[Math] equation $x^2+ax+6a = 0$ has integer roots, Then integer values of $a$ is

algebra-precalculus

If the equation $x^2+ax+6a = 0$ has integer roots, Then integer values of $a$ is

$\bf{My\; Try}::$ Let $\bf{\alpha,\beta}$ be two roots of given equation $x^2+ax+6a = 0$

So $\bf{\alpha+\beta = -a}$ and $\bf{\alpha \cdot \beta = 6a}$ and $\bf{\alpha,\beta \in \mathbb{Z}}$

So $\bf{\alpha \cdot \beta =-6\alpha -6\beta \Rightarrow 6\alpha+\alpha \cdot \beta +6\beta +36 = 36}$

So $\bf{(\alpha+6)\cdot (\beta+6) = 36 = 6\times 6 = 9\times 4 = 18\times 2 = 36\times 1}$ and many more

But I did not understand how can i calculate for all ordered pairs.

Is there is any other method to solve it

Help Required

Thanks

Best Answer

You've done nothing wrong so far. And there's not "and many more", for we may assume wlog. that $\alpha\ge \beta$ and then your enumeration is complete at least for the cases with $\alpha+6,\beta+6\ge 0$. The same factorings with negative numbers are of course possible as well. All in all you get that $$\begin{align}a=-\alpha-\beta&\in\{0+0,-3+2,-12+4,-30+5, 15+10, 24+8, 42+7\}\\&=\{0,-1,-8,-25,25,32,49\}\end{align}$$ and in fact all these are correct solutions.

EDIT: Oops, your "and many more" indeed was justified - slightly - as you had left out the factorization $12\times 3$. This adds $-6+3=-3$ and $18+9=27$ to the list of solutions.

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