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.
 A: 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,
$$rs=4-\frac{\sqrt{42}}2.$$
A: 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
$$2 y^2 + 16y+ 11=0$$
A: $2x^4-8x^3+24^2-32x+11=0$
Let $u = x -1 $
$11-32(u+1)+24(u+1)^2-8(u+1)^3+2(u+1)^4=0$ 
$2u^4+12u^2-3=0$
Let $v = u^2$
$2v^2+12v-3=0$
Can you continue from here?
A: 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 $f(x)=2$.
A: hint: You have: $x+y = 2, x^4+y^4 = 5\implies 4 = (x+y)^2 = x^2+y^2 + 2xy \implies 4t^2 =4(xy)^2 = (x^2+y^2-4)^2 = x^4+y^4+16+2(xy)^2 - 8(x^2+y^2) = 5+16+2t^2 - 8(4-2t) $. Can you take it from here ? You got a quadratic equation in $t = xy$.
