Using Mathematica, it is relatively straightforward to check that the zeroes of $$ f_n(z)\equiv \sqrt{z} J_{n+\frac{1}{2}}(\sqrt{z})+J_{n+\frac{3}{2}}(\sqrt{z}) $$ with $n=0,1,2,\ldots$ and $J_n$ a Bessel function, lie on the positive real axis, that is to say, if $f_n(z)=0$, then $z\in \mathbb{R}^+$. Is there a simple way to prove this? What about more general linear combinations?
Update 1
I have noted that the above linear combination can be written in a nicer form that might be easier to eventually prove the statement asked. Indeed
$$ f_n(z)\equiv F_n(\sqrt{z})\quad\text{with}\quad F_n(x)=\left(n+\frac{5}{2}\right)J_{n+\frac{3}{2}}(x)+x\,J^\prime_{n+\frac{3}{2}}(x)\,. $$
Update 2
This paper https://www.tandfonline.com/doi/epdf/10.1080/00036818608839633?needAccess=true appears to show exactly what I wanted in Corollary 2.3. A similar proof can be found in page 482 of Watson's treatise.