Can a linear combination of even Legendre polynomials have common real root(s) with a linear combination of odd Legendre polynomials?

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I am using the following definition of Legendre Polynomials: $P_0(x)=1$, $P_1(x)=x$ and $$P_{k+1}(x)=\left(\frac{2k+1}{k+1}\right)xP_k(x)−\left(\frac{k}{k+1}\right)P_{k−1}(x)$$

Let $b_0>b_1>\ldots >b_n>\ldots >0$ and define $$f_{2n}(x)=\sum_{k=0}^n b_{2k} P_{2k}(x)$$ $$f_{2n+1}(x)=\sum_{k=0}^n b_{2k+1} P_{2k+1}(x)$$

Q1: Prove that $f_{2n}(x)$ and $f_{2n+1}(x)$ have no common real root.

Updated with the second question!

Let $b_0>b_1>\ldots >b_n>\ldots >0$ and define $$q_{2n}(x)=\sum_{k=0}^n \left(-1\right) ^k b_{2k} P_{2k}(x)$$ $$q_{2n+1}(x)=\sum_{k=0}^{n-1} \left(-1\right) ^{k+1} b_{2k+1} P_{2k+1}(x)$$ Q2: Prove that $q_{2n}(x)$ and $q_{2n+1}(x)$ have no common real root.

The second one is probably a lot harder than the first one.