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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.

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

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