Are these all real?

$\sqrt{x}=x^{2}$ $x=x^{4}$ $x^{4}-x=0$ $x(x^{3}-1)=0$ $x=0\text{ or }x^{3}-1=0$

The first part has one solution, as stated. The second part has three solutions, all of which have a magnitude (or absolute value) of $1$ . It is not necessary to calculate the actual solutions, as long as you can visualize the magnitude. Thus, we have one solution with a magnitude of $0$ and three solutions with a magnitude of $1$ . Thus, the sum of the magnitude of the solutions is $1 \times 0 + 3 \times 1 = \boxed{3}$ .

It can be factored as

$x = x^4$

$x(x^3 - 1) = 0$

Now, three roots are $0, 1, \omega, \omega^2$ where $\omega = \text{ imaginary cube root of unity}$ .

$\text{Now } |\omega| = |\omega^2| = 1$

$\text{Therefore }\color{#3D99F6}{\boxed{0 + 1 + |\omega| + |\omega^2| = 3}}$