Why doesn't squaring the radicand of a square root introduce a plus-minus sign here?

Question

The question I have concerns the following problem:

1. $\sqrt{4x-1} = \sqrt{x+2}-3$
2. $(\sqrt{4x-1})^2 = (\sqrt{x+2}-3)^2$
3. $\sqrt{4x-1}\times\sqrt{4x-1} = (\sqrt{x+2}-3)\times(\sqrt{x+2}-3)$
4. $\sqrt{(4x-1)\times(4x-1)} = (\sqrt{x+2})^2-3\sqrt{x+2}-3\sqrt{x+2}+9$
5. $\sqrt{(4x-1)^2} = \sqrt{(x+2)\times(x+2)}-6\sqrt{x+2}+9$
6. $\underline{4x-1} = \sqrt{(x+2)^2}-6\sqrt{x+2}+9$
7. $4x-1 = \underline{x+2}-6\sqrt{x+2}+9$

How did the underlined expressions in steps 6 and 7 not include $\pm$? I thought that $\sqrt{x^2} = \pm x$.

Answer 1

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 $x$.

Answer 2

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: $\sqrt{x^2} = \lvert x \rvert$.

Answer 3

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.

Answer 4

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 $x+2\ge0$.)

Answer 5

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 $(\sqrt{x})^2 = x$.