Solving a biquadratic.

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 $f(x)=2$.

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,

$$rs=4-\frac{\sqrt{42}}2.$$

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

$$2 y^2 + 16y+ 11=0$$