The given equation is a biquadratic equation: $x^4 - 5x^2 + 4 = 0$.
This can be solved by treating it as a quadratic equation in terms of $x^2$.
Let $y = x^2$. The equation becomes:
$y^2 - 5y + 4 = 0$.
Now, we factorize this quadratic equation. We need two numbers that multiply to 4 and add up to -5. These numbers are -4 and -1.
$(y - 4)(y - 1) = 0$.
This gives two possible values for y:
$y = 4$ or $y = 1$.
Now, we substitute back $x^2$ for $y$.
Case 1: $x^2 = y = 4$.
Taking the square root of both sides, we get $x = \pm\sqrt{4}$, so $x = 2$ and $x = -2$.
Case 2: $x^2 = y = 1$.
Taking the square root of both sides, we get $x = \pm\sqrt{1}$, so $x = 1$ and $x = -1$.
The roots of the equation are 2, -2, 1, and -1.
This matches the set of roots given in option (B).