Prove roots of quadratic are rational 
Given the equation:
  $$(3m-5)x^2-3m^2x+5m^2=0$$ 
  Prove the roots are rational given $m$ is rational.

I have tried finding the discriminate of the quadratic however this has not been fruitful and ends up being very ugly. I have tried setting the discriminate to $a^2$ however I don't know if this is the right thing to do.
 A: Why, just calculate the discriminant!
$$\Delta=(-3m^2)^2-4(3m-5)(5m^2)=9m^4-60m^3+100m^2=(3m^2-10m)^2$$
and so $\sqrt\Delta=|3m^2-10m|$. The remainder of the quadratic formula consists of rational manipulations, so the roots are rational if $m$ is rational.
A: Yes, you can calculate of course the discriminant and to end this problem, 
but I think it's better to use the Viet's theorem.
If $a\neq0$ and $x_1$, $x_2$ are roots of the equation 
$$ax^2+bx+c=0$$
then $$x_1+x_2=-\frac{b}{a}$$ and
$$x_1x_2=\frac{c}{a}.$$
Indeed, easy to see that $m$ is a rational root because $(3m-5)m^2-
3m^2\cdot m+5m^2=0.$
Thus, for $m\neq\frac{5}{3}$ we get more rational root: $\frac{5m}{3m-5}$ because $x_1x_2=\frac{5m^2}{3m-5}.$
If $m=\frac{5}{3}$ then our equation has one rational root and we are done!
A: You could always spot that $x=m$ is a solution. 
To get this you can use the rational root theorem, and note that the coefficient of $x^2$ has no factors - if the roots are rational for every $m$, therefore, there should be a factorisation $\left(\left(3m-5\right)x+a\right)\left(x+b\right)$ and $b$ must be a factor of $5m^2$ so there is not far to look.
Otherwise just use the quadratic formula to find the roots.
