Can we find $t\in \mathbb R$ such that the roots of $tx^2-2x+2t=0$ are rational ?
MY ATTEMPT
For $t\neq 0$ $$tx^2-2x+2t=0$$
$$x=\frac{1\pm\sqrt{1-2t^2}}{t}$$ . Suppose that $x\in \mathbb Q $, thus $x \in \mathbb R$ .
Therefore $$ \frac{1}{\sqrt{2}} \leq t \leq\frac{1}{\sqrt{2}}$$
Thus $1-2t^2$ is always less than $1$. Never equal to 1. (Since $t \neq 0$)
So I am left with the only option (from my perspective) that is $1-2t^2 = \frac{1}{c^2}$ where $c>1$
I am stuck here. I feel Like there is no such $t \in \mathbb R$
Take $x=1$ and get a value of $t$.
Or take $t=\frac{2x}{x^2+2},$ where $x$ is a rational number.