Sum of roots

The condition is $\frac{-35}{2}\leq x \leq\frac{47}{2}$ Let $\sqrt[4]{47-2x}=a$ $\sqrt[4]{35+2x}=b$ Then we get this system of equations $\begin{cases}a^4+b^4=82 \\ a+b=4\end{cases}$ $\Leftrightarrow\begin{cases}a^4+(4-a)^4=82 \\ b=4-a\end{cases}$ $\Leftrightarrow\begin{cases}a^4-1+(4-a)^4-81=0 \\ b=4-a\end{cases}$ $\Leftrightarrow\begin{cases}(a^2-1)(a^2+1)+[(4-a)^2-9][(4-a)^2+9]=0 \\ b=4-a\end{cases}$ $\Leftrightarrow\begin{cases}(a-1)(a+1)(a^2+1)-(a-1)(7-a)[(4-a)^2+9]=0 \\ b=4-a\end{cases}$ $\Leftrightarrow\begin{cases}(a-1)[a^3+a^2+a+1+(a-7)(a^2 -8a+25)]=0\\b=4-a\end{cases}$ $\Leftrightarrow\begin{cases}(a-1)(2a^3-14a^2+82a-174)=0\\b=4-a\end{cases}$ $\Leftrightarrow\begin{cases}(a-1)(a-3)(2a^2-8a+58)=0\\b=4-a\end{cases}$ We get $(a;b)={(1;3);(3;1)}$ $(a;b)=(1;3)\Leftrightarrow x=23$ $(a;b)=(3;1)\Leftrightarrow x=-17$ $\Rightarrow 23-17=6$ Tell me if there are any other solutions