For part A, the only possible option that could represent a rational number is
[tex]b^2[/tex]Because all the other options are irrational numbers.
For part B, the other zero has to be irrational too. To prove this we use the general formula for quadratic equations:
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \end{gathered}[/tex]If x1 is an irrational number, then
[tex]\sqrt[]{b^2-4ac}\text{ }[/tex]