Let $f(x) = a_1 x^2+ b_1x + c_1$ and $g(x) = a_2 x^2+b_2 x+ c_2$, where $a_i,b_i,c_i\in\mathbb{Z}$. Assume there is an $x$ such that $f(x) = \sqrt{2}g(x)$, that is \[ (a_1-\sqrt{2}a_2)x^2 + (b_1-\sqrt{2}b_2)x + (c_1-\sqrt{2}c_2)=0,\quad\text{for some}\quad x. \]Let $\Delta_1=b_1^2 - 4a_1c_1$, $\Delta_2 = b_2^2-4a_2c_2$. Then $a_1,a_2>0$ and $\Delta_i<0$. Furthermore, if $\Delta$ is the discriminant of the quadratic above, we must have $\Delta\le 0$. Thus, the equality implies $\Delta=0$. Solving, we get \[ \Delta_1 + 2\Delta_2 = 2\sqrt{2}\bigl(b_1b_2 - 2a_1c_2-2a_2c_1\bigr). \]The LHS is negative. So, $b_1b_2 - 2a_1c_2-2a_2c_1\ne 0$. But then we have a contradiction as otherwise $2\sqrt{2}\in\mathbb{Q}$, which is absurd.