Q:

How to show an irrational equation have no solution?

There are different cases, one of them is when I have a negative $\sqrt{x}$ so I know there is no solution because $\sqrt{x}$ can't be negative.
However there is another case where I just can't isolate $x$, $x$ is always depending on $\sqrt{x}$ (and $\sqrt{x}$ is always depending on $x$).
Here is an example :
$\sqrt{x-2}=3-2\sqrt{x}$
Squaring both sides :
$x-2=(3-\sqrt{x})^2=9-12\sqrt{x}+4x$
What to do know ? I can try to isolate $\sqrt{x}$ as I usually do :
$12\sqrt{x}=-x+2+9+4x=3x+11$
$\sqrt{x}=\frac{3x+11}{12}$
But $\sqrt{x}$ is depending on $x$. How can I prove that there is no solution in this case ?
I could try to square again but then $x$ would be depending on a $x^2$ equation. Where is it logic enough to affirm that there is no solution ?
I can also try to isolate $x$ directly, but it would end up that $x$ is depending on $\sqrt{x}$, wouldn't change that much.

A:

After you get $12\sqrt x=3x+11$, you can square again both sides, thereby getting the quadratic equation $144x=9x^2+66x+121$. This equation has two roots: $\frac13\left(13+4\sqrt3\right)$ and $\frac13\left(13-4\sqrt3\right)$. But the first of these solutions is too large: if $x=\frac13\left(13+4\sqrt3\right)$, then $x>4$ and therefore $3-2\sqrt x<0$. On the other hand, $\frac13\left(13-4\sqrt3\right)\in(2,3)$, and therefore it is a solution (it's about $2.02393$).

