How many?

The condition of $x$ is $x\in [0;2\sqrt{2}]\cup [-\infty ; -2\sqrt{2}]$

By Cauchy-Schwarz inequality we have $\bigg(x\sqrt{8-x}+\sqrt{x(8-x^2)}\bigg)^2\leq (8-x+x)(8-x^2+x^2)=64$ $\Leftrightarrow x\sqrt{8-x}+\sqrt{x(8-x^2)}\leq 8$ So $x\sqrt{8-x}+\sqrt{x(8-x^2)}\geq 8$ only holds true when $LHS=RHS$ , which means $\sqrt{(8-x)(8-x^2)}=\sqrt{x^3}$ Solving the equation and we get $x=\frac{-1+\sqrt{33}}{2}$ as the answer, so the correct choice is $1$