The number of distinct real roots of $x^4-4x+1=0$ is:
 

$\begin{array}{l}Let f\left(x\right)=x^4-4x+1 \\ {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)=4{\mathrm{x}}^3-4 \\ \\ {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)=0\Rightarrow \mathrm{x}=1 \\ \\ \mathrm{x}=1 \mathrm{is} \mathrm{point} \mathrm{of} \mathrm{minima}.\\ \\ \mathrm{f}\left(0\right)=1,\mathrm{f}\left(1\right)=-2 ,f\left(2\right)=9\\ f\left(0\right)f\left(1\right)<0,f\left(1\right)f\left(2\right)<0\\ \mathrm{f} \mathrm{has} \mathrm{one} \mathrm{root} \mathrm{in}\left( 0,1\right) \mathrm{and} \mathrm{one} \mathrm{root} \mathrm{in} \left(1,2\right)\end{array}$

 Hence f(x)=0 has 2 distinct  real solutions.