For how many real numbers 'b', does $x^2 + bx + 6b = 0$ have one integral root?

Question

The question states.
For how many real numbers 'b', does $ f(x) =x^2 + bx + 6b = 0$ have one integral root ?
My line of thinking :
Let $\alpha , \beta$ be the roots of of $f(x)$.
$\alpha + \beta = -b. $ $\alpha\beta = 6b. $ How to proceed ? A hint would also suffice.

Answer 1

The quadratic has exactly one root if $b^2 - 24b = 0$

And it will be an integer if $b$ is even.

That is the easy part.

Suppose one root is an integer and one is not.

$6b = \alpha\beta\\ b = \frac {\alpha\beta}{6}\\ \alpha + \beta = - \frac {\alpha\beta}{6} \\ \alpha (6+\beta) = -6\beta\\ \alpha = \frac {-6\beta}{6+\beta}$

And now we can choose any integer value for $\beta$, and find a corresponding $\alpha$

$b = \frac{-\beta^2}{6+\beta}$

And this should work for any integer $\beta$