$\sqrt{6}+\sqrt{3}$ is not rational proof I want to prove $\sqrt{6}+\sqrt{3}$ is not rational; here is my attempt:
Assume for the sake of contradiction that $\sqrt{6}+\sqrt{3}$ is rational. Then $(\sqrt{6}+\sqrt{3})^2$ must also be rational. Since $$(\sqrt{6}+\sqrt{3})^2=9+2\sqrt{6}\sqrt{3}=9+2\sqrt{2}\sqrt{3}\sqrt{3}=9+6\sqrt2,$$
we see $9+6\sqrt2$ must be rational. But, since $\sqrt{2}$ is not rational we have a contradiction and hence $\sqrt{6}+\sqrt{3}$ is not rational.
Is it correct?

You can also appeal to the rational root theorem,
$$ \begin{align*} x &= \sqrt{3} + \sqrt{6}\\
x^2 &= 9 + 6\sqrt{2}\\
x^4-18x^2 &+9=0 \end{align*} $$
So if $x = p/q$ with $p,q$ coprime then $p |9, q|1$ or $p = 1,3,9$ and $q=1,-1$, but none of the options qualify as $x$ is not an integer
