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?
If we let $y = \sqrt{x}$, we get
\begin{align*}
\frac{1}{y^{2}} = 3 - 2y &amp; \Longleftrightarrow 1 = 3y^{2} - 2y^{3}\\
&amp; \Longleftrightarrow 2y^{3} - 3y^{2} + 1 = 0\\\\
&amp; \Longleftrightarrow (2y^{3} - 2y^{2}) - (y^{2} - 1) = 0\\\\
&amp; \Longleftrightarrow 2y^{2}(y-1) - (y-1)(y+1) = 0\\\\
&amp; \Longleftrightarrow (2y^{2} - y - 1)(y-1) = 0\\\\
&amp; \Longleftrightarrow \left(y + \frac{1}{2}\right)(y-1)^{2} = 0\\\\
&amp; \Longleftrightarrow \left(y = -\frac{1}{2}\right)\vee(y = 1)
\end{align*}
Since $y = \sqrt{x} &gt; 0$, there is only one possible solution, which is $x = 1$.
Hopefully this helps!
But $x$ cannot be negative, given we must have $\sqrt x &gt;0$, right?
You solved the equation by squaring it, and your solutions solve the squared equation.  But when squaring both sides of an equation, please realize that can introduce an extraneous solution.
Hence you need to check which solution(s) is/are valid: only the solutions for which the original question is defined are valid.  But given the original equation, $x=-1/4$ makes the equation undefined.