Given the two quadratic equations:

(I) $3x^2 + 20x + 25 = 0$

(II) $3y^2 + 14y + 8 = 0$

Determine the correct relationship between $x$ and $y$ where $x$ and $y$ are the roots of equations (I) and (II) respectively.

Question

Given the two quadratic equations:

(I) $3x^2 + 20x + 25 = 0$

(II) $3y^2 + 14y + 8 = 0$

Determine the correct relationship between $x$ and $y$ where $x$ and $y$ are the roots of equations (I) and (II) respectively.

Accepted Answer

**Solution:**

1. Consider the quadratic equations:
   - (I) $3x^2 + 20x + 25 = 0$
   - (II) $3y^2 + 14y + 8 = 0$

2. Find the roots of equation (I):
   - Calculate the discriminant $D_x = b^2 - 4ac = 20^2 - 4 \times 3 \times 25 = 400 - 300 = 100$
   - Roots of (I) are:
     $$
     x = \frac{-20 \pm \sqrt{100}}{2 \times 3} = \frac{-20 \pm 10}{6}
     $$
   - So, $x_1 = \frac{-20 + 10}{6} = \frac{-10}{6} = -\frac{5}{3} \approx -1.67$
   - $x_2 = \frac{-20 - 10}{6} = \frac{-30}{6} = -5$

3. Find the roots of equation (II):
   - Calculate the discriminant $D_y = 14^2 - 4 \times 3 \times 8 = 196 - 96 = 100$
   - Roots of (II) are:
     $$
     y = \frac{-14 \pm \sqrt{100}}{2 \times 3} = \frac{-14 \pm 10}{6}
     $$
   - So, $y_1 = \frac{-14 + 10}{6} = \frac{-4}{6} = -\frac{2}{3} \approx -0.67$
   - $y_2 = \frac{-14 - 10}{6} = \frac{-24}{6} = -4$

4. Now, compare the roots:
   - For equation (I), roots are $-5$ and $-1.67$
   - For equation (II), roots are $-4$ and $-0.67$

5. Since the question does not specify which roots to compare, consider the maximum roots of each:
   - Maximum root of (I) is $-1.67$
   - Maximum root of (II) is $-0.67$

6. Clearly, $-1.67 < -0.67$, so $x < y$ for the maximum roots.

7. Similarly, for minimum roots:
   - Minimum root of (I) is $-5$
   - Minimum root of (II) is $-4$

8. Again, $-5 < -4$, so $x < y$ for the minimum roots as well.

**Answer:** (c) $x < y$

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