The number of the real roots of equation $3^{x+1}+3^{-x+1}=10$ is 

$$\begin{gathered} 3^{x+1}+3^{1-x}=10 \\ {3}^x.3+\frac{3}{3^x}=10 \\ Put 3^x=y \\ 3y+\frac{3}{y}=10 \\ \Rightarrow 3y^2-10y+3=0 \\ \Rightarrow 3y^2-9y-y+3=0 \\ \Rightarrow y=3,\frac{1}{3} \\ \Rightarrow 3^x=3^1,3^{-1} \\ \Rightarrow x=1,-1 \end{gathered}$$