Puzzling solution for $1/x = 3 - 2\sqrt{x}$ What is the solution for
$$\frac{1}{x} = 3 - 2\sqrt{x}$$
When I plot $1/x$ and $3-2\sqrt{x}$ separately, it meets at $x = 1$.
However when I solve the equation $1/x = 3 - 2\sqrt{x}$ algebraically, I get two solutions $-1/4$ and $1$.
What am I missing?
 A: If we let $y = \sqrt{x}$, we get
\begin{align*}
\frac{1}{y^{2}} = 3 - 2y & \Longleftrightarrow 1 = 3y^{2} - 2y^{3}\\
& \Longleftrightarrow 2y^{3} - 3y^{2} + 1 = 0\\\\
& \Longleftrightarrow (2y^{3} - 2y^{2}) - (y^{2} - 1) = 0\\\\
& \Longleftrightarrow 2y^{2}(y-1) - (y-1)(y+1) = 0\\\\
& \Longleftrightarrow (2y^{2} - y - 1)(y-1) = 0\\\\
& \Longleftrightarrow \left(y + \frac{1}{2}\right)(y-1)^{2} = 0\\\\
& \Longleftrightarrow \left(y = -\frac{1}{2}\right)\vee(y = 1)
\end{align*}
Since $y = \sqrt{x} > 0$, there is only one possible solution, which is $x = 1$.
Hopefully this helps!
A: At some point in your calculation you are squaring both sides.  This introduces extraneous solutions.  If you have
$$x=-2$$
and square both sides you get
$$x^2=4$$
which has two solutions.
A: Rewrite the equation as
$$
2\sqrt{x}=3-\frac{1}{x}
$$
This shows that there is an implied condition, besides $x>0$ necessary in order that the equation makes sense: indeed you need
$$
3-\frac{1}{x}\ge0
$$
which becomes, taking into account that $x>0$, $3x-1\ge0$, so $x\ge1/3$.
Now you can safely square and get
$$
4x=9-\frac{6}{x}+\frac{1}{x^2}
$$
and therefore
$$
4x^3-9x^2+6x-1=0
$$
This factors as $(x-1)^2(4x-1)=0$. Since $1/4<1/3$, this is a spurious solution that must be discarded. Hence the only solution is $x=1$.
