Possible values of $b$ for $x^2 + bx + 6 = 0,$ for integer solutions The question is essentially in the title. The actual question is here(and also my working out) if there's any confusion in the title http://m.imgur.com/a/lgNzt

My method was to find the discriminant = 0 and then use trial and error since this question was multiple choice.

Is there a way of doing this properly, as in algebraically without trial and error?
Continuing your's work:
$\sqrt{b^2-24}=s\ \text{, s is integer}\implies b^2-s^2=24\implies (b-s)(b+s)=24$

Now, factor $24\begin{cases}24 &amp;=1\cdot 24\\ 24 &amp;=2\cdot 12\\ 24 &amp;=3\cdot 8\\ 24 &amp;=4\cdot 6\\ &amp;\vdots\end{cases}$  

Of which two products adds to an odd integer will not work for e.g. $(b-s)(b+s)=24=1\cdot 24$ here two product adds to $24+1=25$, odd,  hence kick it out. Similarly $24=8\cdot 3$ will be kicked out.  Leaving rest to verify it yourself.
Let A and B be the roots. So A and B are integers.
(X-A)×(X-B)=X^2+bX+6

X^2+bX+6=X^2-(A+B)X+AB

So AB=6 . And they are integers. So possible value are

1×6 , -1×-6 , 2×3 , -2×-3.

b=-(A+B)

Putting above values , b=1+6 , -1+-6 , 2+3 ,-2,+-3
                                            7 , -7 , 5 , -5