Show that $g(x)=r f(x)$ for some rational number $r$. 
Let $f(x)$ and $g(x)$ be two quadratic polynomials all of whose coefficients are rational numbers. Suppose $f(x)$ and $g(x)$ have a common irrational root. Show that $g(x)=r f(x)$ for some rational number $r$.


My solution (please check if it is correct or not)
Denote this irrational root to be $k=a+b\sqrt{c}$ where $a, b, c \in \mathbb{R}$ and $c$ is not divisible by the square of any prime. Then if $a+b\sqrt{c}$ is a root of the polynomial, so does $a-b\sqrt{c}$ by the Irrational Conjugate Theorem. Then we may write $f(x)=\alpha(x-(a+b\sqrt{c}))(x-(a-b\sqrt{c}))$ and $g(x)=\beta(x-(a+b\sqrt{c}))(x-(a-b\sqrt{c}))$ where $\alpha, \beta \in \mathbb{Q}$. Dividing both sides by $\beta$ gives $f(x)=\frac{\alpha}{\beta}(x-(a+b\sqrt{c}))(x-(a-b\sqrt{c}))$ and $g(x)=(x-(a+b\sqrt{c}))(x-(a-b\sqrt{c}))$; since $\alpha$ and $\beta$ are rational. Let $r=\frac{\alpha}{\beta}$ and hence we are done.
 A: You solution is OK, but it can be even simpler, because you don't really need to write out the root at all.

Let $x_0$ be the irrational root of $f$ and $g$, and let $x_1$ be the irrational conjugate of $x_0$. Then, by the Irrational Conjugate theorem, $x_1$ is a root of $f$ and of $g$. This means that $f(x)=a(x-x_0)(x-x_1)$ and $g(x)=b(x-x_0)(x-x_1)$ for some $a,b\in\mathbb R$.
Because $f$ and $g$ have rational coefficients, $a,b\in\mathbb Q$ because $a$ is the leading coefficient of $f$ and $b$ is the leading coefficient of $g$.
Finally, because $$g(x) = \frac{b}{a}\cdot f(x),$$ we can set $r\frac ba$ and we have $g(x)=rf(x)$.

Or, you can also argue another way, a little more hand-wavy, but for most purposes, still rigorous enough.

If $x_0$ is the irrational root of $f$ and $g$, then so is the irrational conjugate of $x_0$. Because $f$ and $g$ are quadratic polynomials with identical roots, we know that $g(x)=r\cdot f(x)$ for some real number $r$ and all real numbers $x$. Plugging in $x=0$ we get $r=\frac{g(0)}{f(0)}$, meaning $r$ is a rational number (because $g(0)$ and $f(0)$ must be rational numbers).

