Let $p_i,q_i \in \mathbb{Z}[X]$ and Let $P(x) = \sum_{i=1}^n p_i(x) \sqrt{q_i(x)} $. Can we deduce that if $P(x) = 0$ then $x$ is an algebraic number? For some simple examples I can see that it's true, like trivially if $P(x)$ is itself in $\mathbb{Z}[X]$, or if $P_1(x) = \sqrt{q_1(x)}$ then $x$ satisfies $P_1(x)^2 = 0$ where $ P_1(x)^2 \in \mathbb{Z}[X]$. For $P_2(x) = \sqrt{q_1(x)} + \sqrt{q_2(x)} + p_3(x)$ also one can do the following $$ \sqrt{q_1(x)} + \sqrt{q_2(x)} =- p_3(x) \\ q_1(x) + q_2(x) + 2\sqrt{q_1(x)q_2(x)} = p_3(x)^2 \\ 2\sqrt{q_1(x)q_2(x)} = p_3(x)^2 - q_1(x) - q_2(x) \\ 4 \ q_1(x)q_2(x) = \big(p_3(x)^2 - q_1(x) - q_2(x)\big)^2 $$ And so again we can see that $x$ is algebraic. The same technique is applicable to $P_3(x) = \sum_{i=1}^{n} \sqrt{q_i(x)}$ with $n \leq 5$ by moving the radicals to different sides of the equation. But for example I don't have an idea for $P_4(x) = \sum_{i=1}^{n} \sqrt{q_i(x)}$ with $ n \geq 6$, the above technique for $P_2(x)$ and $P_3(x)$ does not seem to me to be applicable in this case: For $P_4(x)$ we might want to put $4$ radicals at on left side of the equation and the other $2$ at the right side. Then raising to the second power would produce $6$ radicals at left and $1$ at right, which counts for total of $7$ radicals, and that's more than the number of radicals we started with. For other partitionings of the radicals in the equation the problem still doesn't get solved with this approach; I see that's basically because $ { n \choose 2} \geq n $ for $n \geq 3$. And this is where I got stuck.
I would appreciate suggestions for proceeding and answers.