Problem: Solve $\sqrt{5-x}=5-x^2$ without taking square from both sides.
The one who sent the problem to me claims that this is possible.
I would like to know if the method I applied below really works.
$\color{black}{\text{Method} \thinspace 1:}$
$$\begin{cases} 5-x\geq 0 \\5-x^2 \geq 0 \end {cases} \Longrightarrow -\sqrt{5}\leq x \leq \sqrt{5}$$
$-\sqrt{5}$ and $\sqrt{5}$ are not solutions. Therefore, we have: $~$ $-\sqrt{5} < x < \sqrt{5}$
Let, $5-x=u$ and $5-x^2=v$, we have :
$$u-v=x^2-x \\ v^2-v =x^2-x \\v^2-v-x^2+x=0 \\ (x-v)(x+v)-(x-v)=0 \\ (x-v)(x+v-1)=0 \\ x_1=v, ~~~ x_2=1-v$$
Then, we have
$$\begin{cases}x=5-x^2 \\ -\sqrt{5} < x < \sqrt{5} \end {cases} \Longrightarrow \begin{cases}x^2+x-5=0 \\ -\sqrt{5} < x < \sqrt{5} \end {cases} \Longrightarrow x=\dfrac{\sqrt {21}-1}{2}$$
$$\begin{cases}x=1-(5-x^2) \\ -\sqrt{5} < x < \sqrt{5} \end {cases} \Longrightarrow \begin{cases}x^2-x-4=0 \\ -\sqrt{5} < x < \sqrt{5} \end {cases} \Longrightarrow x=\dfrac{1- \sqrt {17}}{2}$$
So, we get: $$\color{red}{x= \left\{ \dfrac{\sqrt {21}-1}{2}, \dfrac{1- \sqrt {17}}{2} \right\}}$$
$\color{black}{\text{Method} \thinspace 2:}$
Actually a "copy" of Method $1$. So, this is almost the same.
$$ \underline {\color {blue} {x^2-x=5-x-\left(5-x^2 \right)}} \\ x^2-x =\left(5-x^2 \right)^2-\left(5-x^2 \right) \\ x^2-x-\left(5-x^2 \right)^2+\left(5-x^2 \right)=0 \\ \left(x-\left(5-x^2 \right) \right)\left(x+\left(5-x^2 \right) \right)+\left(5-x^2 \right)-x=0 \\ \left(x-\left(5-x^2 \right) \right)\left(x+\left(5-x^2 \right) \right)-\left(x-\left(5-x^2 \right) \right)=0 \\ \left(x-\left(5-x^2 \right) \right)\left(x+\left(5-x^2 \right)-1 \right)=0 \\\left(x^2+x-5 \right)\left(-x^2+x+4 \right)=0 \\ \left(x^2+x-5 \right)\left(x^2-x-4 \right)=0$$
Finally we have:
$$\color{blue}{\begin{cases}\left(x^2+x-5 \right)\left(x^2-x-4 \right)=0\\ -\sqrt{5} < x < \sqrt{5} \end {cases} \Longrightarrow} \color{red} {\begin{cases} x_1=\dfrac{1- \sqrt {17}}{2} \\ x_2=\dfrac{-1+\sqrt {21}}{2} \end{cases}}$$
$\color{black}{\text{Method} \thinspace 3:}$
$$\displaystyle\sqrt {5-x}=5-x^2$$
$x=5-u^2$
$$|u|=5-\left( 5-u^2\right)^2 \\ |u|-|u|^2=5-|u|^2-\left( 5-|u|^2\right)^2$$
$|u|=v$
$$v-v^2=5-v^2-\left( 5-v^2\right)^2 \\ v^2-v =\left(5-v^2 \right)^2-\left(5-v^2 \right) \\ v^2-v-\left(5-v^2 \right)^2+\left(5-v^2 \right)=0 \\ \left(v-\left(5-v^2 \right) \right)\left(v+\left(5-v^2 \right) \right)+\left(5-v^2 \right)-x=0 \\ \left(v-\left(5-v^2 \right) \right)\left(v+\left(5-v^2 \right) \right)-\left(v-\left(5-v^2 \right) \right)=0 \\ \left(v-\left(5-v^2 \right) \right)\left(v+\left(5-v^2 \right)-1 \right)=0 \\ \left(v^2+v-5 \right)\left(v^2-v-4 \right)=0$$
$$x=5-u^2=5-|u|^2=5-v^2$$
where, $-\sqrt5 <x<\sqrt5.$
Finally,
$$\color{red}{\begin{cases}\left(v^2+v-5 \right)\left(v^2-v-4 \right)=0\\ 5+\sqrt{5} > v^2 > 5-\sqrt{5} \end {cases} \Longrightarrow} \color{red}{\begin{cases} v_1=\dfrac{1+ \sqrt {17}}{2} \\ v_2=\dfrac{-1+\sqrt {21}}{2} \end{cases} \Longrightarrow} \color{blue} {\begin{cases} x_1=\dfrac{1- \sqrt {17}}{2} \\ x_2=\dfrac{-1+\sqrt {21}}{2}. \end{cases}}$$
Is there any completely different method besides these methods and what I do is true? Because, I am not sure that I fulfill the requirement of "not taking square from both sides".
But, I think what I do is different from $$\sqrt {5-x}=5-x^2 \\ 5-x= \left(5-x^2 \right)^2 \\ 5-x=25-10x^2+x^4 \\ \cdots \cdots \cdots $$
Best Answer
Begin by subtracting $x$ from both sides:
$\sqrt{5-x}-x=(5-x)-x^2$
And render the difference of squares factorization
$(\sqrt{5-x}-x)(\sqrt{5-x}+x)=(5-x)-x^2$
By comparison we must have
$(\sqrt{5-x}-x)(\sqrt{5-x}+x)=\sqrt{5-x}-x$
and we are led to two possibilities:
Possibility 1: if the common factor $\sqrt{5-x}-x$ is nonzero we must have
$\sqrt{5-x}+x=1$,
from which
$5-x^2+x=1, x^2-x-4=0, x=(1-\sqrt{17})/2$
where the sign on $\sqrt{17}$is fixed by requiring $x^2\le 5$ because $\sqrt{5-x}=5-x^2$ must be nonnegative.
Possibility 2: The common factor is zero, in which case we simply have
$\sqrt{5-x}=x=5-x^2, x^2+x-5=0,x=(-1+\sqrt{21})/2$
where again $x^2\le 5$ to make $\sqrt{5-x}=5-x^2$ nonnegative.
Thus the solution set is $\{(1-\sqrt{17})/2,(-1+\sqrt{21})/2\}$.