Real root dependency

The equation becomes $\begin{aligned} x^4 + ax^3 - bx^2 + ax + 1 &= \; 0 \\ \big(x^2+ \tfrac12ax + 1\big)^2 &= \; \big(2 + b + \tfrac14a^2\big)x^2 \\ x^2 + \tfrac12ax + 1 &= \; \pm\tfrac12\sqrt{8 + 4b + a^2} \\ x^2 + \tfrac12\big(a \pm \sqrt{8 + 4b + a^2}\big)x + 1 & = \; 0 \end{aligned}$ and hence we will have at east one real root provided that $\begin{aligned} \left(\tfrac12\big(a + \sqrt{8 + 4b + a^2}\big)\right)^2 - 4 & \ge \; 0 \\ a + \sqrt{8 + 4b + a^2} & \ge \; 4 \\ 8 + 4b + a^2 & \ge \; (4-a)^2 \\ b & \ge \; 2(1-a) \end{aligned}$ Thus the region $S$ is $\big\{ (a,b)\,\big| \, \tfrac12 < a < 1\,,\,2(1-a) \le b < 1\big\}$ which is a triangle with area $\boxed{\tfrac14}$