Quadratic weirdness

First, rearrange the quadratic: $abx^2 + (ac+b^2)x + bc=0$

Since both roots are equal, the discriminant of the above quadratic must be zero: $\begin{aligned} (ac+b^2)-4ab^2c &= 0 \\ a^2c^2 +2ab^2c + b^4 - 4ab^2c &= 0 \\ a^2c^2 -2ab^2c + b^4 &= 0 \\ (ac - b^2)^2 &= 0 \\ ac &= b^2 \end{aligned}$

Now, the solution to our quadratic is $\frac{-ac-b^2 \pm \sqrt{0}}{2ab} = \frac{-2b^2}{ab} = -\frac{2b}{a}$ . Since all answer options have integer $a$ , this will be rational if and only if $b$ is rational. Note that if $a$ and $c$ are perfect squares then $b$ is rational (say $a=m^2$ and $c=n^2$ for integers $m$ and $n$ , then $b^2=ac=m^2n^2$ and $b=mn$ ).

This rules out every answer option except $a=8$ and $b=49$ . We can verify by computation that if $b^2 = 8\times49$ then $b=14\sqrt{2}$ , which is not rational. Thus, 8 & 49 is the only listed option that is not possible.

Comparing to ax^2 + bx + c, b^2 - 4ac should be equal to 0 for equal roots. And since they are rational, 4ac should be a perfect square. Thus the only incorrect option would be 8 and 49