Let $$X=\{-5,-4,-3,-2,-1,0,1,2,3,4,5\}$$ and $$S=\{(a,b)\in X\times X:x^2+ax+b\text{ and }x^3+bx+a\text{ have at least a common real zero}\}$$. How many elements are there in $$S$$?

If $$r$$ is a common zero, subtracting $$r(r^2+ar+b)$$ from $$r^3+br+a$$ gives $$a(1-r^2) = 0$$, so either $$a = 0$$ or $$r = \pm 1$$. When $$a = 0$$ a common zero exists exactly for $$b \le 0$$, giving 6 pairs; $$r = 1$$ needs $$a+b = -1$$ and $$r = -1$$ needs $$b = a-1$$, giving 10 pairs each. The only pair common to more than one family is $$(0, -1)$$, so by inclusion and exclusion the count is $$6+10+10-1-1-1+1 = 24$$.