Dividing by $ x^2$ ,we get $ x^2+\frac{1}{x^2}+a(x-\frac{1}{x})+b-2=0$ $ (x-\frac{1}{x})^2+a(x-\frac{1}{x})+b=0$ Put $ t=x-\frac{1}{x}$ $ \Rightarrow$ For every real t there will exist two real x,since Discriminant of $ x^2-tx-1=0$ is positive we get $ t^2-at+b=0$ Since $ a^2>4b$,$ t$ will be real.Also,$ t$ can assume all real values except when x=0 and by putting x=0 in the equation $ x^4 + ax^3 + (b - 2)x^2 - ax + 1 = 0$ we can see that $ 0$ is not the root . Therefore,for every real $ t$ we get two real $ x$.Since we will get 2 real $ t$ therefore we will get 4 different real $ x$.Hence the quartic has 4 distinct real roots