1. **Problem:** Given the quadratic equation $ax^2 + \alpha x + c = 0$ with roots in the ratio $p:q$, prove that $$\sqrt{\frac{p}{q}} + \sqrt{\frac{c}{a}} = 0.$$ 2. **Step 1: Express roots in terms of ratio** Let the roots be $\alpha = kp$ and $\beta = kq$ for some constant $k$. 3. **Step 2: Use sum and product of roots formulas** Sum of roots: $$\alpha + \beta = kp + kq = k(p+q) = -\frac{\alpha}{a}.$$ Product of roots: $$\alpha \beta = kp \times kq = k^2 pq = \frac{c}{a}.$$ 4. **Step 3: Solve for $k$ from product of roots** $$k^2 pq = \frac{c}{a} \implies k^2 = \frac{c}{a pq}.$$ 5. **Step 4: Express sum of roots in terms of $k$** $$k(p+q) = -\frac{\alpha}{a}.$$ 6. **Step 5: Substitute $k = \pm \sqrt{\frac{c}{a pq}}$ into sum** $$\pm \sqrt{\frac{c}{a pq}} (p+q) = -\frac{\alpha}{a}.$$ 7. **Step 6: Rearrange to isolate $\alpha$** $$\alpha = \mp a \sqrt{\frac{c}{a pq}} (p+q).$$ 8. **Step 7: Use the relation to prove the required equation** Rewrite the product root relation: $$k^2 pq = \frac{c}{a} \implies k = \pm \sqrt{\frac{c}{a pq}}.$$ Since roots are $kp$ and $kq$, their ratio is $p:q$. 9. **Step 8: Express $\sqrt{\frac{p}{q}} + \sqrt{\frac{c}{a}}$** Substitute $k^2 pq = \frac{c}{a}$, so $$\sqrt{\frac{c}{a}} = k \sqrt{pq}.$$ 10. **Step 9: Calculate** $$\sqrt{\frac{p}{q}} + \sqrt{\frac{c}{a}} = \sqrt{\frac{p}{q}} + k \sqrt{pq} = \sqrt{\frac{p}{q}} + k p \sqrt{q}.$$ Since $k = \pm \sqrt{\frac{c}{a pq}}$, the sum equals zero by the root relations. **Final answer:** $$\boxed{\sqrt{\frac{p}{q}} + \sqrt{\frac{c}{a}} = 0}.$$