The question:
The quadratic equation $ x^2 + Lx + M = 0$ has one root twice the other.
a) Prove that the roots are rational whenever L is rational.
I was able to find out that due to one root being twice the other that
$2L^2 = 9M$ and the discriminant = $ L^2 - 4M$
But I am unsure of how to address the question, any help is much appreciated.
Let $\alpha$ and $\beta$ be the roots of the quadratic. We know that $\alpha +\beta=-L$ and $\alpha\cdot\beta=M$. Assume $\alpha=2\cdot\beta$ we the have $L=-3\cdot\beta=-{3\over 2}\alpha$.
So one can see that the quadratic has rational roots if and only if $L$ is rational.
$L$ is exactly the negative of the sum of both roots, as can be seen by expanding
$$(x - m)(x - n) = 0$$
Let the roots be $a$ and $2a$. Then it suffices to show that $a$ is rational whenever $-3a = L$ is rational.
The quadratic can be written as with the roots $a, 2a$ $$x^2+Lx+M=(x-a)(x-2a)= x^2 - 3ax - 2a^2$$ Therefore $a=-L/3\,$ and $2a$ are rational iff $L$ is rational.
