[tex](a-2)x^4-2(a+1)x^3-ax^2+2(a+1)x+(a-2) = 0\Big|:x^2 \\ \\ (a-2)x^2-2(a+1)x-a+\dfrac{2(a+1)}{x}+\dfrac{a-2}{x^2} = 0 \\ \\ (a-2)\Big(x^2+\dfrac{1}{x^2}\Big)-2(a+1)\Big(x-\dfrac{1}{x}\Big)-a = 0 \\ \\ x-\dfrac{1}{x} = t,\quad t\in \mathbb{R}\\ \\ x^2+\dfrac{1}{x^2} = t^2+2 \\ \\ (a-2)(t^2+2)-2(a+1)t-a = 0 \\ \\ (a-2)t^2-2(a+1)t+a-4 \\ \\ \Delta \geq 0 \Rightarrow 4(a+1)^2 - 4(a-2)(a-4) \geq 0 \Rightarrow[/tex]
[tex]\Rightarrow 4a^2+8a+4-4(a^2-6a+8) \geq 0 \Rightarrow\\ \\ \Rightarrow 32a-28 \geq 0 \Rightarrow a \geq \dfrac{28}{32} \Rightarrow \boxed{a \geq \dfrac{7}{8}}[/tex]
=> a) corect.
