Given the equation $$Y^2 - 11Y + 30 = 0$$ with roots *m* and *n*, and one root *x = -10* of the equation $$\sqrt{(x + 45 - mn)^4} = 5x + Q,$$ compare the following quantities: **Quantity (I):** $$\frac{Q}{3}$$ **Quantity (II):** $$\frac{Q}{2^2}$$

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**Solution:**

1. Given quadratic equation:
   $$Y^2 - 11Y + 30 = 0$$

2. Find roots *m* and *n*:
   - Factors of 30 that add to 11 are 5 and 6.
   - So, $$m = 5, \quad n = 6$$

3. Calculate product:
   $$mn = 5 	imes 6 = 30$$

4. Given equation:
   $$\sqrt{(x + 45 - mn)^4} = 5x + Q$$

Since the square root of a fourth power is the square of the expression inside:
   $$\sqrt{(x + 45 - mn)^4} = (x + 45 - mn)^2$$

5. Substitute $$x = -10$$ and $$mn = 30$$:
   $$(-10 + 45 - 30)^2 = 5(-10) + Q$$

Simplify inside the bracket:
   $$(-10 + 15)^2 = -50 + Q$$

$$5^2 = -50 + Q$$

$$25 = -50 + Q$$

6. Solve for $$Q$$:
   $$Q = 25 + 50 = 75$$

7. Now compare:
   - Quantity (I): $$\frac{Q}{3} = \frac{75}{3} = 25$$
   - Quantity (II): $$\frac{Q}{2^2} = \frac{75}{4} = 18.75$$

8. Since $$25 > 18.75$$,

**Answer:** Quantity I > Quantity II

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