Solving $\frac{10x}{\sqrt{10x^2}} = \sqrt{10}$. How to deal with the square root? For some reason, I'm forgetting how to work with square roots ... I know, not a great look. Anyway, I'm trying to solve for $x$.
$$\frac{10x}{\sqrt{10x^2}} = \sqrt{10}$$
Since I see a square root, I divide the domain into two cases: (1) $x > 0$ and (2) $x < 0$.
(1)
$$\frac{10x}{\sqrt{10x^2}} = \sqrt{10}$$
$$\frac{10x}{\sqrt{x^2}} = 10$$
Here is where I get confused, do I consider $\sqrt{x^2} = x $ or $-x$?
(2) Similarly,
$$\frac{10(-x)}{\sqrt{10(-x)^2}} = \sqrt{10}$$
$$\frac{10(-x)}{\sqrt{(-x)^2}} = 10$$
 A: $$\frac{10x}{\sqrt{10x^2}} = \sqrt{10}$$
First of all, if $x=0$ then the left-hand side $\frac00$ is undefined; so we reject that as a solution, and assume $x\ne 0$. So $\sqrt{10x^2}$ is positive, and we can multiply both sides by it to get
$$10x=\sqrt{10}\sqrt{10x^2}=10\sqrt{x^2}$$
Dividing by $10$ gives
$$x=\sqrt{x^2}$$
And we know that $\sqrt{x^2}=|x|$, so we get
$$x=|x|$$
You can take it from here.
A: Take $x$ and square it. You will get a non-negative number, independently of whether $x$ is positive or negative.
Now, calculate the square root of the result. You will get a non-negative number. Notice that when it comes to real numbers, you are not allowed to take the square root of a negative number.
When working with non-negative real numbers, taking the square root and squaring are operations that cancel out. Thus for all $x >= 0, \space \sqrt(x^{2}) =x.$
But if $x$ is smaller than $0$, squaring it turns it into a positive number. Then one is allowed to take the square root. But the square root only gives non-negative outputs. Thus for all $x < 0, \space \sqrt(x^{2}) = -x.$  Hence
for all real $x:\sqrt(x^{2}) = |x|.$
As you can see, the process results in "forgetting" the sign of the number one is working with.
A: Your way is perfectly fine up here
$$\frac{10x}{\sqrt{x^2}} = 10 \iff \frac{x}{\sqrt{x^2}} = 1$$
with $x\neq 0$, now we can use that $\sqrt{x^2}=|x|$ indeed

*

*$x>0 \implies \sqrt{x^2}=x=|x|$

*$x<0 \implies \sqrt{x^2}=-x=|x|$
A: *

*None of the other Answers have pointed out your most remarkable
error:

$$\frac{10x}{\sqrt{10x^2}} = \sqrt{10}$$ I divide the domain into two
cases:   (1) $x > 0,\;$ and (2) $x < 0$. $$\text{Case
(2)}\quad \frac{10(-x)}{\sqrt{10(-x)^2}} = \sqrt{10}$$

This is illogical and wrong: writing $$-x$$ is not taking the case in which $x$ is negative—it is merely inverting the sign of $x.$
Think of the minus sign not as an adjective, but as an arithmetic operator; read $-x$ not as “negative $x$”, but as “minus $x$”.
In other words, $-x$ means

*

*neither that $x$ is negative ($x$ can still have positive value $7$ so that minus $x$ is negative),

*nor that $-x$ is negative ($-x$ has positive value $7$ when $x=-7$).



*While it is true that both $-2$ and $2$ are square roots of
$4$—just as $4$ has seven $7^\text{th}$ roots, and so
on—the issue is that $$\sqrt[n] x$$ (the radical sign) has
a very particular and clear meaning: for nonnegative real $x, \;\sqrt[n] x$ is defined as the principal $n^\text{th}$ root of $x,$ i.e., $$\sqrt[n] {y^n}=|y|.$$


*However, when $x$ is negative or has an imaginary component, due to
ambiguity, I avoid using the radical sign altogether: for example,
it is unclear whether $$\sqrt[3] {-1}$$ means
$$-1\quad\text{(real)}$$ or $$e^{i \frac\pi3}\quad\text{(smallest
nonnegative argument)}.$$
