Continuous $f$ has $≥2$ roots if $int_{-1}^{1} f(x)sqrt {1 - x^2} mathrm{d}x = int_{-1}^{1} xf(x) mathrm{d}x = 0$?

Since $sqrt{1-x^2}$ is positive in $(-1,1)$ then $int_{-1}^{1} f(x)sqrt {1 - x^2} dx = 0$ implies that the continuous function $f$ has at least a zero $ain (-1,1)$ (otherwise the product $f(x)sqrt {1 - x^2}$ has the same sign over $(-1,1)$ and, recalling that if $Fgeq 0$ is continuous and $int_a^b F(x),dx=0$ then $F=0$ everywhere in $[a,b]$, we have a contradiction.

Assume that $a$ is the unique root of $f$ in $(-1,1)$, then $f$ should be positive on one side of $a$ and negative on the other side. Moreover $$int_{-1}^{1} f(x)g(x),dx=0$$ where $g(x)=(xsqrt{1 - a^2}-asqrt {1 - x^2})$ is a continuous function which is negative in $[-1,a)$ and positive in $(a,1]$. Hence the product $fg$ has the same sign on $(-1,1)$, and, since its integral is zero, we have a contradiction.