$f(x) = ax^2 + bx +c$ is a quadratic equation, show if one root is rational then the second root is rational I'm a new user to this site and look to use this site to expand my math knowledge.
The question: Let $f(x) = ax^2 + bx + c$ be a quadratic whose coefficient $a, b, c$ are rational. Prove that if $f(x)$ has one rational root, then its other root is also rational.
Disclaimer: I understand the question has more than likely been asked before, but the purpose is not being given the answer but rather if I'm making a correct approach to my problem.
My attempt: 
given the coefficients a,b,c are rational and also one root being rational
we let $(x+d)(x+e)$ be the two roots for the function.
If say $x+d$ is the given root to be rational
then solving the product of the two root:$ x^2 + (d+e)x + d*e $ is the same function as: $ax^2 + bx + c $
$ a = 1 $
$b = d+e \rightarrow  e = d-b$, rational minus rational = rational
$c = d*e \rightarrow e = c/d$, rational divided by a rational = rational$
therefore the second root is also rational 
 A: Hint. If the two roots are $p$ and $q$, then $f(x)$ must be proportional to $(x-p)(x-q)$.  It must therefore be equal to $a(x-p)(x-q)$.  What is $c$ in terms of $a$, $p$, and $q$?
Finally, suppose $p$ is rational.  What is $q$ in terms of $a$, $c$, and $p$?
A: Hint: Use Vieta's formulas and the fact that the coefficients $\{a,b,c\}$ are rational. (Actually it suffices to use just one of Vieta's formulas, namely, the fact that the roots are positioned symmetrically with respect to the quadratic parabola's vertex $-{b\over2a}$.)
For example, if the roots are $\{x_1, x_2\}$ and $x_1$ is rational, then we have
$$
x_1+x_2 = -{b\over a} \quad \Rightarrow \quad x_2 = -x_1-{b\over a}
\quad \Rightarrow \quad x_2 \mbox{ is rational.}
$$
A: Your idea is good but there is something you forgot to consider. You know that $ax^2 + bx + c$ has one rational root. The first thing to do is to justify why this equation has then actually two roots (this can be done using polynomial division or the quadratic formula or in any other way). Once you know that, you can write
$$ ax^2 + bx + c = a(x - d)(x - e) $$
where $d,e$ are the roots (note the minus signs and the fact that $a$ appears on both sides!). Let's say that $d$ is rational. Expanding the equation above, we get
$$ ax^2 + bx + c = a(x^2 - (d + e)x + de) = ax^2 - a(d+e)x + ade. $$
Now you can compare coefficients. Since
$b = -a(d + e)$ we have $d + e = -\frac{b}{a}$ so $d + e$ is rational. Since $d$ is also rational, this implies that $e$ must also be rational.
A: The two roots of $ax^2 + bx + c$ are $\frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$
If $\sqrt{b^2 - 4ac}$ is rational, then both the roots are rational.  If $\sqrt{b^2 - 4ac}$ is irrational then both the roots are irrational.
..... or ....
If $d$ and $e$ are the roots and $d$ is rational and $e$ is irrational then $ax^2 + bx + c = a(x-d)(x-e) = ax^2 + a(-d-e)x + ade$ so $b = a(-d-e)$ which is not a rational number.  So a contradiction.
