An equation

$\text{Let }a=x^2-x\\x^4-2x^3+x=a^2-x^2+x=a^2-a\\a^2-a=\sqrt{2a}\\a^4-2a^3+a^2=2a\\a^4-2a^3+a^2-2a=0\\a(a-2)(a^2+1)=0\\a=0,2\\ x^2-x-2=0\quad\text{or}\quad x^2-2x=0\\x=0,-1,1,2\\\text{Therefore, the answer is 2}$

$\begin{aligned} x^4-2x^3+x & = \sqrt{2(x^2-x)} \\ x^4-x^3-x^3+x^2-x^2+x & = \sqrt{2(x^2-x)} \\ x^2(x^2-x)-x(x^2-x)-(x^2-x) & = \sqrt{2(x^2-x)} \\ (\color{#3D99F6}{x^2-x})^2-(\color{#3D99F6}{x^2-x}) & = \sqrt{2(\color{#3D99F6}{x^2-x})} \quad \quad \small \color{#3D99F6}{\text{Let }y^2 = x^2-x} \\ y^4 -y^2 & = \sqrt{2}y \\ y^4 -y^2 - \sqrt{2}y & = 0 \\ y(\color{#3D99F6}{y^3-y-\sqrt{2}}) & = 0 \quad \quad \small \color{#3D99F6}{\text{We note that }y = \sqrt{2} \text{ is a root.}} \\ y(y-\sqrt{2})(\color{#3D99F6}{y^2+\sqrt{2}y+1}) & = 0 \quad \quad \small \color{#3D99F6}{\text{We note that }y^2+\sqrt{2}y+1 = 0 \text{ has no real root.}} \end{aligned}$

$\implies \begin{cases} y = 0 & \implies \sqrt{x^2-x} = 0 & \implies \begin{cases} 0 \\ 1 \end{cases} \\ y = \sqrt{2} & \implies x^2-x = 2 & \implies \begin{cases} -1 \\ 2 \end{cases} \end{cases}$

Therefore, the sum of roots is $0+1-1+2 = \boxed{2}$ .

Vietnam problem ? :-)

$x^{4}-2x^{3}+x=\sqrt{2(x^{2}-x)}$

On squaring:

$x^{8}+4x^{6}+x^{2}-4x^{7}-4x^{4}+2x^{5}=2x^{2}-2x$

$x^{8}-4x^{7}+4x^{6}+2x^{5}-4x^{4}+x^{2}-2x^{2}+2x=0$

$x^{8}-4x^{7}+4x^{6}+2x^{5}-4x^{4}-x^{2}+2x=0$

On factorizing:

$x(x+1)(x-1)(x-2)(x^{4}-2x^{3}+x^{2}+1)=0$

Solutions:

$x_{1} = 0$

$x_{2} = -1$

$x_{3} = 1$

$x_{4} = 2$

$x_{5} = -0.300243 - 0.624811i$

Sum of real solutions $= 0-1+1+2=\boxed{2}$

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4 roots are 0,1,2,-1

Did the same thing.

Hi brooh... I just use vieta theorems... is can?