Given the two quadratic equations:

(I) $ $x^2 - 11x + 28 = 0$ $

(II) $ $y^2 + 13y + 30 = 0 $$

Find the correct relationship between $x$ and $y$ from the options below.

Question

Given the two quadratic equations:

(I) $ $x^2 - 11x + 28 = 0$ $

(II) $ $y^2 + 13y + 30 = 0 $$

Find the correct relationship between $x$ and $y$ from the options below.

Accepted Answer

**Solution:**

1. Solve the first quadratic equation:
   $$
  $ x^2 - 11x + 28 = 0$
   $$
   Factorizing or using the quadratic formula:
   $$
  $ (x - 7)(x - 4) = 0$
   $$
   So, $ x = 7 $ or $ x = 4 $.

2. Solve the second quadratic equation:
   $$
  $ y^2 + 13y + 30 = 0$
   $$
   Factorizing:
   $$
   (y + 10)(y + 3) = 0
   $$
   So, $ y = -10 $ or $ y = -3 $.

3. Now, compare the possible values of $ x $ and $ y $:
   - $ x $ can be 7 or 4
   - $ y $ can be -10 or -3

4. Since both values of $ x $ are greater than both values of $ y $, the relationship is:
   $$
   x > y
   $$

**Answer:** If $ x > y $

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