Roots of $abc^2x^2 + 3a^2 c x + b^2cx-6a^2 -ab +2b^2 = 0$ are rational

Question

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!

Answer 1

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 $$ (9a^4+b^4+10a^2b^2+24a^3b-8ab^3)=(3a^2+4ab-b^2)^2. $$

Answer 2

HINT: i have got $$x_1=\frac{2a-b}{ac}$$ $$x_2=-\frac{3a+2b}{bc}$$ control you calculations please! with $$a,b,c\ne 0$$

Answer 3

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.