Solving equation with absolute value. Two solutions where only one is right. I have solved this equation: 
$$
\sqrt{3x}-\sqrt{4x-5}=0
$$
And I got that the solutions are:
$$
x=\frac{5}{7}, x=5
$$
My question is, did I do it right, because 
$$
x=\frac{5}{7}
$$ 
dont fit in the equation. I mean, if I put 7/5 in equation, I will not get the correct result.
This is how I solved the equation:
$$
\sqrt{3x}-\sqrt{4x-5}=0\\
\sqrt{3x}=\sqrt{4x-5}\\
(\sqrt{3x})²=4x-5\\
|3x|-4x+5=0\\
for\ x<0\\
-3x-4x=-5\\
-7x=-5\\
7x=5\\
x=\frac{5}{7}\\
for\ x>0\\
3x-4x=-5\\
-x=-5\\
x=5\\
x=5\\
$$
Thank you very much!!!
 A: As long as we are working with real numbers, the expression $\sqrt a$ is by definition the unique nonnegative number $b$ with the property that $b^2=a$, provided it exists (that is, we need $a\ge 0$ in the first place).
Unfortunately you did not join your lines of derivation with either $\Rightarrow$ or $\Leftrightarrow$ to tell the reader if you think that what you did was an equivalence transformation or just a conclusion. However, at one step you went from
$$\tag1 \sqrt{3x} =\sqrt{4x-5}$$ 
to
$$\tag2 |3x| = 4x-5.$$
These lines, as they stand, can unfortunately only connected by "$\Rightarrow$", not by "$\Leftrightarrow$".
According to what I wrote in the first paragraph, you could have written simply
$$\tag{2a}3x=4x-5$$
instead of $(2)$ and it would still be a valid conclusion (but not an equivalence), saving you the trouble of working with absolute values and cases.
If the sqaure roots of two numbers are equal, then these numbres themselves are also equal!
In fact we could turn this step of the derivation into an equivalence by adding the conditions that the oriignal radicands are nonnegative:
$$\tag{2b}3x=4x-5\text{ and }3x\ge 0\text{ and }4x-5\ge0$$
or slicghtly simplified
$$\tag{2c}3x=4x-5\text{ and }\frac54\le x.$$
Then we would indeed have an equivalence $(1)\Leftrightarrow(2c)$.
The introduction of absolute value would be justified if root and squaring occured in different order:
$$ \sqrt{a^2}=|a|\qquad\text{for all }a\in\mathbb R$$
$$ (\sqrt a)^2=a\qquad\text{if }a\ge 0 \text{ (and the left side is undefined if }a<0\text{)}.$$
Of course, it is then still true that $(\sqrt a)^2=|a|$, but this makes things unnecessarily complicated if we have to consider only the case $a\ge0$ anyway.

This is what one could have done:
$$\begin{eqnarray}\sqrt{3x}-\sqrt{4x-5}&=&0\\
\iff\sqrt{3x}&=&\sqrt{4x-5}\\
\implies 3x&=&4x-5\\
\iff 0&=&x-5\\
\iff 5&=&x\end{eqnarray} $$
So we find the only possible solution is $x=5$. Since the derivation involved a mere implication in one step, we need to check: $\sqrt{3\cdot 5}-\sqrt{4\cdot 5-5}=\sqrt{15}-\sqrt{15}=0$, OK.
Alternatively, one could avoid the checking phase by enforcing equivalence in all steps:
$$\begin{eqnarray}\sqrt{3x}-\sqrt{4x-5}&=&0\\
\iff\sqrt{3x}&=&\sqrt{4x-5}\\
\iff 3x&=&4x-5\land 3x\ge 0\land 4x-5\ge0 \\
\iff 0&=&x-5\land \frac54\le x\\
\iff 5&=&x\land \frac54\le x\\
\iff 5&=&x.\end{eqnarray} $$
Then again, it is a good idea anyway to crosscheck all solutions one finds ;)
