<s>[INST] 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?
 [/INST] 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!
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