Why do I get one extra wrong solution when solving $2-x=-\sqrt{x}$?

Solution 1:

This is because the equation $\;\sqrt x=x-2$ is not equivalent to $x=(x-2)^2$, but to $$x=(x-2)^2\quad\textbf{and}\quad x\ge 2.$$ Remember $\sqrt x$, when it is defined, denotes the non-negative square root of $x$, hence in the present case, $x-2 \ge 0$, i.e. $x$ must be at least $2$.

Solution 2:

Squaring can change the set of solutions

Consider $x=3$ and $x^2=9$

Solution 3:

Such an interesting question! let's do a 'backwards' reasoning.

It's true that $x=1$ satisfy $x^2-5x+4=0$, and then clearly $x=\left(x-2\right)^2$, since $1 = (1-2)^2=(-1)^2$.

The problem appears when you take square roots, since it is not still true that $\sqrt{(-1)^2}=-1$, in fact, $\sqrt{x^2}= |x|$ so in this case $\sqrt{(-1)^2}=|-1|=1$, which is not equal to $x-2$ when $x=1$.

Solution 4:

You correctly deduced that $$\sqrt{x}=x-2.$$

You then wrote $$x=(x-2)^2.$$

This is true, but it is not as precise as what you started with. If you were to try to derive the original equation from this statement, you could not correctly do so, because $\sqrt{(x-2)^2}$ is not necessarily $x - 2$. Actually,

$$\sqrt{(x-2)^2} = \lvert x - 2 \rvert.$$

So when you write $x=(x-2)^2$, it implies only that $\sqrt x = \lvert x - 2 \rvert$, which means

$$\sqrt x = \begin{cases} x - 2 & \text{if $x \geq 2$,} \\ 2 - x & \text{if $x < 2$.} \\ \end{cases}$$

Since $0 \leq \sqrt x$ whenever $\sqrt x$ is a real number, the original equation, $\sqrt{x}=x-2$, implies that $x \geq 2$, and the "if $x \geq 2$" case of the equation above applies. In that case the only solution is $x = 4$. But in the other case, "if $x < 2$," you end up solving for $x$ in $\sqrt x = 2 - x$. The result $x = 1$ is in fact a correct solution of that equation:

$$ \sqrt 1 = 2 - 1. $$

It is just not the equation you were supposed to solve.