Showing integer roots of a quadratic equation is countable?

Show that the set of real numbers that are the roots of quadratic equations $ax^2+bx+c=0$ with integer coefficients (i.e. $a, b, c\in\mathbb{Z}$) is countable.

My work so far: $ax^2 +bx + c = 0$ can be written as $(mx+p)(nx+q) = 0$, where $mn = a$ and $pq = c$ and $pn + mq = b$. Then $x_1 = -p/m$ and $x_2 = -q/n$. $x_1$ and $x_2$, which are the roots of the polynomial, are countable since they essentially rational numbers. And I think we learned in class an infinite set of all rational numbers are countable.

My question: Does this approach make any sense ? If not, how can improve/change it to answer the question ? Thanks for your time.

• This approach doesn't quite make sense because not all quadratic equations with integer coefficients, such as $x^2 - 2$, can be factored into a product like there where the coefficients are rational (some quadratic equations wit integer coefficients do not have rational roots). Instead, you could try finding an explicit bijection: can you enumerate the polynomials? Then each polynomial has at most two roots.... – user2055 Nov 19 '13 at 17:04
• Hint: How many quadratic equations with integer coefficients are thee? How mayn real solutions are there at most per equation? – Hagen von Eitzen Nov 19 '13 at 17:05
• – lab bhattacharjee Nov 19 '13 at 17:06
• @labbhattacharjee It's true, but the solution in that exercise set is IMHO awful - I would much prefer Hagen's answer, as it doesn't rely on the quadratic formula and thus generalizes effortlessly to the entirety of the algebraic numbers. – Steven Stadnicki Nov 20 '13 at 8:47

Your approach is wrong, not all the real numbers which are roots of such polynomials are rational numbers. For example $\sqrt2$ is the root of $x^2-2$, but is not a rational number.