The question is asked so that I have multiple choices and need to prove the thing both ways (it's an equivalence).
The problem is, whichever thing I compare to the minimal values of the roots I try to use, whether it be Viete's formula, the Discriminant, or even go the other way and try to solve the equation for $a$, I get one single constraint to $a$, and that is $a>1$, which isn't of any help as all the possible correct answers have some fraction of $9$ or $5$ in them.
What limiting factor am I not taking into consideration? A hint would be nice.
The equation can be written as $$(x-3a)^2 + 2-2a = 0 \implies x = 3a \pm \sqrt{2a-2}$$ This forces $a \geq 1$. Hence, the smaller root is $3a - \sqrt{2a-2}$, which needs to exceed $3$, i.e., we need \begin{align} 3a-\sqrt{2a-2} > 3 & \implies 3(a-1) > \sqrt2 \sqrt{a-1} \implies 9(a-1)^2 > 2(a-1)\\ & \implies (a-1) > \dfrac29 \implies a > \dfrac{11}9 \end{align}