Given the two quadratic equations:

I. $ $x^2 + 12x + 35 = 0 $$

II. $ $y^2 + 7y + 10 = 0 $$

Solve both equations and determine the correct relationship between $ x $ and $ y $.

Question

Given the two quadratic equations:

I. $ $x^2 + 12x + 35 = 0 $$

II. $ $y^2 + 7y + 10 = 0 $$

Solve both equations and determine the correct relationship between $ x $ and $ y $.

Accepted Answer

**Solution:** 1. Solve the first quadratic equation: $$ x^2 + 12x + 35 = 0 $$ - Factorize or use the quadratic formula: $$ $x = \frac{-12 \pm \sqrt{12^2 - 4 \times 1 \times 35}}{2} = \frac{-12 \pm \sqrt{144 - 140}}{2} = \frac{-12 \pm 2}{2}$ $$ - Roots are: $$$ x_1 = \frac{-12 + 2}{2} = \frac{-10}{2} = -5$ $$ $$ $x_2 = \frac{-12 - 2}{2} = \frac{-14}{2} = -7$ $$ 2. Solve the second quadratic equation: $$$ y^2 + 7y + 10 = 0$ $$ - Factorize or use the quadratic formula: $$ $y = \frac{-7 \pm \sqrt{7^2 - 4 \times 1 \times 10}}{2} = \frac{-7 \pm \sqrt{49 - 40}}{2} = \frac{-7 \pm 3}{2}$ $$ - Roots are: $$ $y_1 = \frac{-7 + 3}{2} = \frac{-4}{2} = -2$ $$ $$ $y_2 = \frac{-7 - 3}{2} = \frac{-10}{2} = -5$ $$ 3. Now, compare the roots of $ x $ and $ y $: - $ x $ roots: $ -7, -5 $ - $ y $ roots: $ -5, -2 $ 4. The largest root of $ x $ is $ -5 $ and the smallest root of $ y $ is $ -5 $, but $ y $ also has a root $ -2 $ which is greater than any root of $ x $. 5. Since the question asks for the relationship between $ x $ and $ y $, we consider the possible values: - $ x $ can be $ -7 $ or $ -5 $ - $ y $ can be $ -5 $ or $ -2 $ 6. For $ x > y $ to hold, $ x $ must be greater than $ y $ for all roots. Check: - $ -5 > -5 $ is false (equal) - $ -5 > -2 $ is false - $ -7 > -5 $ is false - $ -7 > -2 $ is false 7. For $ x < y $ to hold, check: - $ -5 < -5 $ false - $ -5 < -2 $ true - $ -7 < -5 $ true - $ -7 < -2 $ true 8. Since $ x $ roots are less than or equal to $ y $ roots, the correct relation is $ x < y $. **Answer:** $ x < =y $ (Option b) **Note:** The initial marking of option (c) as correct was incorrect. The correct answer is option (d).

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