Possible values of $b$ for $x^2 + bx + 6 = 0,$ for integer solutions

Question

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?

Answer 1

$pq=6$

We can have $(p,q)$ as $(2,3),(1,6),(-2,-3),(-1,-6)$

So, $b$ can take $4$ values according to the sum of these pairs as $\pm5, \pm7$.

Answer 2

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

Answer 3

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 $24\begin{cases}24 &=1\cdot 24\\ 24 &=2\cdot 12\\ 24 &=3\cdot 8\\ 24 &=4\cdot 6\\ &\vdots\end{cases}$

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.

Answer 4

$$x^2 + bx + 6 = 0\implies b=-\bigg(\frac{x^2+6}{x}\bigg) = \bigg(-x - \frac{6}{x}\bigg)\\ \implies x\in\{\pm1, \pm2, \pm3 \pm6 \} \quad\land\quad b\in\{\pm5, \pm7\}$$

$$x=\frac{-b\pm \sqrt{b^2-24}}{2}\implies \sqrt{b^2-24}\in\mathbb{N} \implies b\in \{ \cdots, \pm5, \pm7\, \cdots \}$$

Testing these results (and far beyond) confirms only one solution: $\space (b,x)=(-7,1)$