Rational Roots of a Quadratic Equation 
If $a,b,c$ are non zero, unequal rational numbers then prove that the roots of the equation
  $$(abc^2)x^2+ 3a^2cx+b^2cx-6a^2-ab+2b^2=0 $$
  are rational.

Use theory of equations and basic mathematics please. I am in high school only!
 A: Here is a method based on how I was first taught to solve quadratics.
Note that $2b^2-ab-6a^2=(2b+3a)(b-2a)$
Then look for two expressions which have 
product $abc^2(2b+3a)(b-2a)$ [coefficient of $x^2$ times the constant term] 
and whose sum is $3a^2c+b^2c$ [coefficient of $x$]
The sum is homogeneous degree three and divisible by $c$, and there is no $ab$ term, so this needs to cancel. The only possibility is easily seen to be $ac(2b+3a) + bc (b-2a)=3a^2c+b^2c$
So the quadratic can be rewritten $$abc^2x^2+ac(2b+3a)x+bc(b-2a)x+(2b+3a)(b-2a)$$and this becomes $$acx \left(bcx+2b+3a\right)+(b-2a)(bcx+2b+3a)=(acx+b-2a)(bcx+2b+3a)$$
It is sometimes almost automatic to compute the discriminant and show that it is a rational square, or to use the quadratic formula. Here it is quite easy to check whether there is an easy spot for the sum/product procedure and fall back on the more general methods if you don't spot one. This method has easier manipulations than the others suggested.
A: The discriminant is $$(3a^2c-b^2c+4abc)^2$$ which implies the claim
A: Solving the given equation we get
$$x_1=\frac{2a-b}{ac}$$
or
$$x_2=-\frac{3a+2b}{bc}$$ if $ac\ne 0$ and $bc\ne 0$
A: it is easy to show . to prove this kind of thing the point to be remember is that the discriminant is perfect square more generally {b^2-4ac} is perfect suare 
 now the discriminant is {(3a^{2}c−a^{2}c+4abc)^{2}} 
 this sufficient to prove 
