<s>[INST] When going from $(x+2)^2=5$ to $x+2=\pm \sqrt{5}$, why isn't there also a $\pm(x+2)$? Say I am solving the following equation:
$$(x+2)^2 = 5$$
$$x + 2 = \pm \sqrt{5}$$
$$x = -2 \pm \sqrt{5}$$
However, when I took the positive and negative square root of $5$ in the second line, I did not take the positive and negative square root of $(x+2)^2$, which would be $\pm (x+2)$. Why is this?
 [/INST] When solving $(x+2)^2=5$, recall in general that for $x\in\mathbb{R}$ we have $\sqrt{x^2}=|x|$. And since  clearly $x+2\in\mathbb{R}$, we have by the last identity that by taking the square root of both sides of $(x+2)^2=5$ $$\sqrt{(x+2)^2}=|x+2|=\sqrt{5}\tag1$$
Then we have reduced the problem to solving $$|x+2|=\sqrt{5}\tag2$$
Recall once more that in general $$|x|=b>0\implies x=b\text{ or }x=-b\tag3$$
Thus putting $(1)$ and $(3)$ together, $$|x+2|=\sqrt{5}\iff x+2=\sqrt{5}\text{ or } x+2=-\sqrt{5}$$
Futhermore, $$x=\sqrt{5}-2\text{ or } x=-\sqrt{5}-2\iff \boxed{x=-2\pm\sqrt{5}}$$
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