<s>[INST] I'm Puzzled inverting a function This is probably a very silly question, but I got puzzled inverted a function. The function to consider is the following

So, I want to solve for $u$ in terms of $x$ in the following equation
$$xu^{2}=x-u$$
I can get the right answer by the following procedure: 
$\rightarrow4x^{2}u^{2}=4x^{2}-4ux$
  (Multiplying by $4x$
 )
$\rightarrow4u^{2}(1+4x^{2})=4x^{2}-4ux+u^{2}$
  (Adding $u^{2}$
  to both sides)
$\rightarrow u^{2}(1+4x^{2})=\left(2x-u\right)^{2}$
  (factorizing the RHS)
$\rightarrow u\sqrt{1+4x^{2}}=\left(2x-u\right)$
  (Taking the positive square root)
$\rightarrow u=\frac{2x}{1+\sqrt{1+4x^{2}}}$
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The question is: Why using the quadratic formula to solve $xu^{2}=x-u$ for $u$ fails to give the correct answer ?
 [/INST] $xu^2+u-x=0$
$a=x$
$b=1$
$c=-x$
So $u=\frac{-1 \pm \sqrt{1-4(x)(-x)}}{2x}$
Multiply top and bottom by -1 giving:
$u=\frac{1 \mp \sqrt{1+4x^2}}{-2x}$
Rationalize the numerator
$u=\frac{1 \mp \sqrt{1+4x^2}}{-2x} \cdot \frac{1 \pm \sqrt{1+4x^2}}{1 \pm \sqrt{1+4x^2}}$
$u=\frac{1-(1+4x^2)}{-2x(1 \pm \sqrt{1+4x^2}}$
$u=\frac{-4x^2}{-2x(1 \pm \sqrt{1+4x^2})}$
$u=\frac{2x}{1 \pm \sqrt{1+4x^2}}$
Now you must consider domain restrictions.
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