Roots of $abc^2x^2 + 3a^2 c x + b^2cx-6a^2 -ab +2b^2 = 0$ are rational We have to show that roots of $$abc^2x^2 + (3a^2 c  + b^2c)x-6a^2 -ab +2b^2 = 0$$ are rational. 
This can be possible if the discriminant is a perfect square. SO I tried converting it into perfect square but failed:
$$\text{Discriminant}=(3a^2c+b^2c)^2-4abc^2(2b^2-6a^2-ab)\\
c^2(9a^4+b^4+10a^2b^2+24a^3b-8ab^3)$$
I cannot proceed please help! Thanks!
 A: Starting with your discriminant:
$$
c^2(9a^4+b^4+10a^2b^2+24a^3b-8ab^3)
$$
we see that we can easily ignore the $c^2$ since that is already a square.  Now, we look at the degrees of all the terms and notice that they are all degree $4$.  Therefore, the factorization (if it exists) has to be of the form 
$$
(9a^4+b^4+10a^2b^2+24a^3b-8ab^3)=(x_1a^2+x_2ab+x_3b^2)^2.
$$
Moreover, we can assume that $x_1>0$ since otherwise, we can multiply through by $-1$, which doesn't change anything since $(-1)^2=1$.
We can multiply out the RHS to get the following system of equations:
\begin{align}
9&=x_1^2&2x_1x_2&=24&x_3^2&=1\\
2x_2x_3&=-8&2x_1x_3+x_2^2&=10
\end{align}
Therefore, $x_1=3$ from the first equation.  $x_2=4$ from the second equation, $x_3=-1$ from the fourth equation.  We can check that all of the equations are satisfied with these choices, so 
$$
(9a^4+b^4+10a^2b^2+24a^3b-8ab^3)=(3a^2+4ab-b^2)^2.
$$
A: HINT: i have got $$x_1=\frac{2a-b}{ac}$$
$$x_2=-\frac{3a+2b}{bc}$$
control you calculations please!
with $$a,b,c\ne 0$$
A: The product of the roots is
$$
\frac{-6a^2 -ab +2b^2}{abc^2}
= \frac{(b - 2 a) (3 a + 2 b)}{a b c^2}
$$
which immediately suggests what the roots are.
