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.

Let $p=x\sqrt[4] {5-x^4}$, then by squaring both sides of $x+ \sqrt[4] {5-x^4}=2$ we get
$$x^2+ \sqrt{5-x^4}+2p=4$$
that is
$$x^2+ \sqrt{5-x^4}=4-2p.$$
Now take the square again and remember that $4-2p\geq 0$ (the l.h.s is non-negative), i.e. $p\leq 2$.
Hence
$$x^4+ (5-x^4)+2p^2=(4-2p)^2=16-16p+4p^2.$$
that is
$$2p^2-16p+11=0\implies  p_1=4+\frac{\sqrt{42}}{2}\;\text{or}\; p_2=4-\frac{\sqrt{42}}{2}.$$
Since $p_1>2$, we have just ONE acceptable solution $p_2=4-\frac{\sqrt{42}}{2}$. 
P.S. $p=4-\frac{\sqrt{42}}{2}$ works because the problem has at least a solution: $f(x):=x+ \sqrt[4] {5-x^4}$ is continuous in $[0,1]$, $f(0)=\sqrt[4] {5}<2$ and $f(1)=1+ \sqrt{2}>2$, so there is at least a real $x$ such that $f(x)=2$.
