$$\sqrt{x+1}-\sqrt{9-x}=\sqrt{2x-12}$$
I know the equation is pretty simple, but I'm interested in other solutions!
You can't square it! I want to find a good solution without squaring and without derivative
My attempt:
$$\begin{cases} \sqrt{x+1}=a, & a\geq 0\\ \sqrt{9-x}=b, & b\geq 0 \end{cases}\Leftrightarrow \begin{cases} x+1=a^2\\ 9-x=b^2 \end{cases}\Rightarrow a^2-b^2-4=2x-12$$ $$a-b=\sqrt{a^2-b^2-4}\Leftrightarrow \frac{(a-b)(a+b)}{(a+b)}=\sqrt{a^2-b^2-4}\Leftrightarrow $$ $$\Leftrightarrow \frac{1}{a+b}=\frac{\sqrt{a^2-b^2-4}}{a^2-b^2}\Leftrightarrow a+b=\frac{a^2-b^2}{\sqrt{a^2-b^2-4}}$$ $$\begin{cases} a-b=\sqrt{a^2-b^2-4}\\ a+b=\cfrac{a^2-b^2}{\sqrt{a^2-b^2-4}} \end{cases}\Rightarrow 2a=\frac{a^2-b^2}{\sqrt{a^2-b^2-4}}+\sqrt{a^2-b^2-4}\Leftrightarrow $$ $$\Leftrightarrow 2a=\frac{a^2-b^2+a^2-b^2-4}{\sqrt{a^2-b^2-4}}\Leftrightarrow 2\sqrt{x+1}=\frac{4x-20}{\sqrt{2x-12}}\Leftrightarrow 2\sqrt{x+1}\sqrt{2x-12}=4x-20\Leftrightarrow$$ $$\Leftrightarrow 4(x+1)(2x-12)=16x^2-160x+400\Rightarrow x=\left \{ 7,8 \right \}$$
Are there other ways that would be easier?
We have that
$$\sqrt{x+1}-\sqrt{9-x}=\sqrt{2x-12}$$
for real solutions requires $(x+1)\ge 0$ , $(9-x)\ge 0$, $2x-12 \ge 0$ that is $6\le x \le 9$, which leads, by inspection, to the integer solutions $x=7$ and $x=8$.
Now we need to prove that there are not other solutions.
At this aim, let $x+1=y^2$ then we obtain the equivalent
$$y=\sqrt{10-y^2}+\sqrt{2y^2-14}$$
and since RHS is the sum of two concave functions it is also concave and we can't have more than two solutions.