Solving a biquadratic. 
If $x$ is a real number and satisfies $$x+ \sqrt[4] {5-x^4}=2$$  then find the value of $$x\sqrt[4] {5-x^4}$$

My try :
The question is significantly asking for the value of $-x(x-2)$ if we get the root of the equation $$2x^4-8x^3+24x^2-32x+11=0$$
Using this I have reached till 
$$-2x(x-2)(x^2-2x+8)=11$$
$$-x(x-2)=\frac {11}{2(x^2-2x-8)}$$
but I couldn't manipulate it further. 
Also upon some second thought I want to ask whether could it be possible to form a quadratic polynomial with two roots $\alpha$ and $\beta$ such that $\alpha=x$ and $\beta = \sqrt[4] {5-x^4}$
but I still couldn't proceed further. Somebody please share some hints.

Note that
$$2x^4-8x^3+24x^2-32x+11=0 \implies 2 (x^2 - 2 x)^2 + 16 (x^2 - 2 x) + 11=0$$
now take $x^2 - 2 x=y$ and solve
$$2 y^2 + 16y+ 11=0$$
