Showing that $\sqrt \pi$ is transcendental

Suppose $\sqrt \pi$ is algebraic : there is a polynomial $P(X)$ with rational coefficients such that $P(\sqrt \pi)=0$.

Develop the product $P(X)P(-X)$. You will observe that the odd powers of $X$ cancel each other, so you can write it as a polynomial in $X^2$ : There is a polynomial $Q$ (again with rational coefficients) such that $P(X)P(-X) = Q(X^2)$.

Now, $Q(\pi) = Q((\sqrt \pi)^2) = P(\sqrt \pi)P(- \sqrt \pi) = 0$, hence $\pi$ is algebraic.

I found another way to solve it, suppose that $\sqrt \pi$ is a root of $P(x)$ which is defined in the question then we have$$ \left ( \sum_{\text{$j$ even}} a_j \pi^{\frac j 2} +a_0 \right ) +\sqrt \pi \sum_{\text{$i$ odd}} a_i \pi^{i-1} = 0 $$ then $Q(\pi)=-\sqrt \pi G(\pi)$ upon squaring both sides we get $Q^2(\pi)= \pi G^2(\pi)$ then $Q^2(\pi)- \pi G^2(\pi)=0$ is a polynomial in $\pi$ equal to zero which is a contradiction since $\pi$ is transcendental.

Suppose $\sqrt{\pi}$ were algebraic over $\Bbb{Q}$, i.e. $\Bbb{Q}\left(\sqrt{\pi}\right)/\Bbb{Q}$ is algebraic. This gives us the intermediate extension $\Bbb{Q}(\pi)$: $$ \Bbb{Q}\left(\sqrt{\pi}\right)/\Bbb{Q}\left(\pi\right)/\Bbb{Q}. $$ In particular, this implies that both $\Bbb{Q}\left(\sqrt{\pi}\right) / \Bbb{Q}(\pi)$ and $\Bbb{Q}(\pi)/\Bbb{Q}$ are algebraic, which is a contradiction to the fact that $\pi$ is transcendental (every element of $\Bbb{Q}\left(\sqrt{\pi}\right)$ is algebraic, and $\Bbb{Q}(\pi)\subseteq\Bbb{Q}\left(\sqrt{\pi}\right)$).

Edit: The argument about the degrees of the extensions actually also works, after noting awllower's hint in the comments. $$ \infty > n = \left[\Bbb{Q}\left(\sqrt{\pi}\right) : \Bbb{Q}\right] = \left[\Bbb{Q}\left(\sqrt{\pi}\right) : \Bbb{Q}(\pi)\right]\cdot\left[\Bbb{Q}\left(\pi\right) : \Bbb{Q}\right], $$ implying $\Bbb{Q}(\pi)$ is finite, and therefore algebraic.