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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?

José Carlos Santos
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Rejeev Divakaran
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3 Answers3

4

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!

user0102
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  • @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. – user2661923 Feb 06 '21 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. – user2661923 Feb 06 '21 at 23:20
2

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.

B. Goddard
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1

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 $x=1$.

egreg
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