Elementary number theory and quadratic Find the value(s) of $a$ for which the equation 
$ax^2-4x+9=0$ 
has integral roots(i.e. $x$ is an integer).
My attempt: 
By hit and trial I am getting answer as $a=\frac{1}{3}$. No information about nature of $a$ is given.

I propose a graphical understanding of this question:
Consider it as looking for the intersection points of these two curves 
$$\cases{y=4x-9\\y=ax^2}$$


*

*The first one is a fixed straight line on which we define points with integer abscissas $P_k(k,4k-9) \ \ (k \neq 0).$

*the second one is a variable parabola with an "opening" coefficient $a$ we have to constrain in order that the parabola passes through the different points $P_k$. And this is possible for every values of $k$ by equating the ordinates (for all $k \neq 0$)
$$a k^2 = 4 k -9 \ \ \ \Longleftrightarrow \ \ \ a=\dfrac{4k-9}{k^2}$$
