The roots of the equation $\frac{1}{p+q+x}=\frac{1}{p}+\frac{1}{q}+\frac{1}{x}$ are:

Rearranging the given equation:
$$\frac{1}{p+q+x}-\frac{1}{x}=\frac{1}{p}+\frac{1}{q}$$

Taking the common denominators on both sides:

$$\frac{x-(p+q+x)}{x(p+q+x)}=\frac{q+p}{pq}$$

$$\frac{-(p+q)}{x(p+q+x)}=\frac{p+q}{pq}$$

Assuming $p+q\ne 0$, we can divide both sides by $(p+q)$:

$$\frac{-1}{x(p+q+x)}=\frac{1}{pq}$$

$$-pq=x(p+q+x)$$

$$-pq=x^2+px+qx$$

$$x^2+px+qx+pq=0$$

Factoring by grouping:

$$x(x+p)+q(x+p)=0$$

$$(x+p)(x+q)=0$$

Thus, the roots are $x=-p$ and $x=-q$.