<s>[INST] $\sqrt{3-x}-\sqrt{-x^2+8x+10}=1$ Consider the equation
$$
\sqrt{3-x}-\sqrt{-x^2+8x+10}=1.
$$
I have solved it in a dumb way by solving the equation of degree four. So, the
only real solution is $x = -1$. Can you please suggest, maybe there are better or easier ways to solve it?
 [/INST] This is a simple way used when I was in high school.
Basically, all you have to do is draw out the minimal polynomial of the known solutions (which make this even useful with most algebraic solutions, not only integers).
Firstly, match each square root with its real value.
\begin{align} (\sqrt{3-x}-2)-(\sqrt{-x^2+8x+10}-1)=0 \end{align}
Then multiply each with its conjugate.
\begin{align} \frac{-1-x}{\sqrt{3-x}+2}-\frac{-x^2+8x+9}{\sqrt{-x^2+8x+10}+1}=0 
\end{align}
\begin{align} (1+x)\left (\frac{1}{\sqrt{3-x}+2}+\frac{9-x}{\sqrt{-x^2+8x+10}+1} \right )=0 
\end{align}
From the fact that $3-x>0$ and $-x^2+8x+10>0$, it can easily be checked that $9-x>0$
Therefore, the only solution would be $x=-1$.
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