Quadratics Not Intersecting

Setting these two parabolas equal to other yields the quadratic $2x^2 - 4ax + (4-b)=0$ . We require the discriminant:

$(-4a)^2 - 4(2)(4-b) = 16a^2 -32 + 8b < 0 \Rightarrow 2a^2 - 4 + b < 0$

in order for these parabolas to not intersect. If $a,b \in \mathbb{N_{0}}$ , then we have:

$a = 0 \Rightarrow b < 4$ , or $(a,b) = (0,0); (0,1); (0,2); (0,3)$

$a = 1 \Rightarrow b < 2$ , or $(a,b) = (1,0); (1,1)$

$a \ge 2 \Rightarrow b < 0 \Rightarrow$ contradiction.

Thus, the are $\boxed{6}$ possible ordered-pairs for non-negative integers $a,b.$