<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] 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 $x+2\ge0$.)
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