For what integer values of a, $x^2+ax+6=0$ has two integer roots?

Question

For what integer values of a, $x^2+ax+6=0$ has two integer roots? Any ideas of hints about how to begin? Ps.I am new at this, sorry for any mistakes.

Answer 1

Let $p$ and $q$ be our roots.

Thus, $pq=6$ and we have $$(p,q)\in\{(1,6),(6,1),(-1,-6),(-6,-1),(2,3), (-2,-3),(-3,-2),(3,2)\},$$ which gives the answer: $\{\pm7,\pm5\}$.

Answer 2

HINT: Use the discriminant of the quadratic. Recall that in a quadratic $y=ax^2+bx+c$, it has no real solutions if $b^2-4ac$ is negative, $1$ if it is $0$, and $2$ if it is positive. So with your quadratic $y=x^2+ax+6$, we have the discriminant $a^2-24$. Now you can determine when it has two roots, and you need only use the quadratic formula to determine when they are integers.

Just comment if you need another hint!

Answer 3

Product of roots is 6, sum is -a. Try for possible combinations of integers and you'll get the possible values of a.

Answer 4

Suppose that $m\le n$ are the two integer roots. We have $$ (x-m)(x-n)=x^2-(m+n)x+mn=x^2+ax+6 $$ so we must have $$ mn=6 \quad \Rightarrow \quad (m,n) \in \{(2,3),(1,6)(-3,-2),(-6,-1) \} $$ and you can find the possible values for $a=-(m+n)$