# Puzzling solution for $1/x = 3 - 2\sqrt{x}$ [closed]

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?

• Be careful about extraneous solutions. Commented Feb 6, 2021 at 21:53
• Because $\sqrt{x}$ forces $x\geq 0$. So your $-4$ solution should be dropped Commented Feb 6, 2021 at 21:53
• How about complex behavior of the square root? $x\mathop \le\limits^\text{can}0, x\in\Bbb C$ Commented Sep 27, 2021 at 22:15

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!

• @RejeevDivakaran +1 to this answer. I think that it is well to add that although each step in the answer is valid, all the analysis does is provide candidate values of $y = (-1/2), y = (1)$, with no guarantee that either or both candidate values will satisfy the original equation, subject to $y = \sqrt{x}.$ Each candidate value must be manually converted to the corresponding value of $x$, and then each (in effect) candidate value of $x$ must be checked against the original equation. Commented Feb 6, 2021 at 23:15
• @RejeevDivakaran Another way of saying the same thing is that in general, in problems of this type, the implications are one-way implications, not two-way implications. The idea is that if (for generic example) a specific value of $x$ solves equation (1), then it solves equation (2), but not (necessarily) vice-versa. Commented Feb 6, 2021 at 23:20

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.

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 $$x=1$$.