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

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

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

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

Plotting the graph we get

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

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

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

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