1. **Problem Statement:** Given the quadratic equation $px^2 + qx + 9 = 0$ with roots $\alpha$ and $\beta$, prove that $$\frac{\sqrt{p} + \sqrt{q}}{\sqrt{p} - \sqrt{q}} = 0.$$ 2. **Recall the relationships between roots and coefficients:** For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, we have: $$\alpha + \beta = -\frac{b}{a}$$ $$\alpha \beta = \frac{c}{a}$$ 3. **Apply to our equation:** Here, $a = p$, $b = q$, and $c = 9$. So, $$\alpha + \beta = -\frac{q}{p}$$ $$\alpha \beta = \frac{9}{p}$$ 4. **Analyze the expression to prove:** The expression $$\frac{\sqrt{p} + \sqrt{q}}{\sqrt{p} - \sqrt{q}} = 0$$ implies the numerator must be zero (since denominator cannot be zero). So, $$\sqrt{p} + \sqrt{q} = 0$$ 5. **Check if this is possible:** Since $p$ and $q$ are coefficients of the quadratic, they are real numbers. The square roots $\sqrt{p}$ and $\sqrt{q}$ are real only if $p \geq 0$ and $q \geq 0$. The sum of two real square roots cannot be zero unless both are zero, which would make the quadratic degenerate. 6. **Conclusion:** The given expression $$\frac{\sqrt{p} + \sqrt{q}}{\sqrt{p} - \sqrt{q}} = 0$$ cannot hold true for real $p$ and $q$ unless $\sqrt{p} + \sqrt{q} = 0$, which is not possible for positive $p$ and $q$. Therefore, the statement as given is incorrect or incomplete. **Final answer:** The expression $$\frac{\sqrt{p} + \sqrt{q}}{\sqrt{p} - \sqrt{q}} = 0$$ is not true under normal conditions for the quadratic equation $px^2 + qx + 9 = 0$.