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

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.

• What is "integral root"? – StAKmod Apr 17 at 17:29
• @StAKmod A root which is an integer. – Arthur Apr 17 at 17:31
• Do you mean exactly one integral root, or at least one integral root? – Arthur Apr 17 at 17:31
• The language of the question does not confirm only one integral root. I think we should consider both cases : one integer or both integers. – Md Masood Apr 17 at 17:33
• @Arthur Ah,I thought that is another "integral" – StAKmod Apr 17 at 17:33

## 1 Answer

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 $$\beta$$