I` 3x^(2) + 4x + 1 = 0" "II. Y^(2) + 5y + 6 = 0`

Options

A. if `x gt y`
B. if `x ge y`
C. if ` y gt x `
D. if ` y ge x`

Explanation

To solve the given equations step by step and find the relationship between \( x \) and \( y \), we will first solve each equation separately.

### Step 1: Solve the first equation \( 3x^2 + 4x + 1 = 0 \)

1. **Identify the coefficients**:
- \( a = 3 \)
- \( b = 4 \)
- \( c = 1 \)

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = 4^2 - 4 \cdot 3 \cdot 1 = 16 - 12 = 4 \]

4. **Calculate the roots**:
\[ x = \frac{-4 \pm \sqrt{4}}{2 \cdot 3} = \frac{-4 \pm 2}{6} \]
- For \( x_1 \):
\[ x_1 = \frac{-4 + 2}{6} = \frac{-2}{6} = -\frac{1}{3} \]
- For \( x_2 \):
\[ x_2 = \frac{-4 - 2}{6} = \frac{-6}{6} = -1 \]

Thus, the solutions for \( x \) are:
\[ x = -1 \quad \text{and} \quad x = -\frac{1}{3} \]

### Step 2: Solve the second equation \( y^2 + 5y + 6 = 0 \)

1. **Identify the coefficients**:
- \( a = 1 \)
- \( b = 5 \)
- \( c = 6 \)

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = 5^2 - 4 \cdot 1 \cdot 6 = 25 - 24 = 1 \]

4. **Calculate the roots**:
\[ y = \frac{-5 \pm \sqrt{1}}{2 \cdot 1} = \frac{-5 \pm 1}{2} \]
- For \( y_1 \):
\[ y_1 = \frac{-5 + 1}{2} = \frac{-4}{2} = -2 \]
- For \( y_2 \):
\[ y_2 = \frac{-5 - 1}{2} = \frac{-6}{2} = -3 \]

Thus, the solutions for \( y \) are:
\[ y = -2 \quad \text{and} \quad y = -3 \]

### Step 3: Find the relationship between \( x \) and \( y \)

Now, we have the values:
- \( x = -1 \) and \( x = -\frac{1}{3} \)
- \( y = -2 \) and \( y = -3 \)

Since there is no direct equality between the values of \( x \) and \( y \), we can compare the smallest values:
- The smallest value of \( x \) is \( -1 \)
- The smallest value of \( y \) is \( -3 \)

### Conclusion

The relationship between the smallest values is:
\[ -1 > -2 > -3 \]