Is my proof for this claim about vieta roots correct? Let ${ x }_{ 1 },{ x }_{ 2 }\in\mathbb{R}$ be the roots of the quadratic equation $ax^2+bx+c=0$ with rational coefficients such that ($a,b,c\in\mathbb{Q}, a\neq 0$). Prove using the Vieta formulas that ${x}_{1}\in\mathbb{Q}\Leftrightarrow{x}_{2}\in\mathbb{Q}$.
My Proof:
$$\frac { -b+\sqrt { b^{ 2 }-4ac }  }{ 2a } +\frac { -b-\sqrt { b^{ 2 }-4ac }  }{ 2a } =\frac { -2b }{ 2a } =-\frac { b }{ a } $$
$$\Rightarrow { x }_{ 1 }+{ x }_{ 2 }=-\frac { b }{ a } $$
1) ${ x }_{ 1 }\in\mathbb{Q}\Rightarrow { x }_{ 2 }\in\mathbb{Q}$
Assume: ${ x }_{ 2 }\notin\mathbb{Q}$ (by contradiction)
Let: ${ x }_{ 1 }= \frac { m }{ n }$ and ${ x }_{ 2 }=p$ such that $m,n\in\mathbb{Z}$ such that $n\neq0$; $p\notin\mathbb{Q}$
$$\frac { m }{ n } +p=-\frac { b }{ a }$$
$$p=-\frac { b }{ a } -\frac { m }{ n } \Rightarrow p=-1(\frac { bn }{ an } +\frac { ma }{ an } )\Rightarrow p=-\frac { bn+ma }{ an } $$
This is contradictory to $p$ being irrational, so $p$ must be rational; therefore, ${x}_{2}\in\mathbb{Q}$
2) ${ x }_{ 2 }\in\mathbb{Q}\Rightarrow { x }_{ 1 }\in\mathbb{Q}$
Assume: ${ x }_{ 1 }\notin\mathbb{Q}$ (by contradiction)
Let: ${ x }_{ 2 }= \frac { k }{ l }$  such that $k,l\in\mathbb{Z}$ and $l\neq0$; ${x}_{1}=q$ such that $q\notin\mathbb{Q}$
$$q+\frac { k }{ l } =-\frac { b }{ a } $$
$$q=-\frac { b }{ a } -\frac { k }{ l } \Rightarrow q=\frac { bl+ka }{ al } $$
This is contradictory to $q$ being irrational; therefore, $q$ must be rational, so, ${x}_{1}\in\mathbb{Q}$
Is my proof any good? In not, where are its flaws and how can I improve it?
 A: Your proof is good. It proves what you needed to prove, and it does so without making any logical mistakes.

That said, the proof is overcomplicated. 
First of all, you don't need to use a proof by contradiction. 
You can simply say, in step $1$, that if $x_1=\frac mn$, then, because $x_1 + x_2 = -\frac ba$, you know that $x_2 = -\frac ba -\frac mn = -\frac{bn + ma}{an}$, and since  $\frac{bn + ma}{an}\in \mathbb Q$, you conclude that $x_2\in \mathbb Q$.
Secondly, you don't need the two steps. Basically, your first step did not prove the statement 

If the first root is rational, then the second root is rational.

It actually proved the statement

If one of the roots is rational, then the other one is too.


Here is a rule of thumb:
Say I have a statement I want to prove, say something of the shape $A\implies B$.
If I want to prove it by contradiction, I usually assume that $A$ is true, and that $B$ is not true, and then arrive to a contradiction.
However, it is then a good idea to look at your proof again and ask yourself:

Is the proof by contradiction really necesary?

For example, sometimes, your proof looks something like this:


*

*Let's assume $A$ is true.

*Let's assume that $\neg B$ is true.

*Then, because (somethingsomething), and therefore, $B$ is true.

*But since $\neg B$ is also true, that is a contradiction.

*Therefore, $B$ is true.

*So, $A\implies B$ is true.


Now, if the somethingsomething does not contain any reference to $\neg B$, or can be proven even if $\neg B$ is not true, then basically, the proof can be reduced to:
  - Let's assume $A$ is true
  - Then, somethingsomething, and therefore, $B$ is true.
  - So, $A\implies B$ is true
which is much more elegant.
A: The first part of your proof is that $x_1+x_2=-b/a$ and, from the assignment text, I think you don't need it, because $x_1+x_2=-b/a$ is one of Viète’s formulas.
On the other hand, the argument is correct.
For the main part you're doing too much work.
First note that, since $a,b,c\in\mathbb{Q}$ by assumption, also $-b/a\in\mathbb{Q}$.
Since $x_2=(-b/a)-x_1$ and $x_1=(-b/a)-x_2$, by Viète’s formula,


*

*if $x_1\in\mathbb{Q}$, then $x_2=(-b/a)-x_1\in\mathbb{Q}$

*if $x_2\in\mathbb{Q}$, then $x_1=(-b/a)-x_2\in\mathbb{Q}$


This just uses the arithmetic properties of $\mathbb{Q}$.
