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

By definition, if $r$ is a nonnegative real number, then $\sqrt{r}$ denotes the unique nonnegative real number whose square is $r$. Note that $\sqrt{r}$ and $-\sqrt{r}$ both have the property that their square is $r$, but there is only one nonnegative square root.
If $x$ is any real number, then $x^2$ is nonnegative, so it has a square root in the real numbers. By definition, $\sqrt{x^2}$ is the unique nonnegative real number whose square is $x^2$. If $x$ is nonnegative, then $x$ itself satisfies this property: $x = \sqrt{x^2}$. If $x$ is negative, then $x$ cannot be $\sqrt{x^2}$ because the square root must be nonnegative (by definition); however, note that $(-x)^2 = x^2$, so $-x = \sqrt{x^2}$.
Putting these together, we see that $\sqrt{x^2}$ is equal to $x$ if $x \geq 0$, and $-x$ if $x < 0$. But this is exactly the definition of $\lvert x \rvert$, so we can more concisely phrase it as: $\sqrt{x^2} = \lvert x \rvert$.
