real values of $x$ in $\sqrt{5-x} = 5-x^2$.

Calculate the real solutions $x\in\mathbb{R}$ to

$$ \sqrt{5-x} = 5-x^2 $$

My Attempt:

We know that $5-x\geq 0$ and thus $x\leq 5$ and

$$ \begin{align} 5-x^2&\geq 0\\ x^2-\left(\sqrt{5}\right)^2&\leq 0 \end{align} $$

which implies that $-\sqrt{5}\leq x \leq \sqrt{5}$. Now let $y=\sqrt{5-x}$. Then

$$ \tag1 y^2=5-x $$

and the equation converts into

$$ \begin{align} y &= 5-x^2\\ x^2 &= 5-y\\ y^2-x^2 &= 5-x-(5-y)\\ y^2-x^2 &= y-x\\ (y-x)(y+x)-(y-x) &= 0\\ (y-x)(y+x-1) &= 0 \end{align} $$

So either $y=x$ or $x+y=1$.

Case 1 ($y=x$):

We can plug this into $(1)$ to get

$$ \begin{align} y^2 &= 5-x\\ x^2 &= 5-x\\ x^2+x-5 &= 0\\ x &= \frac{-1\pm \sqrt{1+20}}{2} \end{align} $$

Since $-\sqrt{5}\leq x\leq \sqrt{5}$, the only solution is

$$ x = \frac{-1+\sqrt{21}}{2} $$

Case 2 ($y=1-x$):

We can plug this into $(1)$ to get

$$ \begin{align} y^2 &= 5-x\\ (1-x)^2 &= 5-x\\ 1+x^2-2x &= 5-x\\ x^2-x-4 &= 0\\ x &= \frac{1\pm\sqrt{17}}{2} \end{align} $$

Since $-\sqrt{5}\leq x\leq \sqrt{5}$, the only solution is

$$ x = \frac{1-\sqrt{17}}{2} $$

So final solution is

$$ x \in \left\{\displaystyle \frac{1-\sqrt{17}}{2}, \frac{-1+\sqrt{21}}{2} \right\} $$

Is it possible to solve this problem using geometry? For example, could we use the properties of a half-circle and a parabola?

$$ \begin{align} \sqrt{5-x}&=5-x^2\\ 5-x &= \left(5-x^2\right)^2\\ 5-x &= x^4-10x^2+25\\ x^4-10x^2+25-5+x &= 0\\ x^4-10x^2+x+20 &= 0\\ (x^2-x-4)(x^2+x-5) &= 0 \end{align} $$

$$ \begin{align} x^2-x-4=0 &\vee x^2+x-5=0\\ x=\frac{-(-1)\pm\sqrt{(-1)^2-4\cdot 1 \cdot (-4)}}{2\cdot 1} &\vee x=\frac{-1\pm\sqrt{1^2-4\cdot 1 \cdot (-5)}}{2\cdot 1}\\ x=\frac{1\pm\sqrt{17}}{2} &\vee x=\frac{-1\pm\sqrt{21}}{2} \end{align} $$

Two of the $4$ solutions are good:

$$ \begin{align} x_1 &= \frac{1-\sqrt{17}}{2}\\ x_2 &= \frac{-1+\sqrt{21}}{2} \end{align} $$

I don't see the positive thing in using geometry! This is the fastest way.

You can see the solutions as the abscissa of the intersection points of the two curves $y=5-x^2$ (a parabola) and $y=\sqrt{5-x}$ (upper half of a parabola). These twoo parabolas intersect at 4 points, but only two of them lie in the $y>0$ half-plane.