Quadratic equation with one root in $[0,1]$ and other root in $[1,\infty]$

First the two roots need to exist, then $$a^2>8.$$

Then the two conditions are

$$a-\sqrt{a^2-8}\le2,\\a+\sqrt{a^2-8}\ge2,$$ or $$a-2,2-a\le\sqrt{a^2-8}.$$

This is equivalent to

$$(a-2)^2\le a^2-8,\\12\le 4a.$$

This condition is stronger than the first one.