Let X=\ -5,-4,-3,-2,-1,0,1,2,3,4,5\ and S=\ (a, b) ∈ X × X: x^2+a...

Let $\alpha$ be a common root, then $$\begin{aligned} & \alpha^2+a \alpha+b=0 ...(i)\\ & \alpha^3+b \alpha+a=0 ...(ii) \end{aligned}$$ (i) $\times \alpha$ - (ii) gives, $a\left(\alpha^2-1\right)=0 \Rightarrow a=0$ or $\alpha=1$ or $\alpha=-1$ Case 1: When $a=0$ then $b \leq 0$ Total 6 cases. Case II: When $\alpha=1$ then $a+b=-1$ $\Rightarrow a=-5,-4,-3,-2,-1,1,2,3,4$ (9 cases) Case III: When $\alpha=-1$ then $a-b=1$ $\Rightarrow a=5,4,3,2,1,-1,-2,-3,-4$ (9 cases) There exist total 24 ordered pairs $(a, b)$