<s>[INST] 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.
 [/INST] 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 $t = xy$.
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