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 $\sqrt{x^2} = \pm x$.

You have $x^2 = y^2 \implies x = \pm y$.  However, if $x = y$, then $x^2 = y^2$.
Also note that $\sqrt{x^2} = |x|$, since we adopt the convention that $\sqrt{x}$ is the nonnegative square root of $x$.
