Find all the possible integer values of $a$ such that the equation about $x$: $(a+1)x^2-(a^2+1)x+2a^3-6=0$ has all integer roots

Question

Find all the possible integer values of $a$ such that the equation about $x$: $(a+1)x^2-(a^2+1)x+2a^3-6=0$ has all integer roots.

I've been given this as a homework problem and haven't been able to solve it. In class, we solved something very similar to this problem by factoring the expression, but I'm fairly sure this expression can't be factored. At least, I haven't been able to factor it.

Since the equation has all integer roots, the discriminant must be a perfect square, so I've tried taking the discriminant of the equation in x and setting it to a perfect square, but I wasn't able to factor the discriminant to use a difference of squares. Could someone help me out?

Answer 1

for some quadratic: $$\alpha x^2+\beta x+\gamma=0$$ $$x=\frac{-\beta\pm\sqrt{\beta^2-4\alpha\gamma}}{2\alpha}$$ In your case: $\alpha=(a+1),\beta=-(a^2+1),\gamma=2a^3-6$ and so: $$x=\frac{a^2+1\pm\sqrt{(a^2+1)^2-4(a+1)(2a^3-6)}}{2}$$ You could try and solve this, but from this equation notice a few things:

• if we can make the squareroot=0, $(a^2+1)$ must be even, which means $a$ is odd.
• if the square root is non-zero, then the inside must be a square number as we need a rational output.