In each of the following questions, two equations (I) and (II) are given . Solve the equations and mark the correct options :
I. ` 3 x ^(2) + 25 x + 50 = 0 `
II. ` 4y ^(2) + 23 y + 33 = 0 `

Options

A. If ` x gt y `
B. If ` x ge y `
C. if ` x lt y `
D. If ` x le y`

Answer

To solve the given equations and find the relationship between the variables \( x \) and \( y \), we will follow these steps:

### Step 1: Solve the first equation \( 3x^2 + 25x + 50 = 0 \)

1. **Identify coefficients**:
- \( a = 3 \), \( b = 25 \), \( c = 50 \)

2. **Use the quadratic formula**:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

3. **Calculate the discriminant**:
\[ b^2 - 4ac = 25^2 - 4 \cdot 3 \cdot 50 = 625 - 600 = 25 \]

4. **Calculate the roots**:
\[ x = \frac{-25 \pm \sqrt{25}}{2 \cdot 3} = \frac{-25 \pm 5}{6} \]
- First root:
\[ x_1 = \frac{-25 + 5}{6} = \frac{-20}{6} = -\frac{10}{3} \]
- Second root:
\[ x_2 = \frac{-25 - 5}{6} = \frac{-30}{6} = -5 \]

### Step 2: Solve the second equation \( 4y^2 + 23y + 33 = 0 \)

1. **Identify coefficients**:
- \( a = 4 \), \( b = 23 \), \( c = 33 \)

2. **Use the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

3. **Calculate the discriminant**:
\[ b^2 - 4ac = 23^2 - 4 \cdot 4 \cdot 33 = 529 - 528 = 1 \]

4. **Calculate the roots**:
\[ y = \frac{-23 \pm \sqrt{1}}{2 \cdot 4} = \frac{-23 \pm 1}{8} \]
- First root:
\[ y_1 = \frac{-23 + 1}{8} = \frac{-22}{8} = -\frac{11}{4} \]
- Second root:
\[ y_2 = \frac{-23 - 1}{8} = \frac{-24}{8} = -3 \]

### Step 3: Compare the values of \( x \) and \( y \)

1. **Values of \( x \)**:
- \( x_1 = -\frac{10}{3} \approx -3.33 \)
- \( x_2 = -5 \)

2. **Values of \( y \)**:
- \( y_1 = -\frac{11}{4} \approx -2.75 \)
- \( y_2 = -3 \)

### Step 4: Determine the relationships

1. **Compare \( x_1 \) and \( y_1 \)**:
\[ -\frac{10}{3} < -\frac{11}{4} \quad \text{(True)} \]
Thus, \( x_1 < y_1 \).

2. **Compare \( x_1 \) and \( y_2 \)**:
\[ -\frac{10}{3} < -3 \quad \text{(True)} \]
Thus, \( x_1 < y_2 \).

3. **Compare \( x_2 \) and \( y_1 \)**:
\[ -5 < -\frac{11}{4} \quad \text{(True)} \]
Thus, \( x_2 < y_1 \).

4. **Compare \( x_2 \) and \( y_2 \)**:
\[ -5 < -3 \quad \text{(True)} \]
Thus, \( x_2 < y_2 \).

### Conclusion

From the comparisons, we conclude that \( x < y \) for all cases. Therefore, the correct option is:

**Option C: \( x < y \)**

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