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Given the equation $x + (y-1)^2=0$ and solving for $y$ I obtain $\vert y-1\vert = \pm\sqrt{-x}$.

I know that the final answer is $y = 1 \pm \sqrt{-x}$, but what I don't understand is why taking off the absolute value sign is valid. Isn't it important to know what the range of y is before you can do such a thing? What property allows this? Hopefully I'm not grossly misunderstanding something. Thanks.

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  • $\begingroup$ If a=|b|, then a=b or a=-b.. $\endgroup$ – Swapnil Tripathi Apr 14 '14 at 4:12
  • $\begingroup$ Isn't it if a=|b|, then a=b or -a=b? But yes I see your point. Since |y-1| already equals both + and - the sqrt(-x), then its ok to take it off. At least thats how I understand it now. Is that what you mean? $\endgroup$ – alan Apr 14 '14 at 4:17
  • $\begingroup$ Yes. That's what i mean. :) $\endgroup$ – Swapnil Tripathi Apr 14 '14 at 4:28
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What you're trying to address with the absolute value is already addressed in that plus/minus sign on the square root. You don't need the absolute value there in the first place.

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  • $\begingroup$ I see, makes sense now that i think about it, thank you $\endgroup$ – alan Apr 14 '14 at 4:18

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