Solving for unknown inside square root Sorry if this is a very primitive question, but I really not sure if I am right about this kind of situations. Imagine the following equation where $a$ , $b$ and $c$ are known numbers and $x$ is the unknown variable:
$$a\sqrt{bx}=c$$
Is it ok in this case to do it like
$$a^2bx=c^2$$
If not, how to solve such equation?
 A: Yes, this is fine, provided that $a$ and $c$ have the same algebraic sign. When you solve the second equation, you get $$x=\frac{c^2}{a^2b}\;.$$ Now try substituting that into the original equation:
$$a\sqrt{\frac{bc^2}{a^2b}}=a\sqrt{\frac{c^2}{a^2}}=a\left|\frac{c}a\right|\;.\tag{1}$$
If $a$ and $c$ have the same algebraic sign, $\left|\dfrac{c}a\right|=\dfrac{c}a$, and $(1)$ can be simplified to $a\left(\dfrac{c}a\right)=c$, as desired.
If one of $a$ and $c$ is positive and the other negative, the original equation has no solution, since by convention $\sqrt{bx}$ denotes the non-negative square root of $bx$.
A: $a\sqrt{bx} = c$
$\sqrt{bx} = \frac{c}{a}$
$bx = \frac{c^2}{a^2}$
$x = \frac{c^2}{ba^2}$
This is essentially your argument. 
A: In general, you have to be careful to check each "solution" by plugging it in to the original equation: this sort of argument often introduces extraneous roots, because squaring is not a one-to-one function.
For example, try 
$$ \sqrt{x} - 1/\sqrt{x} = 2/\sqrt{3}$$
Squaring both sides and expanding gives you
$$ x - 2 + 1/x = 4/3 $$
which has solutions $x=3$ and $x=1/3$.  But only $x=3$ is a solution of the original equation: $x=1/3$ is instead a solution of $\sqrt{x} - 1/\sqrt{x} = -2/\sqrt{3}$.
