Find the roots of the equation $\frac{1}{p+q+x}=\frac{1}{p}+\frac{1}{q}+\frac{1}{x}$.

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

Taking LCM 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 $(p+q)$:

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

Cross-multiplying:

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

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

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

Factoring by grouping:

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

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

Therefore, the roots are:

$$x=-p\text{and}x=-q$$