Let $f(x)$ and $g(x)$ be two quadratic polynomials all of whose coefficients are rational numbers. Suppose that $f(x)$ and $g(x)$ have a common irrational root. Show that $g(x)=r f(x)$ for some rational number $r$.
Suppose we take $\alpha$ to be the common root. We can set $f(\alpha)=g(\alpha)=0$ and obtain a relation between the rational coefficients of the quadratic polynomials. But then how do I proceed from there?
On
Hint: let $f(x)=a_1x^2+b_1x+c_1$ with $a_1 \ne 0$ and $g(x)=a_2x^2+b_2x+c_2$ with $a_2 \ne 0\,$, then:
$$ \begin{cases} \begin{align} \alpha^2+\frac{b_1}{a_1} \alpha + \frac{c_1}{a_1} &= 0 \\ \alpha^2+\frac{b_2}{a_2} \alpha + \frac{c_2}{a_2} &= 0 \\ \end{align} \end{cases} $$
Subtracting the two equations, canceling out the $\alpha^2$ terms, and collecting:
$$ \left(\frac{b_1}{a_1} - \frac{b_2}{a_2}\right) \alpha + \left(\frac{c_1}{a_1} - \frac{c_2}{a_2}\right) = 0 $$
Since $\alpha$ is irrational and the coefficients in parentheses are rational, the equality can only hold iff:
$$ \frac{b_1}{a_1} - \frac{b_2}{a_2} = \frac{c_1}{a_1} - \frac{c_2}{a_2} = 0 \quad \iff \quad a_2 = ra_1, \,b_2 = r b_1, c_2 = r c_1 \quad \text{where} \quad r = \frac{a_2}{a_1} $$
We have that $$f(x)=r_1(x-\alpha)(x-\beta_1)$$ and $$g(x)=r_2(x-\alpha)(x-\beta_2)$$ since $f$ and $g$ are quadratic polynomials ($\alpha$ is the common root and $\beta_i$ are the other)
Multiplying this out we get
$$f(x)=r_1x^2-r_1(\alpha+\beta_1)x+r_1\alpha\beta_1$$
and similarly for $g$. We know that the coefficients of $f$ and $g$ are rational and whence $r_i$ is rational, $r_i(\alpha+\beta_i)$ is rational and $r_i\alpha\beta_i$ is rational. Since $\alpha$ is irrational, $\beta_i$ must be an irrational number such that both $\alpha+\beta_i$ and $\alpha\beta_i$ are rational for $i=1,2$. From this point on I will be using that since $\alpha,\beta_1,\beta_2$ are irrational, they are non-zero. Our equations are
\begin{align} \alpha+\beta_1&=s_1\quad (1)\\ \alpha+\beta_2&=s_2\quad (2)\\ \alpha\beta_1&=s_3\quad (3)\\ \alpha\beta_2&=s_4\quad (4) \end{align}
where the $s_i$ are rational numbers, $s_3,s_4\neq 0$. (1)-(2) and $\frac{(3)}{(4)}$ give us
\begin{align} \beta_1-\beta_2&=s_1-s_2\quad (5)\\ \frac{\beta_1}{\beta_2}&=\frac{s_3}{s_4}\quad (6) \end{align}
Plugginh (6) into (5) and defining $s_5=\frac{s_3}{s_4}$ and $s_6=s_1-s_2$, which are still rational numbers, we get
\begin{align} \beta_2(1-s_5)=s_6 \end{align}
i.e. the irrational number $\beta_2$ multiplied by the rational number $1-s_5$ equals the rational number $s_6$. This is only possible if $s_5=1$ and $s_6=0$, which implies that $\beta_1=\beta_2$ and thus $f$ and $g$ only differ by multiplication of the rational constant $\frac{r_2}{r_1}$. (Note again that $r_1\neq 0$ because otherwise $f$ would not have an irrational root.
There is probably a way to do this quicker, but this does the job.