# A question about eliminating square roots [duplicate]

If $\sqrt{x^2} = \pm x$, then why does $\sqrt{(x+2)^2} = x+2$ and not $\pm (x+2)$?

This is driving me crazy, so feel free to elucidate. Thanks!

---EDIT---
I'm not sure how the other questions' answers would help answer my own question, but it doesn't really matter now that I've figured it out. After thinking a bit about some of what @Daniel Hast typed on a similar question I asked, I realized that the reason why $\sqrt{(x+2)^2} = x+2$ was that more generally, given that $\sqrt{x} = r$ such that $r^2 = x$, squaring $r$ in $\sqrt{x} = r$ produced $(\sqrt{x})^2 = r^2 = x$ and subsequently $(\sqrt{x})^2 = x$.

## marked as duplicate by Peter Woolfitt, user147263, Bungo, coffeemath, peterwhyDec 24 '14 at 7:09

• It is not that $\sqrt{x^2}=\pm x$; if $x\ge0$, then $\sqrt{x^2}=x$, and if $x\le0$, then $\sqrt{x^2}=−x$. – peterwhy Dec 24 '14 at 7:11
It should be $|x+2|$. And you can find out if $x \geq -2$, then it equals $x+2$, and equals to $-x-2$ otherwise.
The usual definition, of a square root is as follows: if $x \geq 0$, then $y =\sqrt{x}$ is the unique positive number such that $y^2 = x$. So, we have $$\sqrt{x^2} = |x|$$ And similarly, $$\sqrt{(x+2)^2} = |x+2|$$ Along these lines, the solution for $y$ to $y^2 = x^2$ is $y = \pm \sqrt{x^2} = \pm x$, and $\sqrt{x^2}$ is by definition the positive number among these two solutions.
A common misconception in algebra is that $\displaystyle \sqrt{x^2}=\pm x$. However, square roots are actually ALWAYS positive or 0. The $\pm$ comes from solving the equation $x^2=n\implies x=\pm\sqrt n\implies|x|=\sqrt{n}$. Therefore your equation actually states that $|x+2|=x+2,$ only valid for $x\ge-2$