How do you solve $\frac{x-1}{\sqrt{x}+2}=\frac{5}{2}$? I solved it using the quadratic formula and it went like:
\begin{gather}
\frac{x-1}{\sqrt{x}+2}=\frac{5}{2} \\
2(x-1)=5(\sqrt{x}+2) \\
2x-2=5\sqrt{x}+10 \\
2x-12=5\sqrt{x} \\
4x^2+144-48x=25x \tag*{(squaring both sides)}\\
4x^2-73x+144=0
\end{gather}
Then do the usual stuff and the solutions are $16$ and $9/4$. However $9/4$ doesn't work when you apply it to the initial equation. Is that solution correct, if not, where is the mistake?
 A: If you rewrite the equation as
$$
2(x-1)=5(\sqrt{x}+2)
$$
and then as
$$
2x-12=5\sqrt{x}
$$
you can immedately notice that a necessary condition for a solution is $x\ge6$, because the right-hand side is positive.
With this side condition, you can square: $4x^2-48x+144=25x$, which indeed has the roots $16$ and $9/4$, but the latter fails to satisfy the condition.
Indeed, if you substitute it in the left-hand side you get $9/2-12=-15/2$, while the right-hand side is $15/2$.

A different strategy could be to set $t=\sqrt{x}+2$, with the condition $t\ge2$, because $\sqrt{x}\ge0$. Then we have $x=t^2-4t+4$ and the equation becomes
$$
t^2-4t+3=\frac{5}{2}t
$$
or
$$
2t^2-13t+6=0
$$
that has roots $6$ and $1/2$. Only the first one is good, whence $\sqrt{x}=4$ and $x=16$.
A: If you do your derivation backwards (reading bottom to top) and take the square root of $25x$,
the previous line would be $\pm5\sqrt x$.
$\dfrac94$ is a solution of $2x-12=-5\sqrt x$.  
That's why you always have to check that the solutions you get are valid.
