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Suppose I have

$\sqrt{r-x}-\frac{x}{\sqrt{r-x}}-o=0$ with $r,x$ and $o$ real.

To solve I multiply by $\sqrt{r-x}$. I assume therefore $r>x$

$r-x-x-o\sqrt{r-x}=0$

$r-2x=o\sqrt{r-x}$

Squaring

$(r-2x)^2=o^2(r-x)$

This is a second order equation,whose solutions are

$x= \frac{r}{2} -\frac{o^2}{8} \pm \frac{o\sqrt{o^2+8r}}{8}$

However, for instance, for $r=1$ and $o=1/2$ I get two solutions. For both $x<1$, however only one is actually a solution.

How can I avoid to introduce a spurious solution?

Fabio Dalla Libera
  • 275

1 Answers1

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The spurious solution is introduced by the squaring. Before squaring, $r-2x$ is constrained to have the sign of $o$ (as the square root is a positive number). When squaring, you drop this condition.

Alternative resolution:

Let $z:=\sqrt{r-x}>0$.

The equation becomes

$$z-\frac{r-z^2}z-o=0$$

or

$$2z^2-oz-r=0.$$

The only positive solutions in $z$ is

$$z=\frac{o\color{green}+\sqrt{o^2+8r}}4.$$

It exist iff $o^2+8r\ge0$, and $x=r-z^2$ is implicitly $\le r$.