Show that $p(x)=rq(x)$ for some rational number $r$ 
Let $p(x)$ and $q(x)$ be two quadratic polynomials with integer coefficients. Suppose they have a non-rational zero in common. Show that $$p(x)=rq(x)$$ for some rational number $r$.

If the common root is an irrational number then the discriminant must be a positive non square integer. This is because all the coefficients are integers. Apart from this I can't think of anything else.
Is it worth trying to define $f(x)=p(x)-rq(x)$ and investigating the properties of $f(x)$?
 A: Let $x_1$ be the irrational root that is shared by $p$ and $q$. Then, by the
Irrational Conjugate Roots Theorem (#16 on this list of polynomial theorems), the irrational conjugate of $x_1$ must also be a root of both $p$ and $q$.
Explicitly, for any rational $a,b$ and irrational $\sqrt{c}$,  if $$x_1 = a + b\sqrt{c}$$ is a root of $p$, then $$x_2 = a - b\sqrt{c}$$ must also be a root. The same reasoning implies that $x_2$ is a root of $q$.

Since $p$ and $q$ are quadratics whose two roots are the same, they must be proportional, up to a constant. Since we have integer coefficients, the constant of proportionality $r$ must be rational, so we are done.
A: Well, first, since you know that they have a common root, you get that $gcd(p,q) \neq 1$.
Then, there exists $r,g \in F[X]$ with $deg(r),deg(g)<2$ so $p=rq+g$.
Now, write $r=ax+b$. Assume $a \neq 0$. Than since $p,q$ are noted that they both quadratic, you get that $p = rq+g$  where $rq+g$ is a polynomial of degree 3, therefore $a$ must be $0$.
You get that $r = b$ where $b \in \mathbb R$. But since all of the coefficients are integers, thus the coefficient of $x^2$ is integer, you get that $b \in \mathbb Q$ as needed.
Now, let us show that $g=0$. Since $p,q$ have only integer coefficients and $r$ is a rational number, $g$ must also have only integer coefficients.
Let us apply the common root on the equation $p=rq+g$. We get $g=0$ since it has only integer coeffients, thus proving $p=rq$.
