<s>[INST] 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 $\sqrt{x^2} = \pm x$.
 [/INST] I wonder if the problem could be between line $2$ and line $5$. In line $2$, you square while in line $5$ you take the square root of the square. This could introduce some confusion. 
Starting from $$\sqrt{4x-1} = \sqrt{x+2}-3$$ just square to get $$4x-1=(\sqrt{x+2}-3)^2=x+2-6\sqrt{x+2}+9=x+11-6\sqrt{x+2}$$ So, now $$3x-12=-6\sqrt{x+2}$$ Now, square again.
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