Why doesn't squaring the radicand of a square root introduce a plus-minus sign here? The question I have concerns the following problem:


$\sqrt{4x-1} = \sqrt{x+2}-3$
$(\sqrt{4x-1})^2 = (\sqrt{x+2}-3)^2$
$\sqrt{4x-1}\times\sqrt{4x-1} = (\sqrt{x+2}-3)\times(\sqrt{x+2}-3)$
$\sqrt{(4x-1)\times(4x-1)} = (\sqrt{x+2})^2-3\sqrt{x+2}-3\sqrt{x+2}+9$
$\sqrt{(4x-1)^2} = \sqrt{(x+2)\times(x+2)}-6\sqrt{x+2}+9$
$\underline{4x-1} = \sqrt{(x+2)^2}-6\sqrt{x+2}+9$
$4x-1 = \underline{x+2}-6\sqrt{x+2}+9$


How did the underlined expressions in steps 6 and 7 not include $\pm$? I thought that $\sqrt{x^2} = \pm x$.
Note that if your equation includes $\sqrt{4x-1}$, this means that $4x-1\ge0$. Similarly, you are only looking for solutions among numbers that fulfill $x+2\ge0$, since the equation contains the expression $\sqrt{x+2}$.

So you can write $\sqrt{(4x-1)^2}=|4x-1|=4x-1$. But do not forget in the end include only those solutions where $4x-1\ge0$, i.e., $x\ge\frac14$. (And the same for the condition $x+2\ge0$.)
After thinking a bit about some of what @Daniel Hast typed, I realized that the reason why, say, $\sqrt{(4x-1)^2} = 4x-1$ 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 $(\sqrt{x})^2 = x$.