1. **State the problem:** Find the value of $k$ such that $\frac{p}{x+r} + \frac{q}{x-r} = \frac{k}{2x}$.\n\n2. **Write the equation:** $$\frac{p}{x+r} + \frac{q}{x-r} = \frac{k}{2x}$$\n\n3. **Find a common denominator on the left side:** The common denominator is $(x+r)(x-r) = x^2 - r^2$. So, $$\frac{p(x-r)}{(x+r)(x-r)} + \frac{q(x+r)}{(x-r)(x+r)} = \frac{p(x-r) + q(x+r)}{x^2 - r^2}$$\n\n4. **Simplify the numerator:** $$p(x-r) + q(x+r) = px - pr + qx + qr = (p+q)x + (qr - pr)$$\n\n5. **Rewrite the equation:** $$\frac{(p+q)x + (q - p)r}{x^2 - r^2} = \frac{k}{2x}$$\n\n6. **Cross multiply:** $$2x \left[(p+q)x + (q-p)r\right] = k(x^2 - r^2)$$\n\n7. **Expand left side:** $$2x(p+q)x + 2x(q-p)r = kx^2 - kr^2$$ $$2(p+q)x^2 + 2(q-p)rx = kx^2 - kr^2$$\n\n8. **Equate coefficients of powers of $x$:** For the equation to hold for all $x$, coefficients of $x^2$, $x$, and constant terms must be equal on both sides. - Coefficient of $x^2$: $2(p+q) = k$ - Coefficient of $x$: $2(q-p)r = 0$ (since right side has no $x$ term) - Constant term: $0 = -kr^2$ \n9. **From the constant term:** $$0 = -kr^2 \implies k r^2 = 0$$ Since $r^2 \neq 0$, this implies $$k = 0$$\n\n10. **Check coefficient of $x^2$ with $k=0$:** $$2(p+q) = 0 \implies p + q = 0$$\n\n11. **Check coefficient of $x$:** $$2(q-p)r = 0 \implies q = p$$\n\n12. **From $p+q=0$ and $q=p$, we get:** $$p + p = 0 \implies 2p = 0 \implies p = 0$$ Then $q = 0$ as well. \n**Summary:** - $k = 0$ - $p = 0$ - $q = 0$ \n**Final answer:** $$\boxed{k = 0}$$