Find all $\ r\in\mathbb{R}\ $ such that $\ \exists\ x_1, x_2 \in\mathbb{R}\ $ such that:
$\ (1)\qquad x_1 + x_2 = r,\ $ and
$\ (2)\qquad {x_1}^{2} + {x_2}^{2} = r.$
Here is my solution:
$$ \exists\ x_1, x_2\in\mathbb{R}\ \text{ such that } (1), (2)\ \text{ both true } $$
$$ \iff \exists\ x_1, x_2\in\mathbb{R}\ \text{ such that }\ (x_2 = r - x_1)\quad \text{ and }\quad ( {x_1}^{2} + {x_2}^{2} = r ) $$
$$ \iff \exists\ x_1, x_2\in\mathbb{R}\ \text{ such that }\ \left( {x_1}^2 + (r - x_1)^2 = r \right) \quad \text{ and }\quad (x_2 = r - x_1) $$
$$ \iff \exists\ x_1, x_2\in\mathbb{R}\ \text{ such that }\ \left( \left( x_1 - \frac{r}{2}\right)^2 = \frac{r}{2} - \frac{r^2}{4} \right) \quad \text{ and }\quad (x_2 = r - x_1) $$
$$ \overset{*}{\iff} \frac{r}{2} - \frac{r^2}{4} \geq 0 $$
$$ \iff 0\leq r \leq 2. $$
$\ * \Longrightarrow\ $ is justified using: $\ y^2 = z \implies z\geq 0.\quad \Longleftarrow\ $ is justified using: $\ v\geq 0,\ w\in\mathbb{R}\implies \exists\ x:\ (x-w)^2 = v. $
I think my argument is logically valid and sound, but just want to check this. Am I right in thinking we need both the forward implications $\ (\Longrightarrow)\ $ and the backwards implications $\ (\Longleftarrow)\ $ throughout to ensure I have answered the question?
Furthermore, are there quicker solutions that have a purely algebraic argument with a logical structure like the one above, rather than proof using diagrams, (for example, $\ (1) + (2)\ $ gives the equation of a circle, and then you could use diagram arguments, but for this question I'm interested in algebraic arguments only).
TL;DR
If I understand correctly (please ask me to delete the answer if you're not asking about this), you want an alternate way.
Some time back, I was solving a similar equation (which came from being given the distance between the centres of two externally touching circles, and the sum of their areas was given too), and sought for a generalization.
Suppose we have a system of equations $$x+y = A$$and $$x^2+y^2 = B$$then the values of $x$ and $y$ are given by $$\frac{A \pm \sqrt{2B-A^2}}{2}$$ where one variable takes the positive square root value while the other takes the negative square root value.
Substituting $A=B=r$, we get that for $x,y \in \Bbb R$, we must have: $$r \ge 0$$since it's the sum of squares. Also, the square root should return a real value, i.e. $$2B-A^2 \ge 0$$ $$2r - r^2 \ge 0$$ Assuming $r \neq 0$: $$r^2 \le 2r$$ $$r \le 2$$Thus we have $$\color{green}{0 \le r \le 2}$$ The same answer which you got.
Edit:
Here's how I derived the solutions for $x$ and $y$: $$x+y = A$$ $$x^2+y^2 + 2xy = A^2$$ $$2xy = A^2 - B$$ $$x^2 + y^2 - 2xy = A^2 - 2A^2 + 2B = 2B - A^2$$ $$x - y = \pm \sqrt{2B - A^2}$$ $$2x = A \pm \sqrt{2B - A^2}$$ $$x = {A \pm \sqrt{2B - A^2}\over 2}$$ Since the equations were symmetric, if one variable takes the positive square-root value, the other takes the negative square-root one.
About your main question : I think they're fine, but I'm not an expert.