Help with polynomial problem, please I am stuck on the following:

If the equation
  $$x^4-ax^3+2x^2-bx+1=0$$
  has a real root, then show that
  $$a^2 + b^2 \geq 8\,.$$

Can I have a hint to go ?
EDIT
My solution after considerable effort:
Since $\,x=0\,$ is not a root of the polynomial we can divide it by $\,x^2$. This yields
$$x^2-ax+2-\frac bx +\frac 1{x^2}\:=\:0 \\[3ex]
\implies\;\left(x-\frac{a}{2}\right)^2 + \left(\frac{1}{x} - \frac{b}{2}\right)^2\:=\:\frac{a^2}{4}+\frac{b^2}{4}-2\:=\:\frac {a^2+b^2-8}{4} \;\geqslant\;0$$
So
$$a^2+b^2-8\:\geqslant\:0\;\;\iff\;\; a^2+b^2\:\geqslant\:8$$

since $x=0$ isn't  a root of the polynomial  ; we can divide it by $x^2 $  yields 
$$x^2-ax+2-b/x +1/x^2=0$$
$$\left(x-\frac{a}{2}\right)^2 + \left(\frac{1}{x} - \frac{b}{2}\right)^2= \frac{a^2}{4}+\frac{b^2}{4}-2 = \frac {a^2+b^2-8}{4} \;\;\gt\;\;0$$
So
$$\frac {a^2+b^2-8}{4} \ge 0 \;\;\implies\;\; a^2+b^2-8 \ge 0 \;\;\iff\;\; a^2+b^2 \ge 8$$
