If $\int_{-1}^1 f(x) \sqrt{1-x^2}dx = 0$ and $\int_{-1}^1 xf(x) dx = 0$, for $f$ continuous on $[-1,1]$, how can we show that $f$ has at least two roots in $(-1,1)$?
I know one root can be found using Rolle's theorem on the first integral. I tried proving it by contradiction. Assuming $f$ has exactly 1 root $r$, then we can assume $f(x) < 0$ on $(-1,r)$ and $f(x) > 0$ on $(r,1)$. I'm trying to show that $\int_{-1}^1 (x-r)f(x) \sqrt{1-x^2} dx = 0$ which would work but I'm not sure if it means that $\int_{-1}^1 xf(x) \sqrt{1-x^2}dx = 0$.
Sketch: Assume that there is $r\in (-1,1)$ so that $f(x)\geq 0$ for $x\in [r,1]$ and $f(x)\leq 0$ for $x\in [-1,r]$. Now, integrate against the auxiliary function $$ g(x)= x \sqrt{1-r^2} - r \sqrt{1-x^2}$$ to deduce what $f$ looks like.