# Absolute value question when solving with square root function

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.

• If a=|b|, then a=b or a=-b.. – Swapnil Tripathi Apr 14 '14 at 4:12
• 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? – alan Apr 14 '14 at 4:17
• Yes. That's what i mean. :) – Swapnil Tripathi Apr 14 '14 at 4:28