TITLE: Solving a biquadratic.
QUESTION [11 upvotes]: 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.

REPLY [8 votes]: You are asked to find the valueof $rs$, given that $$\begin{cases}r+s=2,\\r^4+s^4=5.\end{cases}$$
Then
$$(r+s)^4-(r^4+s^4)=4r^3s+6r^2s^2+4rs^3=2rs(2r^2+3rs+2s^2)=2rs(2(r+s)^2-rs)$$
or, numerically,
$$\color{green}{11=2rs(8-rs)}.$$ 
You solve for $rs$, giving
$$rs=4\pm\frac{\sqrt{42}}2.$$
Anyway, from the second constraint we can draw that
$$r^4s^4\le\left(\frac52\right)^2$$ and this rules out the $+$ solution (which corresponds to complex $x$). Finally,
$$rs=4-\frac{\sqrt{42}}2.$$