# A possible solution to $\sqrt {5-x}=5-x^2$ (without taking square from both sides)

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

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 $$\sqrt {5-x}=5-x^2 \\ 5-x= \left(5-x^2 \right)^2 \\ 5-x=25-10x^2+x^4 \\ \cdots \cdots \cdots$$

• I can’t answer your question, but redpenblackpen posted a video on this exact problem you might want to check out. Here’s the link - youtu.be/BO1T7ebJlO8 – Giraffes4thewin Apr 6 at 20:42
• *blackpenredpen, apologies – Giraffes4thewin Apr 6 at 20:50
• I'm not sure about rigor of that solution @Giraffes4thewin – Aqua Apr 6 at 21:04
• For sure, but if someone can follow it perhaps they can make it rigorous some how! – Giraffes4thewin Apr 6 at 21:13

Hint: Write $$t=\sqrt{5-x}\geq 0$$ so $$x=5-t^2$$ and now $$t=5-(5-t^2)^2$$ so you have to solve $$f(f(t))=t\;\;\;(*)$$

where $$f(t) = 5-t^2$$. Clearly the fixed points of $$f$$ satisfies the equation $$(*)$$, so solution to $$t^2+t-5=0$$ are two solution to $$(*)$$: $$t_{1,2} = {-1 \pm \sqrt{21}\over 2 }$$ Since $$t\ge 0$$ only $${-1 + \sqrt{21}\over 2 }$$ is valid. Also solution to $$f(t)=1-t$$ are also solutions to $$(*)$$ so $$t^2-t-4=0$$ and so $$t_{3,4}= {1 \pm \sqrt{17}\over 2 }$$

Clearly, only $${1+ \sqrt{17}\over 2 }$$ is valid. Now you can calculate both $$x$$...

• Isn't going from $t=\sqrt{5-x}$ to $x=5-t^2$ the moral equivalent of squaring both sides? – Mark H Apr 11 at 20:17

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 $$\{(1-\sqrt{17})/2,(-1+\sqrt{21})/2\}$$.

• Your solution is very beautiful and ingenious. – HiterDean Apr 7 at 5:38

Another way.

We have $$\sqrt{5-x}=5-x+x-x^2$$ Or $$5-x-\sqrt{5-x}-x(x-1)=0$$ or $$(\sqrt{5-x}-x)(\sqrt{5-x}+x-1)=0$$ and the rest is smooth.

Let $$y=\sqrt{5-x}\tag{1}$$ $$\Rightarrow y^{2}=5-x$$ $$\Rightarrow x=5-y^{2}$$ $$\Rightarrow \quad 5-x^{2}=5-\left(5-y^{2}\right)^{2} \tag{2}$$

From (1) and (2) we get: $$y=5-\left(5-y^{2}\right)^{2}$$

On solving it in photomath I get:

$$\mathrm{y}_{1}=\frac{-1-\sqrt{21}}{2}, \mathrm{y}_{2}=\frac{1-\sqrt{17}}{2}, \mathrm{y}_{3}=\frac{-1+\sqrt{21}}{2}, \mathrm{y}_{4}=\frac{1+\sqrt{17}}{2}$$

Now put these in (1) and get the solution.