way to approach this question. However, I cannot take any further steps. Can someone please give me hints on how to approach this question?

Thanks in advance.

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    $\begingroup$ This is a trick question: a sum of three nonnegative numbers can only vanish if the terms are all zero. In this case, if $(x-3)^2$ etc., are all zero. $\endgroup$ – Angina Seng Oct 3 '17 at 5:31
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Each term on the left side of your equation is a square, so it's positive or zero. The only way for non-negative terms to sum to zero is if they're all zero. So $x=3,$ $y=4$, and $z=5.$

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  • $\begingroup$ Thanks! You're really helpful! $\endgroup$ – ianc1339 Oct 3 '17 at 5:47
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    $\begingroup$ Happy to help :) $\endgroup$ – user474330 Oct 3 '17 at 5:48
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Hint: $(x-3)^2>0$ for all $x\neq 3$ and $(x-3)^2=0$ specifically only in the case of $x=3$.

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  • $\begingroup$ Thank you for your response! I can understand the concept better! $\endgroup$ – ianc1339 Oct 3 '17 at 5:47
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$$(x-3)^2+(y-4)^2+(z-5)^2=0$$

means

$$\left\|\begin{bmatrix} x-3 \\ y-4 \\ z-5\end{bmatrix} \right\|^2=0$$

can you conclude what vector is $\begin{bmatrix} x-3 \\ y-4 \\ z-5\end{bmatrix}$ and solve for $x,y,z$?

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  • $\begingroup$ Thank you! However, I'm not very familiar with vectors... $\endgroup$ – ianc1339 Oct 3 '17 at 5:47

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