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

Let $f\left(x\right)=x^4-4x+1$

$f^{\text{'}}\left(x\right)=4x^3-4$ and $f^{\text{'}}\left(x\right)=0\Rightarrow x=1$

So the extrema of the function is at $x=1$

Plotting the graph we get 

$f^{\text{'}\text{'}}\left(x\right)=12x^2 \therefore f^{\text{'}\text{'}}\left(1\right)>0$ 

And at $x=1, f\left(1\right)=-2<0$

So, graph of $f\left(x\right)$ will cut x-axis at two points,

$\therefore$ Number of real roots of $f\left(x\right)=0$ is equal to $2$